# 第 26867 題：商拓撲

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本頁是已完成第二對話 AI 數理與中英語義覆核的教學詳解，只支持本題明示的模型與前提。沒有人工審查或新 Lean 證明，也不提升原題或 MKO 的 Canonical 真值狀態。

R0052-26867 · q-20261015-advance-20261117-r05-quotient-topology-0092



## 原題與已知條件

finite total surjection、fibers與U open iff q-preimage(U) open責任。操作=quotient-topology-boundary；資料={"mode":"non_surjective","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"]}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。


- G001 — finite total surjection、fibers與U open iff q-preimage(U) open責任。操作=quotient-topology-boundary；資料={"mode":"non_surjective","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"]}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。





## 定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。


- [mko-quotient-topology](https://ocme.evemisslab.com/zh-Hant/mko/mko-quotient-topology/)





## 逐步推導

S001 · implies操作前

source roles = [quotient-topology-boundary]

操作後

derived = [[mode, fibers, surjective, final_topology, final_exists, quotient_map_claim, certificate]]

條件：無

直接列出此操作及mode的完整來源欄序，不需JavaScript；布林、字串與null角色保持區別，不補欄或以答案形狀猜介面。

步驟識別與依賴規則：source_ordered_fields_without_JS · 依賴：G001





S002 · implies操作前

source roles = [quotient ORIGINAL X, [0, 1, 2, 3], [[], [0, 2], [0, 1, 2, 3]], 否]

操作後

derived = [[0, 1, 2, 3], [[], [0, 1, 2, 3], [0, 2]], [[0, 0, 0, 0, 是, 是], [0, 1, 15, 0, 是, 是], [0, 2, 5, 0, 是, 是], [1, 0, 15, 0, 是, 是], [1, 1, 15, 15, 是, 是], [1, 2, 15, 5, 是, 是], [2, 0, 5, 0, 是, 是], [2, 1, 15, 5, 是, 是], [2, 2, 5, 5, 是, 是]]]

條件：無

保留JSON型別，set與open族依來源English序。逐全部open pair驗union及intersection，含空集與全集；有限族的二元閉包給任意union閉包，不由抽樣認定topology。

步驟識別與依賴規則：Topology_full_typed_carrier_and_ALL_pair_axioms · 依賴：G001, S001





S003 · implies操作前

source roles = [[0, 1, 2, 3], [y0, y1, y2, y3, y4], [y0, y1, y2, y3]]

操作後

derived = [[[0, y0], [1, y1], [2, y2], [3, y3]], [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], 否]

條件：無

依原始X次序綁定q再排序，對每個typed Y點取完整fiber。total map保每個x恰屬一fiber；surjective才保每fiber非空。

步驟識別與依賴規則：Quotient_ORIGINAL_domain_alignment_ALL_fibers · 依賴：G001, S002





S004 · implies操作前

source roles = [[0, 1, 2, 3], [[], [0, 1, 2, 3], [0, 2]], [y0, y1, y2, y3, y4], [[0, y0], [1, y1], [2, y2], [3, y3]]]

操作後

derived = [[[0, [], [], 是], [1, [y0], [0], 否], [2, [y1], [1], 否], [3, [y0, y1], [0, 1], 否], [4, [y2], [2], 否], [5, [y0, y2], [0, 2], 是], [6, [y1, y2], [1, 2], 否], [7, [y0, y1, y2], [0, 1, 2], 否], [8, [y3], [3], 否], [9, [y0, y3], [0, 3], 否], [10, [y1, y3], [1, 3], 否], [11, [y0, y1, y3], [0, 1, 3], 否], [12, [y2, y3], [2, 3], 否], [13, [y0, y2, y3], [0, 2, 3], 否], [14, [y1, y2, y3], [1, 2, 3], 否], [15, [y0, y1, y2, y3], [0, 1, 2, 3], 是], [16, [y4], [], 是], [17, [y0, y4], [0], 否], [18, [y1, y4], [1], 否], [19, [y0, y1, y4], [0, 1], 否], [20, [y2, y4], [2], 否], [21, [y0, y2, y4], [0, 2], 是], [22, [y1, y2, y4], [1, 2], 否], [23, [y0, y1, y2, y4], [0, 1, 2], 否], [24, [y3, y4], [3], 否], [25, [y0, y3, y4], [0, 3], 否], [26, [y1, y3, y4], [1, 3], 否], [27, [y0, y1, y3, y4], [0, 1, 3], 否], [28, [y2, y3, y4], [2, 3], 否], [29, [y0, y2, y3, y4], [0, 2, 3], 否], [30, [y1, y2, y3, y4], [1, 2, 3], 否], [31, [y0, y1, y2, y3, y4], [0, 1, 2, 3], 是]], [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []]]

條件：無

逐Y的所有subsets，以完整反像在X是否open定義IFF。反像保空、全集、任意union與有限intersection，故生成topology且此IFF逐subset唯一決定它；此唯一性不需要surjection。

