Re: [証明] 合成函數

[nomorethings (舉步邁出一個閃亮)]

※ 引述《GayerDior (蠟筆小新<(￣.￣)/)》之銘言：
: E:
: 請嚴格証明合成函數不滿足交換律
: 如果有人會寫，
: 請幫我寫出詳細過程，
: 謝謝唷~~
Suppose f:A->B g:B->A be two functions ,then by the definition of
comosition of functions we have :
        f(g(.)):g(B)∩A->B  and g(f(.)):f(A)∩B->A .
Since the domains of f(g(.)) and g(f(.)) are not generally equal ,
so these two functions are not generally equal .
                 This completes the proof.



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BTW, you can also make a counterexample to proof this .

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