Show that a surjective local homeomorphism between compact Hausdorff spaces is a covering map.
Solution
Let \(p:E\to B\) be a surjective local homeomorphism, with \(E\) compact and \(B\) Hausdorff.
1. Every fiber \(p^{-1}(b)\) is finite.
The fiber is closed in compact \(E\), hence compact. For each \(x\in p^{-1}(b)\), local homeomorphy gives an open neighborhood \(U_x\) on which \(p\) is injective, so \(U_x\cap p^{-1}(b)=\{x\}\). Thus the fiber is compact and discrete, hence finite.
2. Fix \(b\in B\) and write\[ p^{-1}(b)=\{x_1,\dots,x_r\}. \]Choose pairwise disjoint open neighborhoods \(U_i\) of the \(x_i\) such that each restriction\[ p|_{U_i}:U_i\longrightarrow p(U_i) \]is a homeomorphism onto an open neighborhood of \(b\).
Local homeomorphy gives such neighborhoods individually, and the Hausdorff property together with finiteness of the fiber allows them to be shrunk to be pairwise disjoint.
3. Put\[ K=E\setminus\bigcup_{i=1}^r U_i. \]Then \(K\) is compact, so \(p(K)\) is compact and therefore closed in Hausdorff \(B\). Since no point of the fiber lies in \(K\), we have \(b\notin p(K)\).
4. Define\[ V=(B\setminus p(K))\cap\bigcap_{i=1}^r p(U_i). \]Then \(V\) is an open neighborhood of \(b\) and\[ p^{-1}(V)=\bigsqcup_{i=1}^r (U_i\cap p^{-1}(V)). \]For each \(i\), the restriction\[ p:U_i\cap p^{-1}(V)\longrightarrow V \]is a homeomorphism.
5. Thus every \(b\in B\) has an evenly covered neighborhood, so \(p\) is a covering map.