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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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    $\begingroup$ 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. $\endgroup$
    – David Cohen
    Commented May 31, 2013 at 18:57
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    $\begingroup$ 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. $\endgroup$
    – Hugo Chapdelaine
    Commented May 31, 2013 at 19:35
  • $\begingroup$ 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. $\endgroup$
    – Maxime Fortier Bourque
    Commented May 31, 2013 at 20:04
  • $\begingroup$ Hi @Maxime! This sounds already like a nice theorem. $\endgroup$
    – Hugo Chapdelaine
    Commented May 31, 2013 at 20:13
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    $\begingroup$ David's counterexample is correct. Every open subset of the plane has free fundamental group with at most countably many generators. $\endgroup$
    – Paul Fabel
    Commented May 31, 2013 at 21:28

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The answer is indeed no as David Cohen has pointed out, and more generally the answer is determined by the complements of the sets U and U'.

The complete solution is effectively due to R.L. Moore (1925), the key fact being every nondegenerate monotone upper semicontinuous decomposition of the 2-sphere yields a 2-sphere.

Thus to decide if two open connected planar sets U and U' are homeomorphic, let C and C' denote the respective complements of U and U' in the extended plane.

Now take the respective topological quotients K and K' of C and C', collapsing each component of C or C' to a point. Then U and U' are homeomorphic iff K and K' are homeomorphic.

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    $\begingroup$ To fill in a few details in the above, suppose K and K' are nonempty compact totally disconnected subsets of the 2-sphere, with respective complements U and U'. Then any homeomorphism between U and U' extends to a homeomorphism of the 2-sphere, and any homeomorphism between K and K' extends to a homeomorphism of the 2-sphere. For the latter direction, see the pf. in Van Mill's book on infinitely dimensional topology that the Cantor set cannot be wildly embedded in the plane. For the former direction apply the Schoenflies theorem repeatedly. $\endgroup$
    – Paul Fabel
    Commented May 31, 2013 at 21:13

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