# Proper map

In mathematics, a function between topological spaces is called proper if inverse images of compact subsets are compact. In algebraic geometry, the analogous concept is called a proper morphism.

## Definition

A function $f\colon X\to Y$  between two topological spaces is proper if the preimage of every compact set in Y is compact in X.

There are several competing descriptions. For instance, a continuous map f is proper if it is closed with compact fibers, i.e. if it is a closed map and the preimage of every point in Y is compact. The two definitions are equivalent if Y is locally compact and Hausdorff.

Partial proof of equivalence

Let $f\colon X\to Y$  be a closed map, such that $f^{-1}(y)$  is compact (in X) for all $y\in Y$ . Let $K$  be a compact subset of $Y$ . We will show that $f^{-1}(K)$  is compact.

Let $\{U_{\lambda }\vert \lambda \ \in \ \Lambda \}$  be an open cover of $f^{-1}(K)$ . Then for all $k\ \in K$  this is also an open cover of $f^{-1}(k)$ . Since the latter is assumed to be compact, it has a finite subcover. In other words, for all $k\ \in K$  there is a finite set $\gamma _{k}\subset \Lambda$  such that $f^{-1}(k)\subset \cup _{\lambda \in \gamma _{k}}U_{\lambda }$ . The set $X\setminus \cup _{\lambda \in \gamma _{k}}U_{\lambda }$  is closed. Its image is closed in Y, because f is a closed map. Hence the set

$V_{k}=Y\setminus f(X\setminus \cup _{\lambda \in \gamma _{k}}U_{\lambda })$  is open in Y. It is easy to check that $V_{k}$  contains the point $k$ . Now $K\subset \cup _{k\in K}V_{k}$  and because K is assumed to be compact, there are finitely many points $k_{1},\dots ,k_{s}$  such that $K\subset \cup _{i=1}^{s}V_{k_{i}}$ . Furthermore the set $\Gamma =\cup _{i=1}^{s}\gamma _{k_{i}}$  is a finite union of finite sets, thus $\Gamma$  is finite.

Now it follows that $f^{-1}(K)\subset f^{-1}(\cup _{i=1}^{s}V_{k_{i}})\subset \cup _{\lambda \in \Gamma }U_{\lambda }$  and we have found a finite subcover of $f^{-1}(K)$ , which completes the proof.

If X is Hausdorff and Y is locally compact Hausdorff then proper is equivalent to universally closed. A map is universally closed if for any topological space Z the map $f\times \operatorname {id} _{Z}\colon X\times Z\to Y\times Z$  is closed. In the case that $Y$  is Hausdorff, this is equivalent to requiring that for any map $Z\to Y$  the pullback $X\times _{Y}Z\to Z$  be closed, as follows from the fact that $X\times _{Y}Z$  is a closed subspace of $X\times Z$ .

An equivalent, possibly more intuitive definition when X and Y are metric spaces is as follows: we say an infinite sequence of points $\{p_{i}\}$  in a topological space X escapes to infinity if, for every compact set $S\subseteq X$  only finitely many points $p_{i}$  are in S. Then a continuous map $f\colon X\to Y$  is proper if and only if for every sequence of points $\{p_{i}\}$  that escapes to infinity in X, the sequence $\{f(p_{i})\}$  escapes to infinity in Y.

## Properties

• A topological space is compact if and only if the map from that space to a single point is proper.
• Every continuous map from a compact space to a Hausdorff space is both proper and closed.
• If $f\colon X\to Y$  is a proper continuous map and Y is a compactly generated Hausdorff space (this includes Hausdorff spaces that are either first-countable or locally compact), then f is closed.

## Generalization

It is possible to generalize the notion of proper maps of topological spaces to locales and topoi, see (Johnstone 2002).