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Let $F:X\rightarrow Y$ a unramified holomorphic function between two compact Riemann surfaces. I don't understand why $F$ is a covering map. By a well-known theorem $F$ is surjective; then since the ramification index at every point is $1$, for every chart $(V,\psi)$ centered at $y\in Y$, exists a chart $(U,\phi)$ centered at $x\in X$ such that $\psi\circ F\circ\phi^{-1}:\phi(U)\rightarrow\psi(V)$ is the identity map. I don't see why $F^{-1}(V)$ breaks into a disjoint union of open sets $M_i\subseteq X$ with the map $F$ sending each of them omeomorphically to $V$.

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First, prove that $F$ is a local homeomorphism. Then show that a surjective local homeomorphism between two compact spaces is automatically a covering map. – Zhen Lin Sep 22 '12 at 16:09

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