Timeline for Any map of a contractible complex to itself has a fixed point
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2013 at 8:11 | comment | added | Greg Friedman | @Tom, that's a nice construction! | |
| Jan 27, 2013 at 0:02 | comment | added | Tom Goodwillie | Embed $X$ in $\mathbb R^n$. It's a retarct of a nbhd $N$. In $N$ is a smaller nbhd $K$ that is a union of cubes with edges parallel to the axes. $X$ is also a retarct of $K$. Now let $C$ be a cube containing $K$, and extend the retraction $K\to X$ to a map $C\to X$ (cell by cell) to get that $X$ is a retract of $C$. | |
| Jan 26, 2013 at 22:39 | comment | added | Ami Paz | @Tom, I read about ENR, and indeed any finite complex is an ENR. But this holds even if the complex is not contractible; where do the contractibility goes into the picture? | |
| Jan 25, 2013 at 23:57 | vote | accept | Ami Paz | ||
| Jan 25, 2013 at 18:18 | comment | added | Tom Goodwillie | I think that ENR, euclidean neighborhood retract, is the key phrase. | |
| Jan 25, 2013 at 18:04 | comment | added | Ami Paz | Thanks. I assume the contractility of $X$ is used to prove that such $D^n$ and $r$ exist. Could you give some guidelines for the proof of this fact? | |
| Jan 25, 2013 at 0:46 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |