# composition with coercive function

###### 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. ∎

Title composition with coercive function CompositionWithCoerciveFunction 2013-03-22 15:20:16 2013-03-22 15:20:16 matte (1858) matte (1858) 6 matte (1858) Theorem msc 54A05