Timeline for Fundamental groups of noncompact surfaces
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2021 at 16:19 | comment | added | Dylan Thurston | There are many good answers above, but it's worth noting that you have to be careful citing uniformization, because uniformization proofs frequently end up citing basic surface topology facts. (I came across this question in patching up one proof of uniformization.) | |
| Nov 7, 2018 at 0:01 | answer | added | JHM | timeline score: 2 | |
| Jul 24, 2018 at 3:31 | answer | added | Ian Agol | timeline score: 11 | |
| Jul 23, 2018 at 20:12 | answer | added | Sergiy Maksymenko | timeline score: 2 | |
| May 25, 2018 at 15:17 | vote | accept | Andy Putman | ||
| May 25, 2018 at 15:07 | answer | added | Andy Putman | timeline score: 12 | |
| S Oct 27, 2015 at 18:25 | history | suggested | Ali Taghavi |
I add two tags
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| Oct 27, 2015 at 18:06 | review | Suggested edits | |||
| S Oct 27, 2015 at 18:25 | |||||
| Mar 31, 2012 at 0:34 | answer | added | Igor Rivin | timeline score: 12 | |
| Mar 30, 2012 at 22:53 | answer | added | Lee Mosher | timeline score: 19 | |
| Mar 18, 2010 at 3:22 | comment | added | Maharana | ah, got it. Was thinking of a very special and easy case! | |
| Mar 17, 2010 at 13:57 | answer | added | Mohan Ramachandran | timeline score: 21 | |
| Mar 17, 2010 at 13:44 | comment | added | Andy Putman | Maharana - I'm not talking about punctured surfaces. I'm talking about general noncompact surfaces, like a closed surface minus a Cantor set or the boundary of a regular neighborhood of an infinite graph embedded in R^3. | |
| Mar 17, 2010 at 13:42 | comment | added | Andy Putman | Igor - Excellent observation! Thanks! | |
| Mar 17, 2010 at 13:04 | comment | added | Igor Belegradek | Andy, one does not need uniformization to show that $S$ is Eilenberg-MacLane. Namely, the universal cover of $S$ is acyclic (it does not have homology above dimension $1$ by Poincare duality and it does not have first homology since it is simply-connected). Now Whitehead theorem gives you that the inclusion of any point into the universal cover is a homotopy equivalence. | |
| Mar 17, 2010 at 6:11 | comment | added | Maharana | If you accept the polygonal representation of a closed surface on the plane, then a punctured surface will be represented by the same polygon with punctures easily managable to be in the interior, but now this picture clearly homotopes to a wedge of circles hence $\pi_1$ is free. | |
| Mar 17, 2010 at 3:15 | answer | added | Igor Belegradek | timeline score: 11 | |
| Mar 17, 2010 at 2:57 | vote | accept | Andy Putman | ||
| May 25, 2018 at 15:17 | |||||
| Mar 17, 2010 at 2:28 | answer | added | John Stillwell | timeline score: 44 | |
| Mar 17, 2010 at 1:24 | history | asked | Andy Putman | CC BY-SA 2.5 |