Timeline for Fundamental groups of noncompact surfaces
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 26, 2018 at 18:46 | comment | added | user_1789 | @LeeMosher Thanks for the correction. I think now that at least the following weaker property holds: the components of a maximal forest map onto a dense subset of the ends. | |
| May 26, 2018 at 17:14 | comment | added | Andy Putman | @LeeMosher: Happy to help! | |
| May 26, 2018 at 12:58 | comment | added | Lee Mosher | @AndyPutman Thanks for pointing this out Andy. Looking back on it, I can see I was overly optimistic in trying for a connected forest. I had hoped to avoid Zorn's Lemma, but maybe that's not realistic........ It does make me want to ask the logicians if this might be another avatar of Zorn's Lemma, i.e. another equivalent property ... | |
| May 26, 2018 at 12:56 | comment | added | Lee Mosher | @user_1789 It's not possible for the map from the components of the forest to the ends of $S$ to be onto. The surface can have uncountably many ends (for example a sphere minus a Cantor set), but there's only countably many vertices hence only countably many components to the forest. | |
| May 25, 2018 at 15:13 | comment | added | Andy Putman | (deleted a link to my note working out the details of the proof; I posted a link to a revised version in an answer above) | |
| May 25, 2018 at 11:19 | comment | added | user_1789 | @AndyPutman Agreed, in fact it seems that the components of any forest (of one-ended trees in S) can be naturally mapped to the ends of S (and the map is onto, if the forest is maximal). | |
| May 23, 2018 at 18:39 | comment | added | Andy Putman | component of which is $1$-ended. You then collapse each component. This is, by the way, what is done in the paper of Whitehead that Igor references in his answer. | |
| May 23, 2018 at 18:39 | comment | added | Andy Putman | Sorry to raise an issue in an ancient answer, but I was explaining this to one of my graduate students and I realized that I don't understand your argument for why you can choose a single one-ended tree. It is true that the radius $r$ neighborhood of $v_1$ stabilizes, but I don't think that means the union is one-ended. While I don't have a counterexample for triangulated surfaces, if you apply your argument to $\mathbb{R}^1$ with its usual triangulation, the tree you end up with has two ends. What I think you should do instead is construct a forest in the dual triangulation each | |
| Mar 31, 2012 at 18:12 | history | edited | Lee Mosher | CC BY-SA 3.0 |
Edited to simplify and clarify the argument
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| Mar 30, 2012 at 23:46 | comment | added | Andy Putman | This is a beautiful proof. Thanks Lee! | |
| Mar 30, 2012 at 22:53 | history | answered | Lee Mosher | CC BY-SA 3.0 |