{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0001","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b308eeddabde1e8b2944b5d2a441057bad200c93f1ef7f16c66fbce67820eec","family_id":"qf-mod48-quotient-fibers","seed":202611175001,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1]], target carrier=[a, b] and aligned total q=[a, a, b, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1]], [b, [2, 3, 4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1]], [b, [2, 3, 4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0002","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5df3db7d5f4f6507d84d661b6b3f302d0be1ddf99b3720c052b76e90a0f78c3e","family_id":"qf-mod48-quotient-fibers","seed":202611175002,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2]], target carrier=[a, b] and aligned total q=[a, b, a, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2]], [b, [1, 3, 4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 2]], [b, [1, 3, 4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0003","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06ac08a5aa9c0f3af285fd988d74cffb8a5be4251a27201bd5437a8a7f0005ca","family_id":"qf-mod48-quotient-fibers","seed":202611175003,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2]], target carrier=[a, b] and aligned total q=[a, a, a, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2]], [b, [3, 4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 2]], [b, [3, 4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0004","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f302b1b107b8a200a5c108410961319963d354ed0e0a117479724b82e55215b","family_id":"qf-mod48-quotient-fibers","seed":202611175004,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3]], target carrier=[a, b] and aligned total q=[a, b, b, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 3]], [b, [1, 2, 4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 3]], [b, [1, 2, 4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0005","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32ff55be06d90c150da26d7419f8fe954628e61dd26f5de04f1ed873da3185f2","family_id":"qf-mod48-quotient-fibers","seed":202611175005,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3]], target carrier=[a, b] and aligned total q=[a, a, b, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 3]], [b, [2, 4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 3]], [b, [2, 4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0006","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f723ba7799e369b4bb2471019b91a3772289045621e1f55caf01ae80d384d44d","family_id":"qf-mod48-quotient-fibers","seed":202611175006,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3]], target carrier=[a, b] and aligned total q=[a, b, a, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 3]], [b, [1, 4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 2, 3]], [b, [1, 4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0007","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e00c00482dfdebe418aa61032159f2dc4c9213a4d630bfef5c8a29f2157524b","family_id":"qf-mod48-quotient-fibers","seed":202611175007,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3]], target carrier=[a, b] and aligned total q=[a, a, a, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 3]], [b, [4, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 3]], [b, [4, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0008","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83ca7ad353ef78d9c554b4e86c54e6683f8676612a5bed80d1a1b0e765a51271","family_id":"qf-mod48-quotient-fibers","seed":202611175008,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 4]], target carrier=[a, b] and aligned total q=[a, b, b, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 4]], [b, [1, 2, 3, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 4]], [b, [1, 2, 3, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0009","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df743fc542861c02322853b5f7c969b3e000b0440fd01072b875c64d66a71ace","family_id":"qf-mod48-quotient-fibers","seed":202611175009,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 4]], target carrier=[a, b] and aligned total q=[a, a, b, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 4]], [b, [2, 3, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 4]], [b, [2, 3, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0010","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2356f30b6be46aab6fd6849dea26afa282b60a238736c25ffacc6815926e67be","family_id":"qf-mod48-quotient-fibers","seed":202611175010,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 4]], target carrier=[a, b] and aligned total q=[a, b, a, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 4]], [b, [1, 3, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 2, 4]], [b, [1, 3, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0011","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72600bd5ad9576a782497d60c18e2d0b7b8deb448f8cc76fb34e51496df26a6e","family_id":"qf-mod48-quotient-fibers","seed":202611175011,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 4]], target carrier=[a, b] and aligned total q=[a, a, a, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 4]], [b, [3, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 4]], [b, [3, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0012","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"faa6f87a69717c85b1c7adf7ebfed6305ab6491ae6c790215a4da7fec311815e","family_id":"qf-mod48-quotient-fibers","seed":202611175012,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3, 4]], target carrier=[a, b] and aligned total q=[a, b, b, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 3, 4]], [b, [1, 2, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 3, 4]], [b, [1, 2, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0013","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30d8f7774444f28aeba8e2766d169b96636be03a81c2791aabfbf0a1bb0401f9","family_id":"qf-mod48-quotient-fibers","seed":202611175013,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 4]], target carrier=[a, b] and aligned total q=[a, a, b, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 3, 4]], [b, [2, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 3, 4]], [b, [2, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0014","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a756b829422bc86d85efab1a11926992a871c2a775e0947d0dde4e68ce41432","family_id":"qf-mod48-quotient-fibers","seed":202611175014,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 4]], target carrier=[a, b] and aligned total q=[a, b, a, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 3, 4]], [b, [1, 5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 2, 3, 4]], [b, [1, 5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0015","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9733499328f8c96fa3f622fe9286f8f91df1a8437033f8dca52c77ababf4fab5","family_id":"qf-mod48-quotient-fibers","seed":202611175015,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4]], target carrier=[a, b] and aligned total q=[a, a, a, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 3, 4]], [b, [5]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 3, 4]], [b, [5]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0016","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de6b02f2070789dbd7731a9870e86d5bf37aba6eddbb5f46ebc2a9619d75eab1","family_id":"qf-mod48-quotient-fibers","seed":202611175016,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 5]], target carrier=[a, b] and aligned total q=[a, b, b, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 5]], [b, [1, 2, 3, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 5]], [b, [1, 2, 3, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0017","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33ddf062e619387ccddbd0a8ea6126a8fb10305681c9d0699542fe14e80fa51b","family_id":"qf-mod48-quotient-fibers","seed":202611175017,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 5]], target carrier=[a, b] and aligned total q=[a, a, b, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 5]], [b, [2, 3, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 5]], [b, [2, 3, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0018","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43515920488bee1bf0c64f6eb0179d55e4208ca66dd41a574e64d4488b35db43","family_id":"qf-mod48-quotient-fibers","seed":202611175018,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 5]], target carrier=[a, b] and aligned total q=[a, b, a, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 5]], [b, [1, 3, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 2, 5]], [b, [1, 3, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0019","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d48fb9921461d3117d85017a91a98ea7adc2c56ef92b87bf05e873afbe7f7955","family_id":"qf-mod48-quotient-fibers","seed":202611175019,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 5]], target carrier=[a, b] and aligned total q=[a, a, a, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 5]], [b, [3, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 5]], [b, [3, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0020","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59b7b326ebdc6d024f34aacae66602bd6720a5e22dab012bc276513ab63eff4c","family_id":"qf-mod48-quotient-fibers","seed":202611175020,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3, 5]], target carrier=[a, b] and aligned total q=[a, b, b, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 3, 5]], [b, [1, 2, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 3, 5]], [b, [1, 2, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0021","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1057b39c061bab3b786136014cbf9b6008f4babfeec2a558032c61687ae36dd1","family_id":"qf-mod48-quotient-fibers","seed":202611175021,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 5]], target carrier=[a, b] and aligned total q=[a, a, b, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 3, 5]], [b, [2, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 3, 5]], [b, [2, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0022","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f321a1c3ed0739e1b1f7979ba0376459ac89a9b3fd6c5099b187807970955bc5","family_id":"qf-mod48-quotient-fibers","seed":202611175022,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 5]], target carrier=[a, b] and aligned total q=[a, b, a, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 3, 5]], [b, [1, 4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 2, 3, 5]], [b, [1, 4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0023","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d48f62e2fdd5a0e8869fd82ca2a40a56c93dd8d12e9c31cf25f4fa0c59dfeb76","family_id":"qf-mod48-quotient-fibers","seed":202611175023,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 5]], target carrier=[a, b] and aligned total q=[a, a, a, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 3, 5]], [b, [4]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 3, 5]], [b, [4]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0024","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"557b2976241b760bb7927692ce4786b145b526b3053281748f1f5797ca03af3d","family_id":"qf-mod48-quotient-fibers","seed":202611175024,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,4,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 4, 5]], target carrier=[a, b] and aligned total q=[a, b, b, b, a, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 4, 5]], [b, [1, 2, 3]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 4, 5]], [b, [1, 2, 3]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0025","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7dbb56721d771fbebab6b956a829c3fa02c9269a36f0bc9c00f95be81d66ef8c","family_id":"qf-mod48-quotient-fibers","seed":202611175025,"semantic_template":"quotient-fibers|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,4,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"A Total Surjection Has Complete Nonempty Fibers Partitioning the Source","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 4, 5]], target carrier=[a, b] and aligned total q=[a, a, b, b, a, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, surjective, fiber_cover, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 4, 5]], [b, [2, 3]]], Yes, Yes, fibers_partition_X)","explanation":"Retain every target fiber, surjectivity and complete source coverage. A missing target fiber blocks the quotient-map convention here. Do not confuse the full fiber with one representative or split a fiber when taking a full preimage. The exact result, retaining the requested order, is ([[a, [0, 1, 4, 5]], [b, [2, 3]]], Yes, Yes, fibers_partition_X)."},"skills":["A Total Surjection Has Complete Nonempty Fibers Partitioning the Source"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0026","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b1dad70d75b32b153f2ee674703c46381ce3963fc4a71883e9e2ab26cbd0f52","family_id":"qf-mod48-quotient-topology-generation","seed":202611175026,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1]], target carrier=[a, b] and aligned total q=[a, a, b, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1]], [b, [2, 3, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1]], [b, [2, 3, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0027","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3561bba5164896dfeaa3e90a7bbfc8beabd8aa8bc849de76dcdcdf1f25e66e1","family_id":"qf-mod48-quotient-topology-generation","seed":202611175027,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2]], target carrier=[a, b] and aligned total q=[a, b, a, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2]], [b, [1, 3, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 2]], [b, [1, 3, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0028","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8f67e19a0100803cb660cd1404213f2034aec986b2ab5315ff57bae4fc0f423","family_id":"qf-mod48-quotient-topology-generation","seed":202611175028,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2]], target carrier=[a, b] and aligned total q=[a, a, a, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2]], [b, [3, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 2]], [b, [3, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0029","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62cfe43d2053bcdb314e95d79e900acd68a6decc2a4fb873dc098d9074c89b0b","family_id":"qf-mod48-quotient-topology-generation","seed":202611175029,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3]], target