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主题:函数的映射

有没有学英文版离散的高手呀?

Suppose that we are given functions g : A→B and  f : B→C.

(a) If f and f o g are one-to-one, does it follow that g is one-to-one? Justify your answer.
(b) If g and f o g are one-to-one, does it follow that f is one-to-one? Justify your answer.

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沙发

设a1,a2A,a1≠a2,由于f是单射的,故有:
b1≠b2∈B且b1=f(a1)≠b2=f(a2)
又因为f.g也是单射,所以有
c1≠c2∈C且c1=f.g(a1)=g(f(a1))≠c2=f.g(a2)=g(f(a2))
推出g(b1)≠g(b2)又因为b1≠b2
所以得证明了
第2个不说了.是一样的
我英文不好,若理解错了,那就不好意思了

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