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Let $f:\mathcal{M} \to \hat{\mathbb{C}}$ where $\mathcal{M}$ is a arbitrary Riemann surface and $f$ is a meromorphic function. Let $A \subset \mathcal{M}$. If $f:A \to B$ then $f:\partial A \to \partial B$.

Does this result hold? Does $A$ have to be compact? What's a nice concise but clear proof?


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2 Answers 2

No. Let $B=\hat{\mathbb{C}}$, and $A$ be anything with nonempty boundary. Then $f$ certainly maps $A$ into $B$, but since $\partial B=\varnothing$, $f$ can't map $\partial A$ into $\partial B$.

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How silly of me. Let me rephrase: $f[A]=B$ then $f[\partial A ] = \partial B$? – Alexander Nov 24 '11 at 21:27

Take a disk $A$ and map it holomorphically as in the third image:




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