Re: [高微] ｆ is injective<=> there exists a func

[armopen (八字-風水-姓名學)]

※ 引述《henry781114 (期王)》之銘言：
: ｆ：Ｘ－＞Ｙ，show that
: ｆ is injective
: <=>there exists a function ｇ：Ｙ－＞Ｘ such that g(f(x))=x for all x∈Ｘ
: 想請問大家，從左邊證到右邊
: 要怎麼證明存在唯一呢?
: 我想應該是要用這個 f(x1)=f(x2) => x1=x2
: 但該如何將證明寫得完整呢?
: 謝謝各位......

Suppose that f is an injective function from X into Y.

Let y' in f(X). Then there is a unique x' in X such that f(x') = y'.

(For if there is also  x'' in X such that f(x'') = y', then f(x'') = f(x')

implies x'' = x' since f is injective.) Choose a fixed x0 in X and

verify the map g: Y -> X defined by

         g(b) = a , if b is in f(X) and f(a) = b

                x0, if b is not in f(X)

is such that g[f(x)] = x for all x in X. The detail leave you as an exercise.

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