{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-bijective"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-bijective-01","title":{"zh-Hant":"雙射：以纖維概況判定給定函數，並與「是否存在某個雙射」分開","en":"Bijectivity: testing given functions by fiber profiles, kept apart from whether some bijection exists"},"summary":{"zh-Hant":"設 A={1,2,3}、B={4,5,6}，函數 f、g:A→B 由表給出：f(1)=5、f(2)=6、f(3)=4；g(1)=4、g(2)=4、g(3)=6。先對一般總函數證明「雙射 ⇔ 每個 b∈B 的纖維恰有一個元素」，再按陪域 4、5、6 的順序列出 f 與 g 的完整纖維概況，判定 f 是雙射、g 既非單射也非滿射。最後回答另一類問題「是否存在某個雙射」：A→B 存在（f 即是）、{1,2,3}→{4,5} 不存在、∅→∅ 恰有一個空雙射。","en":"Let A={1,2,3}, B={4,5,6}, and let f, g:A→B be given by tables: f(1)=5, f(2)=6, f(3)=4; g(1)=4, g(2)=4, g(3)=6. First prove, for any total function, the equivalence: bijective ⇔ the fiber of every b∈B has exactly one element. Then list the complete fiber profiles of f and g in the codomain order 4, 5, 6, and decide that f is bijective while g is neither injective nor surjective. Finally answer the different question of whether some bijection exists: one exists A→B (f is one), none exists {1,2,3}→{4,5}, and ∅→∅ has exactly one, the empty bijection."},"mko_ids":["mko-bijective"]}],"title":"Concept derivation guides"}