# Surjective Fibers, Saturated Opens and Final-Topology Scope

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261015-advance-20261117-r05-quotient-topology-foundations · 2026-10-15 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1]], [b, [2, 3, 4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0001/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 2. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 2]], [b, [1, 3, 4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0002/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 3. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 2]], [b, [3, 4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0003/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 4. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 3]], [b, [1, 2, 4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0004/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 5. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 3]], [b, [2, 4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0005/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 6. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 2, 3]], [b, [1, 4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0006/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 7. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 2, 3]], [b, [4, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0007/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 8. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 4]], [b, [1, 2, 3, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0008/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 9. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 4]], [b, [2, 3, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0009/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 10. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 2, 4]], [b, [1, 3, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0010/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 11. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 2, 4]], [b, [3, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0011/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 12. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 3, 4]], [b, [1, 2, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0012/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 13. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 3, 4]], [b, [2, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0013/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 14. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 2, 3, 4]], [b, [1, 5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0014/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 15. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 2, 3, 4]], [b, [5]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0015/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 16. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 5]], [b, [1, 2, 3, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0016/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 17. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 5]], [b, [2, 3, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0017/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 18. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 2, 5]], [b, [1, 3, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0018/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 19. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 2, 5]], [b, [3, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0019/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 20. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 3, 5]], [b, [1, 2, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0020/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 21. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 3, 5]], [b, [2, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0021/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 22. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 2, 3, 5]], [b, [1, 4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0022/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 23. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 2, 3, 5]], [b, [4]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0023/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 24. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 4, 5]], [b, [1, 2, 3]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0024/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 25. A Total Surjection Has Complete Nonempty Fibers Partitioning the Source

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).

View answer and explanation([[a, [0, 1, 4, 5]], [b, [2, 3]]], Yes, Yes, fibers_partition_X)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0025/)

Quotient topology · qf-mod48-quotient-fibers · candidate





## 26. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0026/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 27. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0027/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 28. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0028/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 29. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0029/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 30. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0030/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 31. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0031/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 32. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0032/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 33. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0033/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 34. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0034/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 35. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0035/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 36. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0036/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 37. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0037/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 38. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0038/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 39. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0039/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 40. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0040/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 41. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0041/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 42. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0042/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 43. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0043/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 44. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0044/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 45. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0045/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 46. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0046/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 47. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0047/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 48. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0048/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 49. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0049/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 50. The Full Preimage Iff Condition Defines the Final Topology

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).

View answer and explanation([[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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0050/)

Quotient topology · qf-mod48-quotient-topology-generation · candidate





## 51. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0051/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 52. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0052/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 53. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0053/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 54. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0054/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 55. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0055/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 56. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0056/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 57. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0057/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 58. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0058/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 59. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0059/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 60. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0060/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 61. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0061/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 62. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0062/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 63. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0063/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 64. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0064/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 65. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0065/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 66. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0066/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 67. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0067/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 68. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0068/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 69. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0069/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 70. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0070/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 71. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0071/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 72. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0072/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 73. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0073/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 74. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0074/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 75. The Final-Topology Criterion Compares h with Its Composite with q

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).

h and h∘q can both be discontinuous while criterion_equivalence is Yes. The equivalence uses the qualified surjection and final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0075/)

Quotient topology · qf-mod48-quotient-final-continuity · candidate





## 76. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0076/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 77. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(open_image_error, [0], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0077/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 78. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0078/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 79. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(fiber_split, [0], [a], [0, 1], No, No, only_saturated_subsets_are_full_preimages)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0079/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 80. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0080/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 81. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(open_image_error, [1], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0081/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 82. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0082/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 83. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(fiber_split, [0], [a], [0, 2], No, No, only_saturated_subsets_are_full_preimages)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0083/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 84. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0084/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 85. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(open_image_error, [0, 1], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0085/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 86. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0086/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 87. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(fiber_split, [0], [a], [0, 1, 2], No, No, only_saturated_subsets_are_full_preimages)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0087/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 88. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0088/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 89. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(open_image_error, [2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0089/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 90. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0090/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 91. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(fiber_split, [0], [a], [0, 3], No, No, only_saturated_subsets_are_full_preimages)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0091/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 92. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0092/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 93. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(open_image_error, [0, 2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0093/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 94. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0094/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 95. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(fiber_split, [0], [a], [0, 1, 3], No, No, only_saturated_subsets_are_full_preimages)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0095/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 96. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0096/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 97. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(open_image_error, [1, 2], Yes, [a], [0, 1, 2], No, No, quotient_maps_need_not_be_open)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0097/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 98. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(coarser_target, [[a, b], []], [[a, b], [a], []], Yes, Yes, No, continuity_alone_does_not_identify_the_final_topology)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0098/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 99. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(fiber_split, [0], [a], [0, 2, 3], No, No, only_saturated_subsets_are_full_preimages)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0099/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate





## 100. Final Topology Can Exist without Surjectivity, while Quotient-Map Qualification Still Fails

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).

The four modes have 7/8/7/7 fields. A nonsurjection still defines a final topology by the preimage iff condition but is not a quotient map under this convention. Continuity alone does not identify the final topology.

View answer and explanation(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)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r05-quotient-topology-0100/)

Quotient topology · qf-mod48-quotient-topology-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261117-r05-quotient-topology-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261117-r05-quotient-topology-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261117-r05-quotient-topology-foundations/validation-report.json)