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
There are several competing definitions of a "proper function". Some authors call a function $$f : X \to Y$$ between two topological spaces  if the preimage of every compact set in $$Y$$ is compact in $$X.$$ Other authors call a map $$f$$ if it is continuous and ; 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.

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 . 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 $$X \times_YZ$$ 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 $$X$$  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 $$Y.$$

Properties

 * 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  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.

Generalization
It is possible to generalize the notion of proper maps of topological spaces to locales and topoi, see.