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A map is called proper if the pre-image of a compact set is again compact.

In the Differential Forms in Algebraic Topology by Bott and Tu, they remark that the image of a proper map $f: \mathbb{R}^n \to \mathbb R^m$ is closed, adding the comment "(why?)".

I can think of a simple proof in this case for continuous $f$:

If the image is not closed, there is a point $p$ that does not belong to it and a sequence $p_n \in f(\mathbb R^n)$ with $p_n \to p$. Since $f$ is proper $f^{-1}(\overline {B_\delta(p)})$ is compact for any $\delta$. Let $x_n$ be any point in $f^{-1}(p_n)$ and wlog $x_n \in f^{-1}(\overline{B_\delta(p)})$. Since in $\mathbb{R}^n$ compact and sequentially compact are equivalent, there exists a convergent subsequence $x_{n_k}$ of $x_n$. From continuity of $f$: $f(x_{n_k}) \to f(x)$ for some $x$. But $f(x_{n_k})=p_{n_k} \to p$ which is not supposed to be in the image and this gives a contradiction.

My problem is that this proof is too specific to $\mathbb{R}^n$ and uses arguments from basic analysis rather than general topology.

So the question is for what spaces does it hold that the image of a proper map is closed, how does the proof work, and is it necessary to pre-suppose continuity?

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    $\begingroup$ Often map already implies continuity. I'd check the text for this. $\endgroup$ Commented Jan 8, 2016 at 12:14

2 Answers 2

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First of all the definition of a proper map assumes continuity by convention (I have not come across texts that say otherwise)

Secondly, here is a more general result -

Lemma : Let $f:X\rightarrow Y$ be a proper map between topological spaces $X$ and $Y$ and let $Y$ be locally compact and Hausdorff. Then $f$ is a closed map.

Proof : Let $C$ be a closed subset of $X$. We need to show that $f(C)$ is closed in $Y$ , or equivalently that $Y\setminus f(C)$ is open.

Let $y\in Y\setminus f(C)$. Then $y$ has an open neighbourhood $V$ with compact closure. Then $f^{-1}(\bar{V})$ is compact.

Let $E=C\cap f^{-1}(\bar{V})$ . Then clearly $E$ is compact and hence so is $f(E)$. Since $Y$ is Hausdorff $f(E)$ is closed.

Let $U=V\setminus f(E)$. Then $U$ is an open neighbourhood of $y$ and is disjoint from $f(C)$.

Thus $Y\setminus f(C)$ is open. $\square$

I hope this helps.

EDIT: To clarify the statement $U$ is disjoint from $f(C)$ -

Suppose $z\in U\cap f(C)$ Then there exists a $c\in C$ such that $z=f(c)$. This means $c\in f^{-1}(U)\subseteq f^{-1}(V)\subseteq f^{-1}(\bar V)$. So $c\in C\cap f^{-1}(\bar V)=E$. So $z=f(c)\in f(E)$ which is a contradiction as $z\in U$.

Further Edit from @Hans: $U$ is disjoint from $f(C)$ is essentially and seen clearer from the following equation $$f(A\cap f^{-1}(B))=f(A)\cap B$$ for arbitrary function $f$ and sets $A$ and $B$. We shall show $\supseteq$ as the other direction is trivial. Suppose $z\in f(A)\cap B$. $\exists a\in A\ni z=f(a).\ z\in B\implies a\in f^{-1}(B)$.

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  • $\begingroup$ Why is $U$ disjoint from $f(C)$? From your definition it is clear that $E \subseteq C$. So $f(E) \subseteq f(C)$. Hence $V \setminus f(C) \subseteq V \setminus f(E) = U$. If the containment is proper then $U$ may contain some element of $f(C)$. Who knows that? Isn't it so @R_D? $\endgroup$ Commented Jan 25, 2019 at 10:26
  • $\begingroup$ Is it fine now? @Dbchatto67 $\endgroup$ Commented Jan 26, 2019 at 3:34
  • $\begingroup$ Yeah @R_D it's now absolutely fine. Thanks so much. $\endgroup$ Commented Jan 26, 2019 at 6:04
  • $\begingroup$ Why is $f(E)$ compact? $\endgroup$ Commented Mar 18, 2020 at 3:26
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    $\begingroup$ @XiuyiYang continuous image of a compact set is compact. $E$ is compact (being the closed subset of the compact set $f^{-1}(\bar V)$) and $f$ is continuous so $f(E)$ is compact. $\endgroup$ Commented Mar 18, 2020 at 14:58
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One may generalize the result in R_D's answer even further:

A proper map $f:X\to Y$ to a compactly generated Hausdorff space is a closed map (A space $Y$ is called compactly generated if any subset $A$ of $Y$ is closed when $A\cap K$ is closed in $K$ for each compact $K\subseteq Y$).
Proof: Let $C\subseteq X$ be closed, and let $K$ be a compact subspace of $Y$. Then $f^{-1}(K)$ is compact, and so is $f^{-1}(K)\cap C =: B$. Then $f(B)=K\cap f(C)$ is compact, and as $Y$ is Hausdorff, $f(B)$ is closed. Since $Y$ is compactly generated, $f(C)$ is closed in $Y$.

A locally compact space $Y$ is compactly generated: If $A\subset Y$ intersects each compact set in a closed set, and if $y\notin A$, then $A$ intersects the compact neighborhood $K$ of $y$ in a closed set $C$. Now $K\setminus C$ is a neighborhood of $y$ disjoint from $A$, hence $A$ is closed.

We need the following proposition for the above proof. $$f(A\cap f^{-1}(B))=f(A)\cap B$$ for arbitrary function $f$ and sets $A$ and $B$. We shall show $\supseteq$ as the other direction is trivial. Suppose $z\in f(A)\cap B$. $\exists a\in A\ni z=f(a).\ z\in B\implies a\in f^{-1}(B)$.

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