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Consider a compact Riemann surface $X$. If $p_1:Y\longrightarrow X $ and $p_2:Z\longrightarrow X$ are two topological coverings of $X$, then it is univocally defined a structure of Riemann surface on both $Y$ and $Z$ such that $p_1$ and $p_2$ are holomorphic. Now if these two coverings are equivalent, then there is a homeomorphism $g: Y\longrightarrow Z$ such that $p_2\circ g=p_1$, but is $g$ a biholomorphism? In other terms two equivalent coverings of a compact Riemann surfaces are isomorphic in the category of Riemann surfaces?

I think that the answer is yes because in classical complex analysis if $g$ and $f$ are two holomorphic functions with $g$ non costant and moreover if $F$ is a continuous function such that $g\circ F=f$, then $F$ is holomorphic. I can't formalize this argument in the case of Riemann surfaces.

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    $\begingroup$ You just need to verify that the homeomorphism is locally holomorphic (and the same for its inverse). $\endgroup$ – Dustan Levenstein Mar 22 '14 at 21:20
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    $\begingroup$ And since the coverings are local biholomorphisms, that isn't hard. $\endgroup$ – Daniel Fischer Mar 22 '14 at 21:22
  • $\begingroup$ Yes, I had missed the local biholomorphism of the coverings. Many thanks $\endgroup$ – Dubious Mar 22 '14 at 21:25
  • $\begingroup$ Being holomorphic is a local property. Everything that is a local property works the same for Riemann surfaces as in classical complex analysis; just pick convenient local charts and do things as usual. $\endgroup$ – Gunnar Þór Magnússon Mar 22 '14 at 21:28

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