39

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.

Martin Sleziak
  • 56,060
JZS
  • 5,126

1 Answers1

48

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.

Ayman Hourieh
  • 40,828
  • 3
    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. – Yai0Phah Apr 20 '14 at 05:22
  • 2
    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 08 '14 at 07:07
  • 2
    @Arundhathi: If $a \in V \cap f(C)$, there exists $b \in C$ such that $f(b) = a$ and $b \in f^{-1}(V)$. Therefore $b \in E$, $a \in f(E)$ and $a \not\in \hat V$. – Miikka Feb 02 '15 at 17:34