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Let $f: X \to Y$ be continuous and proper (a map is proper iff the preimage of a compact set is compact). Furthermore, assume that $Y$ is locally compact and Hausdorff (there are various ways of defining local compactness in Hausdorff spaces, but let's say this means each point $y \in Y$ has a local basis of compact neighborhoods).

Prove that $f$ is a closed map.

I know that this proof cannot require much more than a basic topological argument. But there's just something that I'm missing.

We can start with $C \subseteq X$ closed, and then try to show that $Y \setminus F(C)$ is open (for each $q \in Y \setminus F(C)$, we would want to find an open set $V_q$ with $q \in V_q \subseteq Y \setminus F(C)$).

Hints or solutions are greatly appreciated.

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1 Answer 1

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Let $C \subset X$ be closed. Let $y \in Y - f(C)$. Since $Y$ is locally compact, $y$ has a neighborhood $V$ with compact closure. Since $f$ is proper, $f^{-1}(\overline{V})$ is compact in $X$. Let $E = C \cap f^{-1}(\overline{V})$. $E$ is compact; thus, $f(E)$ is compact. Since $Y$ is Hausdorff, $f(E)$ is closed. Let $\hat V = V - f(E)$. $\hat V$ is a neighborhood of $y$ disjoint from $f(C)$ as desired.

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In fact, locally compact spaces are compactly generated, and a continuous from a topological space to a compactly generated Hausdorff space is proper if and only if it's closed and preimages of singletons are compact. –  Frank Science Apr 20 at 5:22
    
Why is $\hat V$ disjoint from $f(C)$? It is clear that is disjoint from $f(E)$ but not why it is from $f(C)$. Am I missing something? Thanks! –  Arundhathi Dec 8 at 7:07

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