# Is the homeomorphism class of a connected open set of C determined by its fundamental group?

Let $U,U'\subseteq\mathbf{C}$ be two connected open sets such that $\pi_1(U)\simeq\pi_1(U')$.

Q: Does this imply that $U$ is homeomorphic to $U'$?

In the case where the $\pi_1$'s are trivial then the answer is yes, this is a consequence of Riemann's mapping theorem. May be one should try to prove it when the $\pi_1$'s are free groups on $n$ generators...

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The answer is no. Let $K\subset \mathbb{C}$ be a cantor set, and $z$ some point not in $K$, then $U=\mathbb{C}\setminus K$ and $U^{\prime}=\mathbb{C}\setminus (K\cup z)$ are not homeomorphic (if I understand the classification of noncompact surfaces correctly.) However, both have $\pi_{1}$ equal to an infinite free group. –  David Cohen May 31 '13 at 18:57
This seems to be a nice example. Then probably one should impose some "finiteness" conditions on the $\pi_1$'s in order (to have a chance) for the statement to be true. –  Hugo Chapdelaine May 31 '13 at 19:35
If $\pi_1$ is finitely generated, then it is true. For example, U is conformally equivalent to a circle domain (Koebes's theorem), and two such domains with the same $\pi_1$ are easily seen to be homeomorphic. –  Maxime Fortier Bourque May 31 '13 at 20:04
Hi @Maxime! This sounds already like a nice theorem. –  Hugo Chapdelaine May 31 '13 at 20:13
David's counterexample is correct. Every open subset of the plane has free fundamental group with at most countably many generators. –  Paul Fabel May 31 '13 at 21:28