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Suppose $X$ is a Riemann surface, $Y$ is a Hausdorff topological space and $p: Y\to X$ is a local homeomorphism. Then there is a unique complex structure on $Y$ such that $p$ is holomorphic. Now if $\pi:Y\to X$ is a branched cover, can we also lift a complex structure to $Y$, s.t. $\pi$ is holomorphic?

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closed as unclear what you're asking by Moishe Kohan, Yanior Weg, Xander Henderson, Adrian Keister, José Carlos Santos May 3 at 18:17

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    $\begingroup$ In the local homeomorphism case $\pi$ is a local chart making $Y$ a Riemann surface. Then it suffices to check what happens at isolated branch points where the degree is finite. $\endgroup$ – reuns Apr 21 at 18:19
  • $\begingroup$ @reuns sorry can you explain how you check? Can you give an example? $\endgroup$ – Danny Apr 22 at 6:55
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    $\begingroup$ You should first define what a "branched cover" means when $Y$ is a Hausdorff topological space. I know how to define it when $Y$ is a surface.... $\endgroup$ – Moishe Kohan Apr 23 at 1:32