# Sufficient Conditions for a Covering Space

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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.

jedishrfu, Greg Bernhardt and topsquark

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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.

topsquark