设f:A→B,g:BA,f•g=IA (此处A为下角标),证明:f是单射,g是满射
人气:190 ℃ 时间:2020-04-30 09:45:01
解答
题目应该是:
设有两个映射f:A→B,g:B→A.若g*f=IA ,则f是单射,g是满射.
证明
(1)证明映射f是单射.
对任意的b∈B,如果存在a1,a2∈A(a1!=a2),使g(a1)=b,g(a2)=b,即g(a1)=b= g(a2).
因为 a1=IA(a1)=(g*f)(a1)= f(g(a1)) = f(g(a2)) =(g*f)(a2) =IA(a2)= a2 .
所以f是单射的.
(2)证明映射g是满射.
因为(g*f)(A)=IA(A)= A,所以g*f是满射的.
又对任意的c∈A,由g*f是满射的可知,存在a∈A,使(g*f)(a)=c.
那么存在b∈B,使f(a) = b,g(b) = c.
所以存在b∈B,使g(b) = c,
所以g是满射的.
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