作者nomorethings (举步迈出一个闪亮)
看板trans_math
标题Re: [证明] 合成函数
时间Tue Aug 1 18:35:42 2006
※ 引述《GayerDior (蜡笔小新<(￣.￣)/)》之铭言：
: E:
: 请严格证明合成函数不满足交换律
: 如果有人会写，
: 请帮我写出详细过程，
: 谢谢唷~~
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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