carrier=[a, b] and aligned total q=[a, b, b, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 3]], [b, [1, 2, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 3]], [b, [1, 2, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0030","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88ae7d763f0e116bb6d274323c8c2cbeac3ce211bc5a2e581adcbb69a4f0aaf2","family_id":"qf-mod48-quotient-topology-generation","seed":202611175030,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3]], target carrier=[a, b] and aligned total q=[a, a, b, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 3]], [b, [2, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 3]], [b, [2, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0031","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e78e3ee422ec2e5a2e3be359c96b464b2fe289dd96623d95164009c4d59021bf","family_id":"qf-mod48-quotient-topology-generation","seed":202611175031,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3]], target carrier=[a, b] and aligned total q=[a, b, a, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 3]], [b, [1, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 2, 3]], [b, [1, 4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0032","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11d92e3d43845d4ab1a36619c741e2bd09b74cb86e576dccd7a85b2c99a8f70a","family_id":"qf-mod48-quotient-topology-generation","seed":202611175032,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3]], target carrier=[a, b] and aligned total q=[a, a, a, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 3]], [b, [4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 3]], [b, [4, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0033","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c9fd9cab3fa86f68eea673405bcdc888bfa4e8931c0046d4a5acb39cf58a0e9","family_id":"qf-mod48-quotient-topology-generation","seed":202611175033,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 4]], target carrier=[a, b] and aligned total q=[a, b, b, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 4]], [b, [1, 2, 3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 4]], [b, [1, 2, 3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0034","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7fe7abb3dc110a6a1654cf05155f8d5810aacce6f0e3457a8a1ec3442db0e15","family_id":"qf-mod48-quotient-topology-generation","seed":202611175034,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 4]], target carrier=[a, b] and aligned total q=[a, a, b, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 4]], [b, [2, 3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 4]], [b, [2, 3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0035","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0595f73150e84e28ad6f47da4dbfd589627a3a54a50265c69196319c6d61e0da","family_id":"qf-mod48-quotient-topology-generation","seed":202611175035,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 4]], target carrier=[a, b] and aligned total q=[a, b, a, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 4]], [b, [1, 3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 2, 4]], [b, [1, 3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0036","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b143b7ae7b47db0319ca5889aecd771ec8bbc65184ad7419fc4b179ac9796856","family_id":"qf-mod48-quotient-topology-generation","seed":202611175036,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 4]], target carrier=[a, b] and aligned total q=[a, a, a, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 4]], [b, [3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 4]], [b, [3, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0037","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bbd355df3b4d80a6bc9334e069ccd1fb7fa6f7745d751cb9fd31d136f157e47","family_id":"qf-mod48-quotient-topology-generation","seed":202611175037,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3, 4]], target carrier=[a, b] and aligned total q=[a, b, b, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 3, 4]], [b, [1, 2, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 3, 4]], [b, [1, 2, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0038","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a330c5c9ba56c0f047858334552392548547fd42589f08790d21361ffea8ac9","family_id":"qf-mod48-quotient-topology-generation","seed":202611175038,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 4]], target carrier=[a, b] and aligned total q=[a, a, b, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 3, 4]], [b, [2, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 3, 4]], [b, [2, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0039","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dce6e0e90abac0181f4371152262127df571f3f1a3592b2560f44f394d6b4deb","family_id":"qf-mod48-quotient-topology-generation","seed":202611175039,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 4]], target carrier=[a, b] and aligned total q=[a, b, a, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 3, 4]], [b, [1, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 2, 3, 4]], [b, [1, 5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0040","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a0586e11ad9d986416fc5158a6fec403a3513be8712a729766cf89f0d8f33a92","family_id":"qf-mod48-quotient-topology-generation","seed":202611175040,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,4]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4]], target carrier=[a, b] and aligned total q=[a, a, a, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 3, 4]], [b, [5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 3, 4]], [b, [5]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0041","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3d73a4d5dcfad94624b3206b69517510b6ccf386f2bfa2f2cdb9315988964a8","family_id":"qf-mod48-quotient-topology-generation","seed":202611175041,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 5]], target carrier=[a, b] and aligned total q=[a, b, b, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 5]], [b, [1, 2, 3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 5]], [b, [1, 2, 3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0042","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"222c6240ec554be764f43e7c1ace805e6a4fc3bd0202dd0a181d3d8b3f8a564d","family_id":"qf-mod48-quotient-topology-generation","seed":202611175042,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 5]], target carrier=[a, b] and aligned total q=[a, a, b, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 5]], [b, [2, 3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 5]], [b, [2, 3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0043","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a05dd466ea5b08c39bd522e6b3435bbf23d051b488d54c6a51144d4b5af9540","family_id":"qf-mod48-quotient-topology-generation","seed":202611175043,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 5]], target carrier=[a, b] and aligned total q=[a, b, a, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 5]], [b, [1, 3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 2, 