When is the image of a proper map closed?

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?

• Often map already implies continuity. I'd check the text for this. Jan 8, 2016 at 12:14

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 $$z\in U$$.

• 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? Jan 25, 2019 at 10:26
• Is it fine now? @Dbchatto67
– R_D
Jan 26, 2019 at 3:34
• Yeah @R_D it's now absolutely fine. Thanks so much. Jan 26, 2019 at 6:04
• Why is $f(E)$ compact? Mar 18, 2020 at 3:26
• @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.
– R_D
Mar 18, 2020 at 14:58

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.