Suppose that f:A->A and g:A->A. In each case, either prove your claim or give a

Question

Suppose that f:A->A and g:A->A. In each case, either prove your claim or give a counterexample.

Suppose that f(g(f)) is bijective. Does it follow that g is bijective?

Other information:
This is part c of the problem. In part a I proved that if g(f) and f(g) are both bijective, both f and g are bijective. In part b I proved that if f(f) is bijective, f is bijective.
I don't think that g is bijective, but I've been trying to find a counterexample when A={1, 2, 3} and can't seem to make it work.

Explanation / Answer

If f is injective, every member of X has its own unique matching member in X. But since this is the case, every member in X has a matching memeber, since it is a mapping to itself --> it is surjective. *If f is surjective, then every member of X has at least one matching member in X (maybe more than one). But it can't be more than one, since elsewise not every element of X is mapped. Thus it must be exactly one --> it is injective

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