← cgad.ski 2023-03-29

Hausdorff and Compact

It's easy to prove that the inverse of a continuous bijection f ⁣:RRf \colon \R \to \R is continuous—basically, "because ff is monotone." Proving the same for the inverse of a continuous bijection f ⁣:RnRnf \colon \R^n \to \R^n is quite a bit harder.

What are some general situations where it's easy prove that a continuous bijection has a continuous inverse? The following is probably my favorite short result from point-set topology.

Proposition: A continuous map from a compact space to a Hausdorff space is closed.

The proof is just a collection of definitions: a closed set in a compact space is compact, the continuous image of a compact set is compact, and a compact set in a Hausdorff space is closed. However, this is enough to prove that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

What if our domain is not compact? Can we use the compact-to-Hausdorff idea to get anything for free? I was thinking about this today and came up with the following.

Proposition. A continuous, proper (preimage of compacts are compact) bijection from anything into a locally compact Hausdorff space is a homeomorphism.

Proof. Let f ⁣:ABf \colon A \to B be our map. Proving that f1f^{-1} is continuous is the same as proving that ff is closed. If AA is compact then we're done by the above. Otherwise, lift to a map f ⁣:ABf^* \colon A^* \to B^* between Alexandroff extensions, defining f()=f^*(\infty) = \infty. (We take A=A⨿{}A^* = A \amalg \{ \infty \}, with the additional open sets U{}U \cup \{ \infty \} for all open UU such that AUA \setminus U is compact.)

We need three facts about the Alexandroff extension.

  1. The lift ff^* is continuous exactly when ff is proper. (Basically by definition.)
  2. AA^* is compact. (This doesn't depend on anything—the Alexandroff extension is always compact, essentially by definition.)
  3. BB^* is Hausdorff. (This relies both on BB being locally compact and Hausdorff to guarantee that any point can be separated from \infty.)

Now the proof is easy—again a collection of definitions.

Let CC be a closed set in AA. Suppose that it was not previously compact. It extends to a compact set C=C{}C^* = C \cup \{ \infty \} in AA^*. Since ff is proper, ff^* is continuous and f(C)=f(C){}f^*(C) = f(C) \cup \{ \infty \} is compact in BB^*. Since BB^* is Hausdorff, f(C)f^*(C) is closed, and hence f(C)f(C) is closed relative to BB. □

← cgad.ski