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The Riemann mapping theorem in complex analysis implies that any two simply connected open subsets of $\mathbb C$ are homeomorphic.

Does anybody know if there is a more general theorem along the lines of: "Two connected open subsets of $\mathbb C$ are homeomorphic iff they have the same fundamental groups"?

I have never seen this written down, and yet, it is very difficult to come up with a counterexample. The examples I'm coming up with all look like an annulus with multiple holes - for these examples, the statement is obviously true.

Kenny Wong
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This is false. Consider $S=R^2 - {\mathbb Z}\times \{0\}$ as one surface; as the other surface consider $S - \{\frac{1}{n}: n\in {\mathbb N}\}$. Both surfaces have free groups of countably infinite rank, but they are not homeomorphic. (To prove the latter you need the theory of ends. You can also take a look at the references here). On the other hand, if you consider only domains in $R^2$ with finitely generated fundamental groups then indeed the fundamental group (or even just the 1st Betti number) determines the surface up to homeomorphism.

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Moishe Kohan
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  • Thank you for this answer! – Kenny Wong Feb 24 '17 at 18:34
  • are you sure that the second example has countable fundamental group ? – mercio Feb 24 '17 at 19:22
  • @mercio: Yes, every CW complex with countably many cells has countable fundamental group. In the case of surfaces: Each open connected surface is homotopy equivalent to a bouquet of at most countably many circles. I think you are confusing this example with Hawaiian earrings. – Moishe Kohan Feb 24 '17 at 19:42
  • are you saying it's not homotopy equivalent to a hawaiian earring ? (yeah i'm possibly very confused) – mercio Feb 24 '17 at 21:08
  • @mercio: Yes, it is not h.e. to H.E.. – Moishe Kohan Feb 24 '17 at 21:20
  • I just now realized it was $S$ minus stuff and not $R^2$ minus stuff, and since $(0,0)$ is removed..... yeah. – mercio Feb 24 '17 at 21:24