Proof that a composition of surjections is surjective

Proof

We check the definition of surjectivity directly.

1. Let $f \colon A \to B$ and $g \colon B \to C$ be surjective functions.
2. Let $c \in C$. Since $g$ is surjective, there is $b \in B$ with $g(b) = c$.
3. Since $f$ is surjective, there is $a \in A$ with $f(a) = b$.
4. By the definition of composition and steps 2 and 3, $(g \circ f)(a) = g(f(a)) = g(b) = c$.
5. So each $c \in C$ equals $(g \circ f)(a)$ for some $a \in A$. Hence $g \circ f$ is surjective.