Compact universal covering spaces

Question

Let $X$ be a topological compact space admitting a universal covering $C$. When is $C$ again compact?

Thanks.

Answer 1

Assuming $X$ is Hausdorff, $C$ is compact if and only if $\pi_1(X)$ is finite. If $C$ is compact, then (because points of $X$ are closed), $p^{-1}(x)$ is closed, and so compact (and discrete!), so finite. Because $|p^{-1}(x)| = |\pi_1(X)|$, we see that $\pi_1(X)$ is finite.

Conversely, if $X$ is compact Hausdorff, any finite covering space is compact. See here for a sketch of a proof; you should fill in the details yourself.

Answer 2

Let $X$ be hausdorff and connected.

Claim: The universal covering $\tilde X \to X$ is compact if and only if $\pi_1(X)$ is finite.

Proof: If it is compact it follows that $\pi_1X$ is finite, since we could cover it otherwise with infinitely many trivializations around the basepoint which don't admit a subcover.

If $\pi_1X$ is finite, then it is a finite covering, hence compact.