Sufficient Conditions for a Covering Space

[Euge]

Prove that every surjective local homeomorphism $\pi : \tilde{X} \to X$ from a compact Hausdorff space $\tilde{X}$ to a Hausdorff space $X$ is a covering space.

 

[Euge]

Spoiler

Fix $x\in X$. The fiber $\pi^{-1}(x)$ is a closed subset of the compact space $\tilde{X}$ so it is compact. Further, since $\pi$ is a local homeomorphism, $\pi^{-1}(x)$ is discrete. Therefore $\pi^{-1}(x)$ is a finite set, say $\{\tilde{x}_1,\ldots, \tilde{x}_n\}$. For each index $i\in \{1,\ldots, n\}$, there are open neighborhoods $U_{i} \ni \tilde{x}_i$ and $V_{i}\ni x$ such that $\pi$ restricts to a homeomorphism from $U_{i}$ onto $V_{i}$. Since $\tilde{X}$ is Hausdorff, we may assume the $U_{i}$ are disjoint. Then $O := \bigcup_{i = 1}^n V_i - \pi(\tilde{X}\setminus \bigcup_{i = 1}^n U_i)$ is a covering neighborhood of $x$, as desired.