Let $f : X \to Y$ and $g : Y \to Z$ each be a D466: Surjective map such that

(i)	$g \circ f : X \to Z$ is a D527: Composite map of $f$ with $g$

Due to result R4628: Map to empty set is a surjection, we may assume that $Z$ is nonempty. Fix $z \in Z$. Since $g$ is a surjection, there exists $y \in Y$ such that $g(y) = z$. And since $f$ is a surjection, there exists $x \in X$ such that $f(x) = y$. Hence, $z = g(y) = g(f(x)) = h(x)$, which shows that $h$ is a surjection. $\square$