Why nonconstant holomorphic map between compact riemann surfaces is surjection? I don't understand: if open-closed connected subset X of connected space Y, then X is all Y??
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4This is not research-level. The image $f(X)$ of a compact open set $X$ (such as the original Riemann surface) under a non-constant holomorphic map $f:X \rightarrow Y$ must be compact and open, therefore clopen. So the complement $Y - f(X)$ must also be open (compact sets are closed in Hausdorff spaces), so $Y$ is disconnected by definition. Contradiction. – Jul 05 '14 at 19:24
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Remember that nonconstant holomorphic maps are open (any open set of a compact Riemann surface is a union of open sets that are each holomorphically equivalent to an open disc in the complex plane; then use the open mapping theorem and the fact that taking images under a function preserves unions), and that continuous maps from a compact space to a Hausdorff space are closed. Hence the image $f(X)$ is both open and closed in $Y$. If your definition of compact Riemann surface includes connectedness, then the image must be all of $Y$.
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Dear user: an open annulus is not holomorphically equivalent to an open disc! – Bruno Joyal Jul 05 '14 at 20:02
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@BrunoJoyal But an open annulus is a union of open discs. (Perhaps the English threw you off?) – user43208 Jul 05 '14 at 20:12
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1Oops! I simply misread. Your answer is fine (and so is my English, I think...) – Bruno Joyal Jul 05 '14 at 20:15