5]], [b, [1, 3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0044","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aec688594ecf857b5908b8bef9023ca6b5fa062c6ed0d6a56027cec6ffe3fa56","family_id":"qf-mod48-quotient-topology-generation","seed":202611175044,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 5]], target carrier=[a, b] and aligned total q=[a, a, a, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 5]], [b, [3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 5]], [b, [3, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0045","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f9f6d58f19e672a79099ef85b773b1f0955b79f8d9265fa9529c2f3a433dc35","family_id":"qf-mod48-quotient-topology-generation","seed":202611175045,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3, 5]], target carrier=[a, b] and aligned total q=[a, b, b, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 3, 5]], [b, [1, 2, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 3, 5]], [b, [1, 2, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0046","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"506a1b726e0e9d7c4157e940e24e8830ebca0708db1125675ccd378ffc37d4b0","family_id":"qf-mod48-quotient-topology-generation","seed":202611175046,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 5]], target carrier=[a, b] and aligned total q=[a, a, b, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 3, 5]], [b, [2, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 3, 5]], [b, [2, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0047","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"146f50b1316b9c154e61ec99e860125a796c50716b14082abcf209df6c869931","family_id":"qf-mod48-quotient-topology-generation","seed":202611175047,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 5]], target carrier=[a, b] and aligned total q=[a, b, a, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 2, 3, 5]], [b, [1, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 2, 3, 5]], [b, [1, 4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0048","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f95b257aa4b88c1a919bdb21ecabf75ae029afcf1ee959d36b129387e4ae861","family_id":"qf-mod48-quotient-topology-generation","seed":202611175048,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 5]], target carrier=[a, b] and aligned total q=[a, a, a, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 2, 3, 5]], [b, [4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 2, 3, 5]], [b, [4]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0049","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef94d736b1951ef730a564fd80a23a6ed1cf3dc832bcac25e2c74683b48a20c5","family_id":"qf-mod48-quotient-topology-generation","seed":202611175049,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,4,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 4, 5]], target carrier=[a, b] and aligned total q=[a, b, b, b, a, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 4, 5]], [b, [1, 2, 3]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 4, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 4, 5]], [b, [1, 2, 3]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 4, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0050","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db63dd61c6f6ce10f94595f61b6a67bab82ae144e56b3916f28c2aae2fe9001c","family_id":"qf-mod48-quotient-topology-generation","seed":202611175050,"semantic_template":"quotient-topology-generation|{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,4,5]],y_points:[\"a\",\"b\"]}"},"fields":{"title":"The Full Preimage Iff Condition Defines the Final Topology","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 4, 5]], target carrier=[a, b] and aligned total q=[a, a, b, b, a, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Give (fibers, final_topology, saturated_source_opens, open_count_correspondence, quotient_certificate, certificate).","answer":{"kind":"expression","value":"([[a, [0, 1, 4, 5]], [b, [2, 3]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 4, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)","explanation":"Enumerate every target subset whose full preimage is source-open, certify the resulting topology and retain the saturated source opens. Their correspondence uses saturation, not the image of an arbitrary source open. The surjection qualifies this final topology as the quotient topology in this node. The exact result, retaining the requested order, is ([[a, [0, 1, 4, 5]], [b, [2, 3]]], [[a, b], [a], []], [[], [0, 1, 2, 3, 4, 5], [0, 1, 4, 5]], Yes, Yes, U_open_iff_q_preimage_U_open)."},"skills":["The Full Preimage Iff Condition Defines the Final Topology"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0051","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7b2b0bc26ae4a641826673e7ba47e0e096a9955989691c484d95f562ba570bc","family_id":"qf-mod48-quotient-final-continuity","seed":202611175051,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1]], target carrier=[a, b] and aligned total q=[a, a, b, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0052","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e01b7d68861d1503cb32506bb5ad077a9f315d753db7ec818573e278887d1ae","family_id":"qf-mod48-quotient-final-continuity","seed":202611175052,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2]], target carrier=[a, b] and aligned total q=[a, b, a, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 1], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 1], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0053","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8fd3b01d7f0b9a0a2e2315f3a9a5e5c5a56603c7603418fc6c33ee352b29b79c","family_id":"qf-mod48-quotient-final-continuity","seed":202611175053,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2]], target carrier=[a, b] and aligned total q=[a, a, a, b, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 1], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 1], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0054","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f73a884efb8faf2a749aff18fe027603abc1e8f1fe1723af82bad144f402224","family_id":"qf-mod48-quotient-final-continuity","seed":202611175054,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3]], target carrier=[a, b] and aligned total q=[a, b, b, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0055","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2870d7f12d99ef80d0c783d26a9d34137f6d8c3138598dca30d09d89800d715","family_id":"qf-mod48-quotient-final-continuity","seed":202611175055,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3]], target carrier=[a, b] and aligned total q=[a, a, b, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0056","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11611f9a66a041482f6a3d5ccc32711ef9629bee49a4cd89616134c545150c88","family_id":"qf-mod48-quotient-final-continuity","seed":202611175056,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3]], target carrier=[a, b] and aligned total q=[a, b, a, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0057","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04f59190c88e42f6df8fd81fcff0d1d368c41e09829882996037802c577e6ae9","family_id":"qf-mod48-quotient-final-continuity","seed":202611175057,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3]], target