composition with coercive function

Theorem 1.

Suppose $X\mathrm{,}Y\mathrm{,}Z$ are topological spaces, $f\mathrm{:}X\mathrm{\to }Y$ is a bijective proper map, and $g\mathrm{:}Y\mathrm{\to }Z$ is a coercive map. Then $g\mathrm{\circ }f\mathrm{:}X\mathrm{\to }Z$ 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
Canonical name	CompositionWithCoerciveFunction
Date of creation	2013-03-22 15:20:16
Last modified on	2013-03-22 15:20:16
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	6
Author	matte (1858)
Entry type	Theorem
Classification	msc 54A05