Proper map

This article is about the concept in topology. For the concept in convex analysis, see proper convex function.

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 : XY 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 locally compact 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 × idZ: 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 {pi} in a topological space X escapes to infinity if, for every compact set SX only finitely many points pi are in S. Then a continuous map f : XY is proper if and only if for every sequence of points {pi} that escapes to infinity in X, {f(pi)} 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 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

Generalization

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

See also

References

  1. Palais, Richard S. (1970). "When proper maps are closed" (PDF). Proc. Amer. Math. Soc. 24: 835–836. doi:10.1090/s0002-9939-1970-0254818-x.
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