carrier=[a, b] and aligned total q=[a, a, a, a, b, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 0], [4, 1], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0058","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"742d8818b59425c24e71c9525c7c0f82585e1993a56d210d649aac2b79cd279f","family_id":"qf-mod48-quotient-final-continuity","seed":202611175058,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 4]], target carrier=[a, b] and aligned total q=[a, b, b, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0059","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4be78e6100892b309246149e878a00ec9da283bc2dce7652b71a789778f424b","family_id":"qf-mod48-quotient-final-continuity","seed":202611175059,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 4]], target carrier=[a, b] and aligned total q=[a, a, b, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0060","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a7cc50dcadb57c63cb930e26c8a1ab4943b5c7bede0035c423c50b3cf7a4a0e","family_id":"qf-mod48-quotient-final-continuity","seed":202611175060,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 4]], target carrier=[a, b] and aligned total q=[a, b, a, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0061","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"401d8b1c9dfa9c96f7422e8fd218c7693a25fe88fafdce38b6e8c03cf465ad87","family_id":"qf-mod48-quotient-final-continuity","seed":202611175061,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 4]], target carrier=[a, b] and aligned total q=[a, a, a, b, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 1], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0062","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd7f09b8d136b79886bae0bbcb43e8c4601f6a2324f851f8abe229d3cce97bc3","family_id":"qf-mod48-quotient-final-continuity","seed":202611175062,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3, 4]], target carrier=[a, b] and aligned total q=[a, b, b, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0063","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be25b89a1c674e21d94818b593b3c49c0f02a85c33b8b4971d07e1051f135866","family_id":"qf-mod48-quotient-final-continuity","seed":202611175063,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 4]], target carrier=[a, b] and aligned total q=[a, a, b, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0064","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f4a709c707a1301485ae5c98cbfea53fa6f36bdb0a5e20d0f8726dbaad8391c","family_id":"qf-mod48-quotient-final-continuity","seed":202611175064,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 4]], target carrier=[a, b] and aligned total q=[a, b, a, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0065","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ca654d72d229a2e34dcc04102c8e51a0eced89eb762fb9d0e45c894c3cc63ff","family_id":"qf-mod48-quotient-final-continuity","seed":202611175065,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,4]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4]], target carrier=[a, b] and aligned total q=[a, a, a, a, a, b]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 0], [4, 0], [5, 1]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0066","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43ad0f4a17e08cd27f138807067d89da2d0f3be24dd63ea3811907e819f42f23","family_id":"qf-mod48-quotient-final-continuity","seed":202611175066,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 5]], target carrier=[a, b] and aligned total q=[a, b, b, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0067","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ca1b04b0eb26498eb822cb971e24312cbe00b1b01ded43a016d2edbeacadeb3","family_id":"qf-mod48-quotient-final-continuity","seed":202611175067,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 5]], target carrier=[a, b] and aligned total q=[a, a, b, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0068","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7645a9b34ae7c70542274256ce682e63553a8b8493a09b7883d88e43b8840092","family_id":"qf-mod48-quotient-final-continuity","seed":202611175068,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 5]], target carrier=[a, b] and aligned total q=[a, b, a, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0069","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"118afb59374e453ac50c9a09894291abd2bba6f93fcfccb64867d784552e33c7","family_id":"qf-mod48-quotient-final-continuity","seed":202611175069,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 5]], target carrier=[a, b] and aligned total q=[a, a, a, b, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 1], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0070","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b162932a5a63655fcc49d680636f3ebfe17f987f00e6416383e36fcc849e8d0c","family_id":"qf-mod48-quotient-final-continuity","seed":202611175070,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,3,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 3, 5]], target carrier=[a, b] and aligned total q=[a, b, b, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0071","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"642430cb6152718d86dd0bac352555f7f5753b0cc7ff042abd7ca98d6c363d10","family_id":"qf-mod48-quotient-final-continuity","seed":202611175071,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,3,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 3, 5]], target carrier=[a, b] and aligned total q=[a, a, b, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0072","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"673f045b17c60af4b38c1cdd665071bc54e0c7467643c356a997e5de6252eaa5","family_id":"qf-mod48-quotient-final-continuity","seed":202611175072,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,2,3,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 2, 3, 5]], target carrier=[a, b] and aligned total q=[a, b, a, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 0], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0073","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c18d3257b2825842e994bb16994f346a35ff8614090d62142f1978d2721a60ca","family_id":"qf-mod48-quotient-final-continuity","seed":202611175073,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 5]], target carrier=[a, b] and aligned total q=[a, a, a, a, b, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 0], [3, 0], [4, 1], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0074","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce37364ef26b6a17ebc20061bff3343e3a3f47007551bc281bd44f47e06798ba","family_id":"qf-mod48-quotient-final-continuity","seed":202611175074,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,4,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 4, 5]], target carrier=[a, b] and aligned total q=[a, b, b, b, a, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 1], [4, 0], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 1], [2, 1], [3, 1], [4, 0], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0075","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96aecaf9022bed4640c5a884737f25a3fc8b54432429e14a842992ddcc2fc783","family_id":"qf-mod48-quotient-final-continuity","seed":202611175075,"semantic_template":"quotient-final-continuity|{h_map:[[\"a\",0],[\"b\",1]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"a\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,4,5]],y_points:[\"a\",\"b\"],z_points:[0,1],z_topology:[[],[0,1],[1]]}"},"fields":{"title":"The