# 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. – Henno Brandsma Jan 8 '16 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? – Dbchatto67 Jan 25 '19 at 10:26
• Is it fine now? @Dbchatto67 – R_D Jan 26 '19 at 3:34
• Yeah @R_D it's now absolutely fine. Thanks so much. – Dbchatto67 Jan 26 '19 at 6:04

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.