Proper map

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 and only 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. For a proof of this fact see the end of this section. More abstractly, "f" is proper if it is a closed map, and for any space "Z" the
_"Z"): "X" × "Z" → "Y" × "Z"is closed. These definitions are equivalent to the previous one if the space "X" is locally compact.

An equivalent, possibly more intuitive definition 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 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".

Proof of fact

Let $f:\; X\; o\; Y$ 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 Harv|Johnston|2002.

See also

* Perfect map
* Topology glossary

References

* | year=1998
*, esp. section C3.2 "Proper maps"