← cgad.ski
2023-03-29

It's easy to prove that the inverse of a continuous bijection $f \colon \R \to \R$ is continuous—basically, "because $f$ is monotone." Proving the same for the inverse of a continuous bijection $f \colon \R^n \to \R^n$ 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 $f \colon A \to B$ be our map. Proving that $f^{-1}$ is continuous is the same as proving that $f$ is closed. If $A$ is compact then we're done by the above. Otherwise, lift to a map $f^* \colon A^* \to B^*$ between Alexandroff extensions, defining $f^*(\infty) = \infty$. (We take $A^* = A \amalg \{ \infty \}$, with the additional open sets $U \cup \{ \infty \}$ for all open $U$ such that $A \setminus U$ is compact.)

We need three facts about the Alexandroff extension.

- The lift $f^*$ is continuous exactly when $f$ is proper. (Basically by definition.)
- $A^*$ is compact. (This doesn't depend on anything—the Alexandroff extension is always compact, essentially by definition.)
- $B^*$ is Hausdorff. (This relies both on $B$ 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 $C$ be a closed set in $A$. Suppose that it was not previously compact. It extends to a compact set $C^* = C \cup \{ \infty \}$ in $A^*$. Since $f$ is proper, $f^*$ is continuous and $f^*(C) = f(C) \cup \{ \infty \}$ is compact in $B^*$. Since $B^*$ is Hausdorff, $f^*(C)$ is closed, and hence $f(C)$ is closed relative to $B$. □