Timeline for Whitehead theorem for cohomotopy
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2014 at 2:47 | comment | added | Jeff Strom | @RicardoAndrade -- This is pretty cool. I ran across the "hypoabelian" condition in the work leading to "Theorem M" in the edit to my answer above. | |
| May 15, 2014 at 0:45 | comment | added | Ricardo Andrade | (continuation) Then observe that this generalized Postnikov tower provides obstruction groups for extending $f$ to a map $CX\to Y$ (where $CX$ is the cone on $X$). These obstruction groups are identified with relative cohomology groups of the pair $(CX,X)$ with local coefficients. Finally, since $X$ is acyclic, all these relative cohomology groups are zero (i.e. the map $X\to CX\simeq\ast$ is acyclic). The condition on the fundamental group of $Y$ is there just to deal with the very first lift to $K(\pi_1 Y,1)$, which does not correspond to any cohomology/obstruction group. | |
| May 14, 2014 at 23:18 | comment | added | Ricardo Andrade | @Jeff Strom, I guess you mean a Postnikov tower of principal fibrations. By the way, if you want a more general result, I think the following holds: if $X$ is acyclic and $Y$ has hypoabelian fundamental group (i.e. the fundamental group has no non-trivial perfect subgroup), then any map $f:X\to Y$ is null. I think you can prove it by using a generalized Postnikov tower for $Y$ originally given by Robinson (projecteuclid.org/euclid.ijm/1256052280). (to be continued...) | |
| May 14, 2014 at 17:35 | comment | added | Jeff Strom | Well, non-nilpotent spaces, right? | |
| May 14, 2014 at 14:50 | comment | added | Tom Goodwillie | What's a space without a Postnikov tower? | |
| May 14, 2014 at 13:21 | comment | added | Jeff Strom | Good, so the theorem is: if $\Sigma X \simeq *$, then $[X,Y] = *$ for any space with a Postnikov tower. I didn't know that before. | |
| May 14, 2014 at 0:39 | comment | added | Tom Goodwillie | I have added some details. | |
| May 14, 2014 at 0:38 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
added 629 characters in body
|
| May 13, 2014 at 23:23 | comment | added | Jeff Strom | I've puzzled over this for some time now. I know it is true if $X$ is $2$-dimensional, for then $\pi^k = 0$ for $k > 2$, $\pi^1 (X)= \widetilde H^1(X;\mathbb{Z})$ because $S^1 = K(\mathbb{Z}, 1)$ and $\pi^2 = \widetilde H^2(X;\mathbb{Z})$ by Hopf. But if $\mathbf{dim}(X) > 2$, how do you approach the middle cohomology? | |
| May 13, 2014 at 22:52 | comment | added | Tom Goodwillie | I was thinking of using the Postnikov tower of the sphere. | |
| May 13, 2014 at 21:47 | comment | added | Mingcong Zeng | Would you mind to give more hint on the answer? I thought it is because when we look at the AHSS and set $E = S^0$ since we have no homology the spectral sequence is trivial but I think this might only be the stable case: when we use $S^0$ as a cohomology theory we already suspend it enough times. So I think I must miss the reason why trivial reduced homology implies trivial cohomotopy and the reason should be very simple. | |
| May 13, 2014 at 0:53 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |