Timeline for On a special case of Alexander duality
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 19, 2011 at 13:25 | answer | added | John Klein | timeline score: 1 | |
| Aug 4, 2011 at 12:35 | vote | accept | Hugo Chapdelaine | ||
| Aug 2, 2011 at 18:14 | comment | added | Ben Wieland | If you don't assume smooth (or something weaker, like Lipschitz or "tame"=tubular neighborhood), you should not expect an easy proof because closely related results are difficult or false. The Jordan curve theorem is related to codimension 1. The double suspension theorem shows that a circle in a large sphere need not have simply connected complement, so general position arguments are tricky for topological manifolds. en.wikipedia.org/wiki/Double_suspension_theorem | |
| Aug 2, 2011 at 17:38 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 2, 2011 at 17:36 | comment | added | t22 | What do you mean by "subspace of real codimension >= 2?" Does it mean that dim(K)<= n-2, where 'dim' is the covering dimension? | |
| Aug 2, 2011 at 16:55 | comment | added | Igor Rivin | Or Milnor's "topology from a differentiable viewpoint", which is more fun to read than G&P. | |
| Aug 2, 2011 at 16:53 | answer | added | Moosbrugger | timeline score: 4 | |
| Aug 2, 2011 at 16:42 | comment | added | Ryan Budney | If everything is smooth, you get a simple proof that the complement is path connected from transversality -- generically a path in the sphere does not intersect the set $K$. If the co-dimension was at least $3$, you could similarly deduce that the complement is simply connected. These techniques are well written up in Guillemin and Pollack's "Differential Topology" text. | |
| Aug 2, 2011 at 16:29 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |