Continuous proper map into compactly generated Hausdorff space is closed

Question

Throughout, 'proper' means 'pulls back compacts to compacts'.

I've read here and there some claims about properness implying closedness. I want to check whether my attempt at a generalization is correct.

Proposition. Suppose $Y$ is compactly generated Hausdorff. Let $f:X\rightarrow Y$ be a continuous function. Then $f$ is proper $\implies f$ is closed.

Proof. Since $Y$ is compactly generated $f(X)$ is closed in $Y$ iff $f_\ast f^\ast (K)$ is closed in $K$ for all compacts $K$ in $Y$. Since $Y$ is Hausdorff, $K$ is compact Hausdorff and hence $f_\ast f^\ast (K)$ is closed in $K$ iff its compact in $K$. $f^\ast(K)$ is compact since $f$ is proper and its image is compact as a continuous image of a compact set. Since $K$ is arbitrary this proves $f(X)$ is closed in $Y$ as desired.

Is my proof correct?

Answer 1

To prove that a map $f:X\to Y$ is closed it is not enough to show that $f(X)$ is closed in $Y$. You have to show that for every closed set $C$ of $X$ , $f(C)$ is closed in $Y$.

Now if, $C\subseteq X$ is closed, and let $K$ be a compact subspace of $Y$. Then $f^{-1}(K)$ is compact, and so is $f^{-1}(K)\cap C =: A$. Then $f(A)=K\cap f(C)$ is compact, and as $Y$ is Hausdorff, $f(A)$ is closed. Since $Y$ is compactly generated, $f(C)$ is closed in $Y$