Timeline for A homotopy commutative diagram that cannot be strictified
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2011 at 3:15 | comment | added | Tom Goodwillie | @Tyler: Dihedral of order $10$. | |
| Dec 3, 2011 at 5:21 | comment | added | Tyler Lawson | Basically, I wanted to use that for groups $G$ and $H$, $[BG,BH]$ is the set of homomorphisms mod conjugation. This should give two examples: (1) a cubical diagram that can't be strictified, and (2) in the homotopy category, the automorphism group of $BG$ is $Out(G)$; this action can be viewed as a diagram, and it lifts to an actual diagram if and only if the surjection $Aut(G) \to Out(G)$ splits. I'm embarassed to say that I don't actually remember an example of a group for which this doesn't split. | |
| Dec 3, 2011 at 5:19 | comment | added | Tyler Lawson | @Akhil: Sorry for this silliness. I don't feel like I want to edit this post now that Jeff's pretty much finished it off; I've been waffling because I wanted to come up with an example from group theory, but haven't gotten around to doing anything about it. | |
| Dec 3, 2011 at 4:18 | comment | added | Akhil Mathew | @Tyler: Thanks for your help with this question (and earlier ones), and I apologize for not saying anything sooner, having been unexpectedly busy. Though I don't understand it fully yet, I am quite happy with this answer (which seems to be complemented by Jeff Smith's as well). | |
| Dec 3, 2011 at 4:17 | vote | accept | Akhil Mathew | ||
| Nov 26, 2011 at 22:52 | history | edited | Tyler Lawson | CC BY-SA 3.0 |
added 798 characters in body; added 6 characters in body
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| Nov 26, 2011 at 19:52 | comment | added | Tom Goodwillie | In Tyler's example, a lifting is given after restriction to a certain subcategory of $I$, and what does not exist is a lifting on all of $I$ such that on the subcategory it is equivalent (in some homotopy-coherence sense) to the given lifting. | |
| Nov 26, 2011 at 19:49 | comment | added | Tom Goodwillie | I think that this example answers a slightly different question form what was asked. I think the question is: "What is an example of a functor $X:I\to Ho(Top)$ such that there is no functor $I\to Top$ such that the composed functor $I\to Top\to Ho(Top)$ is isomorphic to $X$?" | |
| Nov 25, 2011 at 6:44 | comment | added | Tyler Lawson | I guess I was thinking that the range category (topological spaces) is not particularly finite in any sense. | |
| Nov 24, 2011 at 16:22 | comment | added | Tim Porter | I forgot to say I liked your example a lot. | |
| Nov 24, 2011 at 16:22 | comment | added | Tim Porter | Tyler, what did you mean when you said you ignored the finite category restriction as your cube looks highly finite to me! | |
| Nov 23, 2011 at 20:03 | comment | added | Tyler Lawson | I ignored the "finite category" restriction. A useful exercise might be to write down a chain-level version of this diagram. | |
| Nov 23, 2011 at 20:02 | history | answered | Tyler Lawson | CC BY-SA 3.0 |