Timeline for Is every connected space equivalent to some B(Aut(X))?
Current License: CC BY-SA 3.0
10 events
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| Sep 28, 2015 at 20:49 | comment | added | Ilias A. | I think you can solve the problem locally using MatanP's nice answer and Kan-Thurston theorem in the following sense. First, for any connected space $X$, there is a discrete group $G$ such that $BG\rightarrow X$ induces an isomorphism in homology with local coefficients. Second, by MatanP's argument you get that for any connected space $X$ there is a map $BAut(BS)\rightarrow X$ which induces an isomorphism in homology with local coefficients. | |
| Sep 11, 2015 at 9:19 | comment | added | Mark Grant | The rational homotopy version of this question is a well-known open problem. See here for instance: arxiv.org/abs/1502.05625 | |
| Nov 21, 2013 at 2:52 | vote | accept | Peter LeFanu Lumsdaine | ||
| Nov 19, 2013 at 12:58 | comment | added | Tom Goodwillie | Or "Does every connected homotopy type occur as a component of the space of all spaces?" | |
| Nov 18, 2013 at 20:39 | comment | added | Ben Wieland | Maybe it's a matter of taste, but this seems like an awkward phrasing to me. I'd rather first say "Is every topological group equivalent to the automorphism group of a space?" which sounds like a natural question; and only then clarify with the technical details of what I mean by "topological group" and "automorphism group of a space." | |
| Nov 18, 2013 at 14:19 | answer | added | MatanP | timeline score: 12 | |
| Nov 18, 2013 at 1:49 | answer | added | John Klein | timeline score: 7 | |
| Nov 16, 2013 at 3:38 | history | edited | Peter LeFanu Lumsdaine | CC BY-SA 3.0 |
some context
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| Nov 15, 2013 at 22:40 | comment | added | Ilias A. | If $B$ is of the form $\Sigma Y$ then $\Omega B$ is a free $A_{\infty}$-space. That means that $Aut(X)$ is a free $A_{\infty}$-space. Which gives a hint that the clam should be wrong. | |
| Nov 15, 2013 at 22:04 | history | asked | Peter LeFanu Lumsdaine | CC BY-SA 3.0 |