Composition of surjections is surjective

Statement

Let $f \colon A \to B$ and $g \colon B \to C$ be surjective functions. Then the composition $g \circ f \colon A \to C$ is surjective.

Remarks

Proof: let $c \in C$. Since $g$ is surjective, $c = g(b)$ for some $b \in B$. Since $f$ is surjective, $b = f(a)$ for some $a \in A$. Then $c = (g \circ f)(a)$. A partial converse holds: if $g \circ f$ is surjective, then $g$ is surjective. The function $f$ need not be surjective. Together with the result for injections, this shows that a composition of bijections is a bijection.

Explanations

L1 human

To reach a point $c$ of $C$, first find a point of $B$ that $g$ sends to $c$. Then find a point of $A$ that $f$ sends to that point of $B$.