 Proper Map
Get Proper Map essential facts below. View Videos or join the Proper Map discussion. Add Proper Map to your PopFlock.com topic list for future reference or share this resource on social media.
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).