Proof
(related to Proposition: Composition of Surjective Functions is Surjective)
 Let $c\in C$.
 Since $g:B\to C$ is surjective, there is a $b\in B$ with $g(b)=c$.
 Since $f:A\to B$ is surjective, there is an $a\in A$ with $f(a)=b$.
 Since $(g\circ f)$ is a composition, we have $(g\circ f)(a)=g(f(a))=g(b)=c$.
 Therefore, there is an $a\in A$ with $(g\circ f)(a)=c$.
 Thus, $(g\circ f)$ is surjective.
∎
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References
Bibliography
 Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016