Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 21 down vote accepted

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.

share|cite|improve this answer
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 '14 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 '14 at 7:07
@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 2 '15 at 17:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.