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Problem of the Week #203 - April 19, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,894
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,894
No one answered this week's problem. You can read my solution below.


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.
 
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