Proper map

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In mathematics, a continuous 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 : X → 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 a closed map and the pre-image of every point in Y is compact. The two definitions are equivalent if Y is compactly generated and Hausdorff. For a proof of this fact see the end of this section. More abstractly, f is proper if f is universally closed, i.e. if for any topological space Z the map

f × id_Z: X × Z → Y × Z

is closed. These definitions are equivalent to the previous one if X is Hausdorff and Y is locally compact Hausdorff.

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 ⊂ X only finitely many points p_i are in S. Then a continuous map f : X → Y is proper if and only if for every sequence of points {p_i} that escapes to infinity in X, {f(p_i)} escapes to infinity in Y.

This last sequential idea looks like being related to the notion of sequentially proper, see a reference below.

Proof of fact

Let $f:X\to Y$ be a continuous 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.

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 : X → Y is a proper continuous map and Y is a compactly generated Hausdorff space (this includes Hausdorff spaces which 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 Template:Harv.

See also

• Perfect map
• Topology glossary

References

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• Brown, R. "Sequentially proper maps and a sequential compactification", J. London Math Soc. (2) 7 (1973) 515-522.