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Given A,B,C are sets and g: A -> B, f: B -> C.

I've been at this for a while now while also focusing on similar problems. I'm thinking a contradiction what be easiest suited.

So far,

Suppose f o g is surjective, g is not surjective and with the view to contradiction, suppose f is injective.

Since g is not surjective, we know there exists some b in B such that for all a in A, g(a) != b.

Moreover, since f is injective, for all b, b' in B such that b != b' => f(b) != f(b').

let c be in C, such that c = f(b), but there exists no g(a) = b, and given there exists a c that has no a such that (f o g)(a) = c, this opposes our assumption that f o g is surjective.

This is a contradiction, so f is not injective.

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It is correct. I would do a direct approach: if $f\circ g$ is surjective and $g$ is not surjective, you want to prove that $f$ is not injective. Take $b\in B$ wich doesn't belong to the image of $g$ and take $a\in A$ such that $f\bigl(g(a)\bigr)=f(b)$. But $b\neq g(a)$. Therefore, $f$ is not injective.

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without using elements:

the image $f \circ g(A) \subseteq f(B)$. So, if $f \circ g (A)=C$, then $f$ better be surjective as well. If $f$ were a bijection, then there is an inverse $f^{-1}$. Then $f(g(A))=C=f(B)$. Apply the inverse to both sides of this equation.

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