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Let $S$ be a Riemann surface of genus $g \geq 0$ with $n$ punctures, i.e., with $n$ distict points removed. Let $f: S \rightarrow R$ be a quasi-conformal map. Then $R$ is also Riemann surface of genus $g$ with $n$ punctures.

Assume $S'$ and $R'$ are surfaces arising from $S$ and $R$ by filling in the $n$ punctures. Is it always possible to extend $f$ quasi-conformally to $f':S' \rightarrow R'$ such that the filled in points are fixed but not necessarily pointwise.

The question is true for replacing punctures by boundary components. Since then ideal boundary is not empty and there is always an extension for the ideal boundary.

But in case of punctures the ideal boundary is empty.

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up vote 4 down vote accepted

Isolated points are removable singularities for quasiconformal maps, see for instance Theorem 17.3 in Vaisala's book (where higher-dimensional case is proven too).

J. Vaisala, Lectures on n-dimensional quasiconformal mappings, Lecture Notes in Mathematics 229, Springer, Berlin-Heidelberg-New York, 1971.

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Thank you for your help. –  berl13 Mar 7 '12 at 19:15
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