Proving surjection across sets

Question

I've been challenged by a rather straightforward question. I need to brush up on my set theory. Anyways, here's the problem I'm attempting to solve:

Given sets $A, B, C$ sets, and surjections $f:A\to B, g:B\to C$, prove that $h=g\circ f:A\to C$ is also a surjection.

So far I've only given the definition of surjective functions, but I'm not sure how to prove that $h$ is surjective.

Thanks

Answer 1

Given $c\in C$, there exists $b\in B$ with $g(b)=c$ because $g$ is surjective. For this (or such) $b$, there exists $a\in A$ with $f(a)=b$ because $f$ is surjective. Hence for our original $c$, we have exhibited $a\in A$ with $g(f(a))=c$. Unavoidably done.