步驟識別與依賴規則：Quotient_ALL_target_subsets_PREIMAGE_IFF_unique_final_topology · 依賴：G001, S003





S005 · implies操作前

source roles = [quotient GENERATED final topology, [y0, y1, y2, y3, y4], [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []], 否]

操作後

derived = [[y0, y1, y2, y3, y4], [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []], [[0, 0, 31, 31, 是, 是], [0, 1, 31, 15, 是, 是], [0, 2, 31, 21, 是, 是], [0, 3, 31, 5, 是, 是], [0, 4, 31, 16, 是, 是], [0, 5, 31, 0, 是, 是], [1, 0, 31, 15, 是, 是], [1, 1, 15, 15, 是, 是], [1, 2, 31, 5, 是, 是], [1, 3, 15, 5, 是, 是], [1, 4, 31, 0, 是, 是], [1, 5, 15, 0, 是, 是], [2, 0, 31, 21, 是, 是], [2, 1, 31, 5, 是, 是], [2, 2, 21, 21, 是, 是], [2, 3, 21, 5, 是, 是], [2, 4, 21, 16, 是, 是], [2, 5, 21, 0, 是, 是], [3, 0, 31, 5, 是, 是], [3, 1, 15, 5, 是, 是], [3, 2, 21, 5, 是, 是], [3, 3, 5, 5, 是, 是], [3, 4, 21, 0, 是, 是], [3, 5, 5, 0, 是, 是], [4, 0, 31, 16, 是, 是], [4, 1, 31, 0, 是, 是], [4, 2, 21, 16, 是, 是], [4, 3, 21, 0, 是, 是], [4, 4, 16, 16, 是, 是], [4, 5, 16, 0, 是, 是], [5, 0, 31, 0, 是, 是], [5, 1, 15, 0, 是, 是], [5, 2, 21, 0, 是, 是], [5, 3, 5, 0, 是, 是], [5, 4, 16, 0, 是, 是], [5, 5, 0, 0, 是, 是]]]

條件：無

保留JSON型別，set與open族依來源English序。逐全部open pair驗union及intersection，含空集與全集；有限族的二元閉包給任意union閉包，不由抽樣認定topology。

步驟識別與依賴規則：Topology_full_typed_carrier_and_ALL_pair_axioms · 依賴：G001, S004





S006 · implies操作前

source roles = [[[], [0, 1, 2, 3], [0, 2]], [[0, y0], [1, y1], [2, y2], [3, y3]]]

操作後

derived = [[[], [0, 1, 2, 3], [0, 2]], 否]

條件：無

source opens依原canonical序逐一核A=q^-1(q(A))；只有saturated opens才對應完整fiber聯集。surjective時q^-1在Y subsets上單射，與這些opens互逆，不能把所有source opens當商開集。

步驟識別與依賴規則：Quotient_ALL_source_open_saturation_correspondence · 依賴：G001, S005





S007 · implies操作前

source roles = [[[0, y0], [1, y1], [2, y2], [3, y3]], [y0, y1, y2, y3, y4], [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []]]

操作後

derived = [是, 否]

條件：無

缺少fiber的Y點不破壞反像IFF所定義topology的存在或唯一性；只是本題quotient map慣例要求surjection，所以該名稱不成立。

步驟識別與依賴規則：Quotient_nonSURJ_unique_final_exists_but_naming_claim_NO · 依賴：G001, S006









## 逐步反向核對


- S007 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
- S006 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
- S005 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
- S004 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
- S003 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
- S002 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。
- S001 · premise_and_result_audit — 反查原題的完整資料、型別、欄序與本步條件；有限核對不升格為無限定理的形式證明。





## 結論

(non_surjective, [[y0, [0]], [y1, [1]], [y2, [2]], [y3, [3]], [y4, []]], 否, [[y0, y1, y2, y3, y4], [y0, y1, y2, y3], [y0, y2, y4], [y0, y2], [y4], []], 是, 否, preimage_defined_final_topology_exists_but_map_is_not_a_quotient_map_by_this_convention)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。





## 本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。

來源雜湊與覆核憑證
- input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
- original question SHA-256: 153c7f7fdce825999a14ea230a6ce3fbeaaa0d186d48c1b3bf75957e644ae64b
- candidate record SHA-256: 393cb43465743a8dfb40b34f0e2dc1b78553b70c1afb79be7581ecbeee896e39
- semantic review receipt SHA-256: 729b25cbf6bf5cb1d07a21a511ae216ec5fdd54a4a7aa2f7c21156b81ff37252
- question-first witness SHA-256: d1ab788770eaad04f986986e65afb0d55791e8a8c0d747e9a2b499f79fe78df2
- Lean: NOT_RUN · human review: NOT_PERFORMED

[同源完整 JSON](https://ocme.evemisslab.com/data/question-derivations/questions/q-20261015-advance-20261117-r05-quotient-topology-0092.json)