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Proper map

Mathematical map between topological spaces

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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.

01Definition

There are several competing definitions of a "proper function". Some authors call a function $f:X\to Y$ between two topological spaces proper if the preimage of every compact set in $Y$ is compact in $X.$ Other authors call a map $f$ proper if it is continuous and closed with compact fibers; that is if it is a continuous 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 $f:X\to Y$ 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 $Y.$ It remains to show that $f^{-1}(K)$ is compact. Let $\left\{U_{a}:a\in A\right\}$ 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 every $k\in K,$ there exists a finite subset $\gamma _{k}\subseteq A$ such that $f^{-1}(k)\subseteq \cup _{a\in \gamma _{k}}U_{a}.$ The set $X\setminus \cup _{a\in \gamma _{k}}U_{a}$ is closed in $X$ and its image under $f$ is closed in $Y$ because $f$ is a closed map. Hence the set $V_{k}=Y\setminus f\left(X\setminus \cup _{a\in \gamma _{k}}U_{a}\right)$ is open in $Y.$ It follows that $V_{k}$ contains the point $k.$ Now $K\subseteq \cup _{k\in K}V_{k}$ and because $K$ is assumed to be compact, there are finitely many points $k_{1},\dots ,k_{s}$ such that $K\subseteq \cup _{i=1}^{s}V_{k_{i}}.$ Furthermore, the set $\Gamma =\cup _{i=1}^{s}\gamma _{k_{i}}$ is a finite union of finite sets, which makes $\Gamma$ a finite set. Now it follows that $f^{-1}(K)\subseteq f^{-1}\left(\cup _{i=1}^{s}V_{k_{i}}\right)\subseteq \cup _{a\in \Gamma }U_{a}$ and we have found a finite subcover of $f^{-1}(K),$ which completes the proof.

Partial proof of equivalence

Let $f:X\to Y$ 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 $Y.$ It remains to show that $f^{-1}(K)$ is compact.

Let $\left\{U_{a}:a\in A\right\}$ 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 every $k\in K,$ there exists a finite subset $\gamma _{k}\subseteq A$ such that $f^{-1}(k)\subseteq \cup _{a\in \gamma _{k}}U_{a}.$ The set $X\setminus \cup _{a\in \gamma _{k}}U_{a}$ is closed in $X$ and its image under $f$ is closed in $Y$ because $f$ is a closed map. Hence the set $V_{k}=Y\setminus f\left(X\setminus \cup _{a\in \gamma _{k}}U_{a}\right)$ is open in $Y.$ It follows that $V_{k}$ contains the point $k.$ Now $K\subseteq \cup _{k\in K}V_{k}$ and because $K$ is assumed to be compact, there are finitely many points $k_{1},\dots ,k_{s}$ such that $K\subseteq \cup _{i=1}^{s}V_{k_{i}}.$ Furthermore, the set $\Gamma =\cup _{i=1}^{s}\gamma _{k_{i}}$ is a finite union of finite sets, which makes $\Gamma$ a finite set.

Now it follows that $f^{-1}(K)\subseteq f^{-1}\left(\cup _{i=1}^{s}V_{k_{i}}\right)\subseteq \cup _{a\in \Gamma }U_{a}$ 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 $f\times \operatorname {id} _{Z}:X\times Z\to Y\times Z$ 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 $f:X\to Y$ is proper if and only if for every sequence of points $\left\{p_{i}\right\}$ that escapes to infinity in $X,$ the sequence $\left\{f\left(p_{i}\right)\right\}$ escapes to infinity in $Y.$

02Properties

Every continuous map from a compact space to a Hausdorff space is both proper and closed.
Every surjective proper map is a compact covering map.
- A map $f:X\to Y$ is called a compact covering if for every compact subset $K\subseteq Y$ there exists some compact subset $C\subseteq X$ such that $f(C)=K.$
A topological space is compact if and only if the map from that space to a single point is proper.
If $f:X\to Y$ 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.

03Generalization

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

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Sources and credits

This article is adapted from the Wikipedia article “Proper map”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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Proper map

01Definition

02Properties

03Generalization

Sources and credits

Related topics

Almost open map

Open and closed maps

Perfect map