# Problem of the Week #203 - April 19, 2016

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#### Euge

##### MHB Global Moderator
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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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