Timeline for The classifying space of an infinite totally ordered set is contractible
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| May 17, 2015 at 19:35 | comment | added | Zhen Lin | It's probably not true without hypotheses on the diagram. It suffices that the functor $\mathbf{Top} \to \mathbf{sSet}$ preserve the colimit in question. I guess that happens if the arrows in the diagram are closed $T_1$-embeddings. | |
| May 17, 2015 at 19:05 | comment | added | Todd Trimble | @ZhenLin Yes, thanks; got it. And I expect that might have been all Mostafa meant in his first comment, although he made a statement that was more general, that directed colimits of contractible spaces are contractible. Any feeling about that? | |
| May 17, 2015 at 18:17 | comment | added | Zhen Lin | @ToddTrimble Filtered colimits (of simplicial sets) preserve weak homotopy equivalences, hence filtered colimits of weakly contractible simplicial sets are weakly contractible. Since the space in question is the geometric realisation of a simplicial set, this proves the claim. | |
| May 17, 2015 at 15:47 | answer | added | Emanuele Dotto | timeline score: 1 | |
| May 17, 2015 at 15:46 | comment | added | Todd Trimble | @Mostafa In an earlier comment I used the phrase "directed colimit", which I believe is pretty standard. (I thought you might have meant this. The nLab ncatlab.org/nlab/show/directed+colimit notes that "direct limit" is also sometimes used, so I guess my earlier comment was overly harsh, although I do find that "direct limit" is confusing because of the synonymous usage noted above). | |
| May 17, 2015 at 15:07 | comment | added | Mostafa - Free Palestine | @ToddTrimble What I mean by "direct limit" is a colimit over a inductive ordered set. It seems it's not standard, what is the standard word for this concept? | |
| May 17, 2015 at 14:30 | comment | added | Todd Trimble | @Mostafa It's possible to adduce some such argument, perhaps, but it would help if you were more careful with language. "Direct limit" to me is a synonym for "colimit" and it's not true that homotopy group functors preserve all colimits (see e.g. math.stackexchange.com/questions/320812/…). Aside from that, the statement in your first comment that I was asking about was stated more generally than for spaces having the homotopy type of a CW complex, which is why I asked. (If you don't think it's true in that extra generality, that's fine.) | |
| May 17, 2015 at 14:08 | comment | added | Mostafa - Free Palestine | @ToddTrimble For a direct limit of connected pointed spaces, I think the limit commutes with homotopy groups and in the case of contractible CW-complexes it implies that the limit is also contractible, true? | |
| May 17, 2015 at 14:02 | comment | added | Todd Trimble | @Mostafa Do you have a reference for the claim that a directed colimit of contractible spaces is contractible? Am I missing something obvious? | |
| May 17, 2015 at 13:27 | comment | added | Ricardo Andrade | Here is a proof sketch. First, a lemma: if a $T_1$-space $Z$ is a filtered colimit of subspaces $Z_i$ which are closed under intersection, and each $Z_i$ contains only finitely many other $Z_j$'s, then every compact subspace of $Z$ is contained in some $Z_i$. Corollary: Under these conditions, if the $Z_i$'s are contractible, then $Z$ must be weakly contractible, since every sphere in $Z$ factors through some $Z_i$. In conclusion, the space $BX$ is weakly contractible since it is a filtered colimit of contractible simplices. Hence, the CW-complex $BX$ is contractible. | |
| May 17, 2015 at 12:20 | review | Close votes | |||
| May 17, 2015 at 15:42 | |||||
| May 17, 2015 at 11:53 | answer | added | Neil Strickland | timeline score: 7 | |
| May 17, 2015 at 10:54 | comment | added | მამუკა ჯიბლაძე | Any filtered category has contractible classifying space. First source is, I guess, Quillen's "Higher algebraic K-theory" (in the Springer LNM 341), with very nice proof. | |
| May 17, 2015 at 10:52 | comment | added | Mostafa - Free Palestine | A finite totally ordered set has a maximum so it has a final object as a category and its classifying space is contractible. On the other hand $X$ can be considered as a direct limit of its finite (ordered) subsets and so its classifying space is a direct limit of contractible spaces hence contractible. | |
| May 17, 2015 at 10:46 | review | First posts | |||
| May 17, 2015 at 11:42 | |||||
| May 17, 2015 at 10:45 | history | asked | Tatjana Popow | CC BY-SA 3.0 |