Timeline for When a compact topological manifold with boundary is a ball?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2015 at 13:49 | comment | added | mme | @IgorBelegradek: I see your point, thank you. This is the 4-dimensional annulus theorem which Quinn proved in 1982. | |
| Nov 13, 2015 at 13:47 | comment | added | Igor Belegradek | @MikeMiller: I do not not follow the sentence "Now deleting the balls we should have just proved the desired result". The question is this: if you have an embedding $f$ of $D^4$ into $Int D^4$ such that $f(\partial D^4)$ is bicollared, why is the closure of $D^4-f(D^4)$ homeomorphic to $S^3\times I$? I also do not see how to use Freedman's works, which I think applies one dimension up (not that I know much about it). | |
| Nov 13, 2015 at 13:34 | comment | added | mme | @IgorBelegradek: Perhaps I'm being silly, but (by Perelman's work we may as well assume we have) an h-cobordism between two 3-spheres; capping off one end with a disc provides a contractible manifold with 3-sphere boundary; so capping off the other side must provide a simply connected homology sphere; and by Freedman's work this is homeomorphic to $S^4$. Now deleting the balls we should have just proved the desired result. (Even if I've missed some subtlety about h-cobordisms, Freedman's work does imply the desired result about compact manifolds with sphere boundary here.) | |
| Nov 13, 2015 at 12:09 | comment | added | Igor Belegradek | @MikeMiller: so who proved the topological h-cobordism theorem for cobordisms between 1-connected 3-manifolds (i.e. 3-spheres) ? | |
| Nov 13, 2015 at 6:59 | comment | added | Andreas Thom | The collar neighborhood theorem is true in the topological setting. I thought one would not need it though, since we only need to argue that the boundary is simply connected. | |
| Nov 13, 2015 at 6:35 | comment | added | Marco Golla | Aren't you using the collar neighbourhood theorem to argue this? Does it hold in the topological category as well? (I hope I don't sound hostile -- I just don't understand if there are subtleties we're overlooking or if I'm just being slow) | |
| Nov 13, 2015 at 4:59 | comment | added | Andreas Thom | @MarcoGolla. $\mathbb R^n$ is simply connected at infinity, this implies that $\delta X$ is simply connected. | |
| Nov 13, 2015 at 1:12 | comment | added | Marco Golla | Don't you need to know that $\partial X$ is simply connected, in order to apply the $h$-cobordism theorem? | |
| Nov 12, 2015 at 23:45 | comment | added | mme | We're only working topologically, yes? Topological h-cobordism is true in all dimensions. | |
| Nov 12, 2015 at 19:14 | comment | added | Andreas Thom | I do not know, I think this is open. | |
| Nov 12, 2015 at 18:02 | comment | added | Pupkin | What about n=4? | |
| Nov 12, 2015 at 16:54 | history | answered | Andreas Thom | CC BY-SA 3.0 |