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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?

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    $\begingroup$ What exactly do you mean by "pullback" here? You are using in a sense I have not seen before and can't find in a light search. In particular, how are you defining $f^*(K)$ and $f_*f^*(K)$? $\endgroup$ Jan 30, 2016 at 5:20
  • $\begingroup$ @PaulSinclair I mean inverse image. $f^\ast{B}=f^{-1}(B),f_\ast{A}= f(A)$. $\endgroup$
    – Arrow
    Jan 30, 2016 at 10:35

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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 $Y$

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  • $\begingroup$ Oops, silly blunder on my part showing only the image is closed! $\endgroup$
    – Arrow
    Jan 30, 2016 at 23:00

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