Why Are Holomorphic Mappings on Compact Riemann Surfaces Constant?

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Here is this week's POTW:

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Show that the holomorphic mappings on a compact connected Riemann surface are constant.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one answered this week's problem. You can read my solution below.

Spoiler

The proof is made simple by applying the open mapping theorem for Riemann surfaces. Let $X$ be a compact connected Riemann surface, and let $f : X\to \Bbb C$ be a holomorphic mapping. By the open mapping theorem, $f(X)$ is open in $\Bbb C$. Since $f$ is continuous and $X$ is compact, $f(X)$ is a compact, hence closed, subset of $\Bbb C$. As $f(X)$ is a nonempty, open and closed subset of the connected set $\Bbb C$, $f(X) = C$. This is a contradiction since $\Bbb C$ is not compact.