Timeline for Proof Sketch: The pullback of the inclusion of the 0th vertex into the standard n-simplex by a right fibration is a deformation retract (450 point bounty if answered by 2am EST)
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
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| Jul 4, 2010 at 7:18 | vote | accept | Harry Gindi | ||
| Jul 4, 2010 at 7:18 | history | bounty ended | Harry Gindi | ||
| Jul 3, 2010 at 20:14 | comment | added | Tom Goodwillie | I will accept the bounty in the spirit in which it is offered. | |
| Jul 3, 2010 at 6:35 | vote | accept | Harry Gindi | ||
| Jul 4, 2010 at 7:18 | |||||
| Jul 2, 2010 at 19:51 | comment | added | Harry Gindi | Dear Tom, not only will I give you the points, I will be happy to do so =). | |
| Jul 2, 2010 at 15:04 | comment | added | Tom Goodwillie | Harry, if on Thursday morning I had bothered to say 'I'll get back to you with a fuller explanation tonight' then you would have had no need to offer a bounty. In an odd way I'm being rewarded for unresponsiveness. You should keep the points. Also, as an old person to a young person let me put in a good word for patience. | |
| Jul 2, 2010 at 3:19 | comment | added | Tom Goodwillie | Lower triangle says pH=h(Id x p). Apply h(Id x p) to 0 x X. Id x p takes 0 x X into 0 x Delta^n. h takes 0 x Delta^n into 0th vertex. So pH takes 0 x X into 0th vertex, so H takes 0 x X into Y. | |
| Jul 2, 2010 at 3:13 | vote | accept | Harry Gindi | ||
| Jul 3, 2010 at 6:35 | |||||
| Jul 2, 2010 at 3:13 | comment | added | Harry Gindi | I'm stupid, nevermind. Thanks a lot. I will award the bounty as soon as possible! | |
| Jul 2, 2010 at 3:01 | comment | added | Harry Gindi | (sorry for splitting up the comments like this). I guess I don't see how that implies that we're mapping into $Y$. | |
| Jul 2, 2010 at 2:58 | comment | added | Harry Gindi |
In particular, the lower triangle says that $pH|_{\{0\}\times X}=h(Id\times p)$, where $H$ is the new diagonal map and $h$ was our original deformation.
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| Jul 2, 2010 at 2:57 | comment | added | Tom Goodwillie | You've got it now, right? 0xX goes into Y because it's related to this other map that takes 0xDelta^n into 0th vertex. I would have answered earlier, but I've been busy all day. | |
| Jul 2, 2010 at 2:51 | comment | added | Harry Gindi |
Dear Tom, that's literally exactly how I did it when I was trying to prove it myself. What I couldn't show is that the restriction to $\{0\}\times X$ maps into $Y$.
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| Jul 2, 2010 at 2:46 | history | answered | Tom Goodwillie | CC BY-SA 2.5 |