Theorem 1   Suppose $X,Y,Z$ are topological spaces, $f\colon X\to Y$ is a bijective proper map, and $g\colon Y\to Z$ is a coercive map. Then $g\circ f\colon X\to Z$ is a coercive map.

Proof. Let $J\subseteq Z$ be a compact set. As $g$ is coercive, there is a compact set $K\subseteq Y$ such that $$ g(Y\setminus K) \subseteq Z\setminus J. $$ Let $I=f^{-1}(K)$ , and since $f$ is a proper map $I$ is compact. Thus $$ (g\circ f) (X\setminus I) = g(Y\setminus K) \subseteq Z\setminus J $$ and $g\circ f$ is coercive.