Timeline for Contractibility of regular CW sphere minus open star
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 30, 2015 at 2:26 | comment | added | Tyler Lawson | @ViditNanda Great, that's much simpler. | |
| May 30, 2015 at 0:18 | comment | added | Vidit Nanda | @TylerLawson Or you could use the minimal CW circle that glues the boundary of an interval to a single vertex. Then the open star of that vertex is everything and its complement is therefore empty. | |
| May 30, 2015 at 0:08 | comment | added | Tyler Lawson | @JohnPardon: Start with a square, identify the left and right edges to a single edge $\sigma$ so you have a tube, then attach disks as top and bottom caps to make the boundary of a cylinder (homeomorphic to $S^2$). The open star of $\sigma$ is the complement of the disjoint top and bottom disks. | |
| May 29, 2015 at 23:47 | comment | added | John Pardon | Do you know a counterexample if you drop the assumption of regularity? (to save people time: a regular CW complex is one in which all attaching maps are homeomorphisms onto their images) | |
| May 29, 2015 at 21:14 | history | edited | Vidit Nanda | CC BY-SA 3.0 |
added 180 characters in body
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| May 29, 2015 at 20:29 | comment | added | Mingcong Zeng | If my understanding is correct, by Prop 1 in appendix A of Hatcher, a CW complex is compact if and only if it has finitely many cells. | |
| May 29, 2015 at 20:24 | history | asked | Vidit Nanda | CC BY-SA 3.0 |