54A05-CompositionWithCoerciveFunction.tex

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\pmtitle{composition with coercive function}
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\begin{document}
\begin{thm} 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.
\end{thm}

\begin{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.
\end{proof}
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