Let \(p: \tilde{X} \to X\) be a covering space with \(p^{-1}(x)\) finite and nonempty for all \(x \in X\). Show that \(\tilde{X}\) is compact Hausdorff if \(X\) is compact Hausdorff.
Solution
Assume \(X\) is compact Hausdorff and every fiber of \[ p:\widetilde X\to X \] is finite and nonempty.
1. The space \(\widetilde X\) is Hausdorff.
Let \(\widetilde x\ne\widetilde y\).
If \[ p(\widetilde x)\ne p(\widetilde y), \] choose disjoint open neighborhoods \(U,V\subseteq X\) of these two image points, using Hausdorffness of \(X\). Then \[ p^{-1}(U),\qquad p^{-1}(V) \] are disjoint open neighborhoods of \(\widetilde x\) and \(\widetilde y\).
If \[ p(\widetilde x)=p(\widetilde y)=x, \] choose an evenly covered neighborhood \(U\) of \(x\). Distinct points in the same fiber lie in distinct sheets over \(U\) because each sheet maps injectively to \(U\). Those sheets are disjoint open neighborhoods of \(\widetilde x\) and \(\widetilde y\).
Thus any two distinct points of \(\widetilde X\) have disjoint open neighborhoods.
2. For every \(x\in X\), there are open sets\[ x\in V_x\subseteq\overline{V_x}\subseteq U_x \]with \(U_x\) evenly covered.
Choose an evenly covered open neighborhood \(U_x\) of \(x\). A compact Hausdorff space is regular, so there is an open neighborhood \(V_x\) of \(x\) whose closure satisfies \[ \overline{V_x}\subseteq U_x. \]
3. For each \(x\), the set\[ p^{-1}(\overline{V_x}) \]is compact.
Write the evenly covered decomposition \[ p^{-1}(U_x)=\coprod_{\lambda\in\Lambda_x}W_\lambda, \] with each \[ p|_{W_\lambda}:W_\lambda\to U_x \] a homeomorphism.
The number of sheets is exactly the cardinality of the fiber over \(x\): each sheet contains exactly one point over \(x\). By hypothesis this fiber is finite, so \(\Lambda_x\) is finite.
For each sheet, \[ W_\lambda\cap p^{-1}(\overline{V_x}) \] is homeomorphic via \(p\) to \(\overline{V_x}\). Since \(X\) is compact Hausdorff, the closed subset \(\overline{V_x}\) is compact. Therefore \(p^{-1}(\overline{V_x})\) is a finite union of compact sets and is compact.
4. Finitely many of the sets \(V_x\) cover \(X\).
The family \[ \{V_x:x\in X\} \] is an open cover of the compact space \(X\). Hence there are points \(x_1,\dots,x_m\) such that \[ X=V_{x_1}\cup\cdots\cup V_{x_m}. \]
5. The space \(\widetilde X\) is compact.
From 4, \[ \widetilde X =p^{-1}(X) =\bigcup_{i=1}^m p^{-1}(V_{x_i}) \subseteq \bigcup_{i=1}^m p^{-1}(\overline{V_{x_i}}) \subseteq\widetilde X. \] Thus equality holds: \[ \widetilde X = \bigcup_{i=1}^m p^{-1}(\overline{V_{x_i}}). \] Each set on the right is compact by 3, so their finite union is compact.
6. Hence \(\widetilde X\) is compact Hausdorff.
Hausdorffness is 1 and compactness is 5.