MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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
up vote 10 down vote accepted

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.

share|cite|improve this answer
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. – Paul Fabel May 31 '13 at 21:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.