Timeline for Is the homeomorphism class of a connected open set of C determined by its fundamental group?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2013 at 18:33 | vote | accept | Hugo Chapdelaine | ||
| Jun 1, 2013 at 4:08 | comment | added | Etienne | Of Course! I had indeed in mind ``conformally equivalent". | |
| May 31, 2013 at 23:45 | comment | added | Malik Younsi | @EtienneMatheron : The unit disk is homeomorphic to $\mathbb{C}$, for example via $f(z)=z/(1-|z|^2)$. These sets are not conformally equivalent, though. | |
| May 31, 2013 at 21:28 | comment | added | Paul Fabel | David's counterexample is correct. Every open subset of the plane has free fundamental group with at most countably many generators. | |
| May 31, 2013 at 20:44 | answer | added | Paul Fabel | timeline score: 10 | |
| May 31, 2013 at 20:16 | comment | added | Maxime Fortier Bourque | David Cohen : I don't think the two domains you gave have the same $\pi_1$... For example, every loop in $\mathbb{C} \setminus K$ goes around infinitely many points of $K$ since the latter has no isolated points. | |
| May 31, 2013 at 20:13 | comment | added | Hugo Chapdelaine | Hi @Maxime! This sounds already like a nice theorem. | |
| May 31, 2013 at 20:04 | comment | added | Maxime Fortier Bourque | 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. | |
| May 31, 2013 at 19:35 | comment | added | Hugo Chapdelaine | 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. | |
| May 31, 2013 at 18:57 | comment | added | David Cohen | 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. | |
| May 31, 2013 at 18:14 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |