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For any finite subset $B\subset \mathbf{P}^1$, the fundamental group of the Riemann surface $\mathbf{P}^1-B$ is finitely generated.

Is this true if we replace $\mathbf P^1 $ by a higher genus compact connected Riemann surface?

More precisely, let $B\subset X$ be a finite subset of a compact connected Riemann surface $X$ of genus $g>0$. Is the fundamental group of $X-B$ finitely generated?

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I don't think this is true. But it is true that they're countable. –  mixedmath Mar 5 '12 at 17:55

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up vote 4 down vote accepted

Yes this is finitely generated. On the level of homotopy, removing a finite set is equivalent to removing finitely many open disks. There is a well-known classification of surfaces with boundary. They are all homeomorphic to a disk with several bands attached, which is homotopy equivalent to a wedge of circles, so the fundamental group is actually a free group of finite rank if there is at least one puncture. If there are no punctures, then the group is no longer free but still finitely generated.

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The complement of a non-empty finite set in a closed (real!) manifold of dimension $2$ has the homotopy type of a finite graph, so the answer is yes.

This is easy to see if you know that all such manifolds can be obtained by identifying the sides of a plane polygon.

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