Harman Patil (Editor)

Proper map

Updated on
Edit
Like
Comment
Share on FacebookTweet on TwitterShare on LinkedInShare on Reddit

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.

Contents

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 preimage 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 × ZY × 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 Y be a closed map, such that f 1 ( y ) is compact (in X) for all y Y . Let K be a compact subset of Y . We will show that f 1 ( K ) is compact.

Let { U λ | λ     Λ } be an open cover of f 1 ( K ) . Then for all k   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   K there is a finite set γ k Λ such that f 1 ( k ) λ γ k U λ . The set X λ γ k U λ is closed. Its image is closed in Y, because f is a closed map. Hence the set

V k = Y f ( X λ γ k U λ ) is open in Y. It is easy to check that V k contains the point k . Now K k K V k and because K is assumed to be compact, there are finitely many points k 1 , , k s such that K i = 1 s V k i . Furthermore the set Γ = i = 1 s γ k i is a finite union of finite sets, thus Γ is finite.

Now it follows that f 1 ( K ) f 1 ( i = 1 s V k i ) λ Γ U λ 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 : XY 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 (Johnstone 2002).

    References

    Proper map Wikipedia


    Similar Topics