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...



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