Final-Topology Criterion Compares h with Its Composite with q","stem":"Use complete source points=[0, 1, 2, 3, 4, 5], topology=[[], [0, 1, 2, 3, 4, 5], [0, 1, 4, 5]], target carrier=[a, b] and aligned total q=[a, a, b, b, a, a]. Normal quotient families require q surjective. Target opens are exactly the subsets with open full preimage; continuity alone is weaker. Target Z=[0, 1], topology=[[], [1], [0, 1]], aligned h=[0, 1]. Give (h_continuous, composite_continuous, criterion_equivalence, h_rows, composite_rows, certificate).","answer":{"kind":"expression","value":"(No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 0], [5, 0]], quotient_final_property_for_surjective_q)","explanation":"Audit every target-open preimage for h and h∘q with their complete aligned tables. Both maps can be discontinuous while the equivalence holds. The qualified surjection and full final topology precede this criterion. The exact result, retaining the requested order, is (No, No, Yes, [[a, 0], [b, 1]], [[0, 0], [1, 0], [2, 1], [3, 1], [4, 0], [5, 0]], quotient_final_property_for_surjective_q)."},"skills":["The Final-Topology Criterion Compares h with Its Composite with q"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0076","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2d5b5676c93d3e295453a47978eb6b0eb657b35b073572bab5a44d8bdfcf2ec","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175076,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[0]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[0],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y4], [y0], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y4], [y0], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0077","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a32d8d630b30118254ac9348f4aafa5e786d2e18add9110f1143644d19be31c7","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175077,"semantic_template":"quotient-topology-boundary|{mode:\"open_image_error\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"]],subset:[0],x_points:[0,1,2,3,4],x_topology:[[],[0,1,2,3,4],[0]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=open_image_error and complete model={\"x_points\":[0,1,2,3,4],\"x_topology\":[[],[0],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, source_open, image, image_preimage, image_open, automatic_open_map_claim, boundary).","answer":{"kind":"expression","value":"(open_image_error, [0], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (open_image_error, [0], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0078","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"072dd6f6cf00039cc91ef1ce1a1939f852361745454eb708caa102809aba2f15","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175078,"semantic_template":"quotient-topology-boundary|{mode:\"coarser_target\",model:{candidate_topology:[[\"a\",\"b\"],[]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=coarser_target and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5],[0,1,2]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\",\"b\"],\"candidate_topology\":[[],[\"a\",\"b\"]]}. Give (mode, candidate_topology, final_topology, q_continuous, properly_coarser, final_topology_identification_claim, boundary).","answer":{"kind":"expression","value":"(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0079","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1f4699ff4ce6b4a89af21a377ef8c6d6210878a1ebab322ce658dd3b62f666f","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175079,"semantic_template":"quotient-topology-boundary|{mode:\"fiber_split\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],subset:[0],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=fiber_split and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"b\",\"b\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, image, saturation, saturated, full_preimage_claim, boundary).","answer":{"kind":"expression","value":"(fiber_split, [0], [a], [0, 1], No, No, only_saturated_subsets_are_full_preimages)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (fiber_split, [0], [a], [0, 1], No, No, only_saturated_subsets_are_full_preimages)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0080","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dcae644702b28b1eb6f6509d4bf52409c5d7ca87ab503c3359b84a90f9227caa","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175080,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[1]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[1],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y1, y4], [y1], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y1, y4], [y1], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0081","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe4fdcf238aa2581c0729251a3f8c4af36addb2559484c0da4db1410ab823b47","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175081,"semantic_template":"quotient-topology-boundary|{mode:\"open_image_error\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"]],subset:[1],x_points:[0,1,2,3,4],x_topology:[[],[0,1,2,3,4],[1]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=open_image_error and complete model={\"x_points\":[0,1,2,3,4],\"x_topology\":[[],[1],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[1]}. Give (mode, source_subset, source_open, image, image_preimage, image_open, automatic_open_map_claim, boundary).","answer":{"kind":"expression","value":"(open_image_error, [1], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (open_image_error, [1], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0082","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f25a56366812cb60a7dbcaf15ec85fe9254a77768faf656bfc2f095aadf98171","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175082,"semantic_template":"quotient-topology-boundary|{mode:\"coarser_target\",model:{candidate_topology:[[\"a\",\"b\"],[]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=coarser_target and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5],[0,1,2,3]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"a\",\"b\",\"b\"],\"candidate_topology\":[[],[\"a\",\"b\"]]}. Give (mode, candidate_topology, final_topology, q_continuous, properly_coarser, final_topology_identification_claim, boundary).","answer":{"kind":"expression","value":"(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0083","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f447999fa3e7c45b6dd533e79b0ddcdbca42df754f113eae959af306a6cf62cd","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175083,"semantic_template":"quotient-topology-boundary|{mode:\"fiber_split\",model:{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],subset:[0],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=fiber_split and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"b\",\"a\",\"b\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, image, saturation, saturated, full_preimage_claim, boundary).","answer":{"kind":"expression","value":"(fiber_split, [0], [a], [0, 2], No, No, only_saturated_subsets_are_full_preimages)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (fiber_split, [0], [a], [0, 