Suppose \(Y\) is Hausdorff, path-connected, locally path-connected, and semilocally simply-connected, and that its universal cover \(\widetilde Y\) is compact. Prove that \(\pi_1(Y)\) is finite.
Solution
Under the stated hypotheses \(Y\) has a universal covering \(p\colon\widetilde Y\to Y\), with \(\widetilde Y\) connected and simply connected. Fix \(y_0\in Y\) and let \(F=p^{-1}(y_0)\).
1. \(F\) is finite.
\(\{y_0\}\) is closed because \(Y\) is Hausdorff, so \(F\) is closed in the compact space \(\widetilde Y\) and hence compact. Over an evenly covered neighborhood of \(y_0\), each sheet meets \(F\) in exactly one point, so \(F\) is discrete. A compact discrete space is finite, since its cover by singletons has a finite subcover.
2. \(\abs{F}=\abs{\pi_1(Y,y_0)}\).
For a connected covering, the number of sheets is the index of \(p_*\pi_1(\widetilde Y,\tilde y_0)\) in \(\pi_1(Y,y_0)\), and this subgroup is trivial because \(\widetilde Y\) is simply connected.
3. Q.E.D.
Steps 1 and 2 show that \(\pi_1(Y,y_0)\) is finite.