2], No, No, only_saturated_subsets_are_full_preimages)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0084","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e18156a5b531703bf4b06f79339544617b7eba2f3859bd3507b914e1f0bde5aa","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175084,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[0,1]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[0,1],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y1, y4], [y0, y1], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y1, y4], [y0, y1], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0085","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad4ff6b769d864678746035258b9723c5778a4cb37deb8d4f3f5b402e6d91382","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175085,"semantic_template":"quotient-topology-boundary|{mode:\"open_image_error\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"]],subset:[0,1],x_points:[0,1,2,3,4],x_topology:[[],[0,1,2,3,4],[0,1]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=open_image_error and complete model={\"x_points\":[0,1,2,3,4],\"x_topology\":[[],[0,1],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[0,1]}. Give (mode, source_subset, source_open, image, image_preimage, image_open, automatic_open_map_claim, boundary).","answer":{"kind":"expression","value":"(open_image_error, [0, 1], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (open_image_error, [0, 1], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0086","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f30d7213ca03d74f29ad22c18cd51c8e72967db133d5e7c338926646b9ad6861","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175086,"semantic_template":"quotient-topology-boundary|{mode:\"coarser_target\",model:{candidate_topology:[[\"a\",\"b\"],[]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,4]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=coarser_target and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5],[0,1,2,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"a\",\"b\"],\"candidate_topology\":[[],[\"a\",\"b\"]]}. Give (mode, candidate_topology, final_topology, q_continuous, properly_coarser, final_topology_identification_claim, boundary).","answer":{"kind":"expression","value":"(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0087","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df7433c5d63ee0f1eb6e8679843394a9e4ef61bd3bc92993d8c86d871f5c60d0","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175087,"semantic_template":"quotient-topology-boundary|{mode:\"fiber_split\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"b\"]],subset:[0],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=fiber_split and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, image, saturation, saturated, full_preimage_claim, boundary).","answer":{"kind":"expression","value":"(fiber_split, [0], [a], [0, 1, 2], No, No, only_saturated_subsets_are_full_preimages)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (fiber_split, [0], [a], [0, 1, 2], No, No, only_saturated_subsets_are_full_preimages)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0088","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3caba32046de3289228b7ef2d9f45151ee3294cae9ab3c0c2de0ce6932bc3f6","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175088,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[2]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[2],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y2, y4], [y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y2, y4], [y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0089","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f32ebe20216200a11e2b181f24ef0983d7c18c9183f3f7ea5bfba40a3a115c81","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175089,"semantic_template":"quotient-topology-boundary|{mode:\"open_image_error\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"]],subset:[2],x_points:[0,1,2,3,4],x_topology:[[],[0,1,2,3,4],[2]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=open_image_error and complete model={\"x_points\":[0,1,2,3,4],\"x_topology\":[[],[2],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[2]}. Give (mode, source_subset, source_open, image, image_preimage, image_open, automatic_open_map_claim, boundary).","answer":{"kind":"expression","value":"(open_image_error, [2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (open_image_error, [2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0090","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dea3d0a6bfc285861ca84d2dd45c5a09d0d126c9248fe1a4bdcd0c9f609f912e","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175090,"semantic_template":"quotient-topology-boundary|{mode:\"coarser_target\",model:{candidate_topology:[[\"a\",\"b\"],[]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"a\"],[5,\"b\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,4]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=coarser_target and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"a\",\"a\",\"b\"],\"candidate_topology\":[[],[\"a\",\"b\"]]}. Give (mode, candidate_topology, final_topology, q_continuous, properly_coarser, final_topology_identification_claim, boundary).","answer":{"kind":"expression","value":"(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0091","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e077b97ab617fe878c6682d9784bbf0c85c48b81ea032885f5e39d3d7983510d","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175091,"semantic_template":"quotient-topology-boundary|{mode:\"fiber_split\",model:{q_map:[[0,\"a\"],[1,\"b\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],subset:[0],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=fiber_split and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"b\",\"b\",\"a\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, image, saturation, saturated, full_preimage_claim, boundary).","answer":{"kind":"expression","value":"(fiber_split, [0], [a], [0, 3], No, No, only_saturated_subsets_are_full_preimages)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (fiber_split, [0], [a], [0, 3], No, No, only_saturated_subsets_are_full_preimages)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0092","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"153c7f7fdce825999a14ea230a6ce3fbeaaa0d186d48c1b3bf75957e644ae64b","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175092,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[0,2]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[0,2],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0093","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22b854f42f356479f37efbe7b11711d221043e2517edde6b25331735e8ab6b1c","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175093,"semantic_template":"quotient-topology-boundary|{mode:\"open_image_error\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"]],subset:[0,2],x_points:[0,1,2,3,4],x_topology:[[],[0,1,2,3,4],[0,2]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=open_image_error and complete model={\"x_points\":[0,1,2,3,4],\"x_topology\":[[],[0,2],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[0,2]}. Give (mode, source_subset, source_open, image, image_preimage, image_open, automatic_open_map_claim, boundary).","answer":{"kind":"expression","value":"(open_image_error, [0, 2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (open_image_error, [0, 2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0094","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bea3e1301a37c907bc66375efbd8269c19f1ecbb29dbb2335e9f1190efde1622","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175094,"semantic_template":"quotient-topology-boundary|{mode:\"coarser_target\",model:{candidate_topology:[[\"a\",\"b\"],[]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=coarser_target and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5],[0,1,2,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\",\"a\"],\"candidate_topology\":[[],[\"a\",\"b\"]]}. Give (mode, candidate_topology, final_topology, q_continuous, properly_coarser, final_topology_identification_claim, boundary).","answer":{"kind":"expression","value":"(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0095","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06be1569504bd6f98f6c77fcf8e543113d9f24f2b503714fa283fd61cc8ffd15","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175095,"semantic_template":"quotient-topology-boundary|{mode:\"fiber_split\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"b\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],subset:[0],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=fiber_split and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"b\",\"a\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, image, saturation, saturated, full_preimage_claim, boundary).","answer":{"kind":"expression","value":"(fiber_split, [0], [a], [0, 1, 3], No, No, only_saturated_subsets_are_full_preimages)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (fiber_split, [0], [a], [0, 1, 3], No, No, only_saturated_subsets_are_full_preimages)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0096","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ffbe4c0b8767ad05ca3d8b01e0bb8a3f3028403164c6b60249fd3d32cbadf399","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175096,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[1,2]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[1,2],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y1, y2, y4], [y1, y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y1, y2, y4], [y1, y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0097","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd169f23dfec5a4ea0bae8008fbf432fa8b9bbe5c307a644db70b6d3da120bd7","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175097,"semantic_template":"quotient-topology-boundary|{mode:\"open_image_error\",model:{q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"b\"],[4,\"b\"]],subset:[1,2],x_points:[0,1,2,3,4],x_topology:[[],[0,1,2,3,4],[1,2]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=open_image_error and complete model={\"x_points\":[0,1,2,3,4],\"x_topology\":[[],[1,2],[0,1,2,3,4]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[1,2]}. Give (mode, source_subset, source_open, image, image_preimage, image_open, automatic_open_map_claim, boundary).","answer":{"kind":"expression","value":"(open_image_error, [1, 2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (open_image_error, [1, 2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0098","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9263f8510295089746b48a0a530d446f8a59caf05763aa9aefe540817a86c4b0","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175098,"semantic_template":"quotient-topology-boundary|{mode:\"coarser_target\",model:{candidate_topology:[[\"a\",\"b\"],[]],q_map:[[0,\"a\"],[1,\"a\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"a\"]],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5],[0,1,2,3,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=coarser_target and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5],[0,1,2,3,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"a\",\"a\",\"a\",\"b\",\"a\"],\"candidate_topology\":[[],[\"a\",\"b\"]]}. Give (mode, candidate_topology, final_topology, q_continuous, properly_coarser, final_topology_identification_claim, boundary).","answer":{"kind":"expression","value":"(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0099","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37442434e7e56eea335f7cf84c4f6f20ab588e45c1c696f826c3553335883253","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175099,"semantic_template":"quotient-topology-boundary|{mode:\"fiber_split\",model:{q_map:[[0,\"a\"],[1,\"b\"],[2,\"a\"],[3,\"a\"],[4,\"b\"],[5,\"b\"]],subset:[0],x_points:[0,1,2,3,4,5],x_topology:[[],[0,1,2,3,4,5]],y_points:[\"a\",\"b\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=fiber_split and complete model={\"x_points\":[0,1,2,3,4,5],\"x_topology\":[[],[0,1,2,3,4,5]],\"y_points\":[\"a\",\"b\"],\"q_map\":[\"a\",\"b\",\"a\",\"a\",\"b\",\"b\"],\"subset\":[0]}. Give (mode, source_subset, image, saturation, saturated, full_preimage_claim, boundary).","answer":{"kind":"expression","value":"(fiber_split, [0], [a], [0, 2, 3], No, No, only_saturated_subsets_are_full_preimages)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (fiber_split, [0], [a], [0, 2, 3], No, No, only_saturated_subsets_are_full_preimages)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261117-r05-quotient-topology-0100","batch_id":"batch-20261015-advance-20261117-r05-quotient-topology-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7bab59e9598ff91f445ce313789f48febbfe2821635568c20eb53ee37bcff56d","family_id":"qf-mod48-quotient-topology-boundary","seed":202611175100,"semantic_template":"quotient-topology-boundary|{mode:\"non_surjective\",model:{q_map:[[0,\"y0\"],[1,\"y1\"],[2,\"y2\"],[3,\"y3\"]],x_points:[0,1,2,3],x_topology:[[],[0,1,2,3],[0,1,2]],y_points:[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"]}}"},"fields":{"title":"Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails","stem":"Use mode=non_surjective and complete model={\"x_points\":[0,1,2,3],\"x_topology\":[[],[0,1,2],[0,1,2,3]],\"y_points\":[\"y0\",\"y1\",\"y2\",\"y3\",\"y4\"],\"q_map\":[\"y0\",\"y1\",\"y2\",\"y3\"]}. Give (mode, fibers, surjective, final_topology, final_topology_exists, quotient_map_claim, boundary).","answer":{"kind":"expression","value":"(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y1, y2, y4], [y0, y1, y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)","explanation":"A nonsurjection still defines a final topology by the full preimage iff condition, but it is not a quotient map under this surjective convention. Open images need not be open; a properly coarser target can keep q continuous without being final. A fiber-splitting subset is not a full preimage. The preimage-if-and-only-if definition uniquely determines the final topology even for a nonsurjection; continuity alone does not. Retain the separate surjective quotient-map naming convention. The exact result, retaining the requested order, is (non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], No, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y1, y2, y4], [y0, y1, y2], [y4], []], Yes, No, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)."},"skills":["Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}