Timeline for Contractibility of the category of cosimplicial resolutions
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2020 at 22:24 | comment | added | Zhen Lin | The smallest category for which this argument works is indeed the image of the functor $Q$ in the slice category over $F$. But it also works for any other subcategory of the slice over $F$ that contains the image of $Q$. | |
| Sep 6, 2020 at 20:03 | comment | added | display llvll | This is more or less what is done in Hirschhorn theorem 14.5.4 | |
| Sep 6, 2020 at 20:01 | comment | added | display llvll | Probably to complete your proof one needs to show that the category you built is a deformation retract of the category of resolutions. | |
| Sep 6, 2020 at 16:13 | comment | added | display llvll | I guess I still have a problem. What you are doing is take the image of the functor $Q$ and take the slice over $F$ and show is contractible. I’m not sure this is the category of resolutions. Because not all maps of resolutions come from functorial factorization. The fact that lets you say that $p$ is a natural transformation is that all maps in your category come from functorial factorization; but this is not the case in the category of resolutions, where there are more maps. | |
| Sep 6, 2020 at 6:11 | comment | added | Zhen Lin | Yes, you take a functorial cofibrant replacement of a cofibrant replacement. | |
| Sep 6, 2020 at 4:34 | comment | added | display llvll | Do you agree or I’m still not understanding ? | |
| Sep 6, 2020 at 3:04 | comment | added | display llvll | Sorry I didnt see your reply. So you are taking a cofibrant replacement of an object which is already cofibrant? $F'$ must be cofibrant otherwise it's not even in the category we want to prove is contractible, this is what I was saying. | |
| Sep 6, 2020 at 3:00 | comment | added | display llvll | As I understand it, what you have done is say that for any resolution $\Gamma,$ you have a wk $\Gamma \to X,$ (what I called $X$ in my question) and you claim that this is natural w.r.t. maps $\Gamma_1 \to \Gamma_2,$ which I don't see how. | |
| Sep 6, 2020 at 2:40 | comment | added | Zhen Lin | The intermediate is $Q F'$. $F'$ is cofibrant for the purposes of this question but it doesn’t really matter. | |
| Sep 6, 2020 at 2:32 | comment | added | display llvll | Or maybe you are taking $F' $ already a cofibrant replacement? | |
| Sep 6, 2020 at 2:27 | comment | added | display llvll | I thought what you were trying to do was to connect any other cofibrant replacement to $QF$ through $F.$ But $F$ lies outside the category, so it can't be what you were doing. | |
| Sep 6, 2020 at 2:20 | comment | added | display llvll | And where is the zig-zag? | |
| Sep 6, 2020 at 2:19 | comment | added | display llvll | In your diagram, you are just restating the fact that what you call $p$ is a natural transformation from $Q$ to the identity on $\mathcal{M}.$ What I do not understand is how does this say that there is "a zigzag of natural transformations connecting the identity functor on $\mathcal{Q} (F)$ and a constant functor". What constant functor? | |
| Sep 6, 2020 at 2:08 | comment | added | Zhen Lin | Constant on $Q F$. | |
| Sep 6, 2020 at 2:06 | comment | added | display llvll | Sorry if I come back on this but I was looking at the proof with more care and I realized I don't actually understand. I don't see how you built the zig-zag of natural transformations from the identity to the constant functor. For starters, who would be the constant functor? On which object it is constant? | |
| Aug 31, 2020 at 1:05 | comment | added | Zhen Lin | Apply the nerve functor. The natural transformations literally become simplicial homotopies. If you prefer to work with actual topological spaces then apply the geometric realisation functor as well. | |
| Aug 31, 2020 at 1:00 | comment | added | display llvll | I see the analogy in the sense that if we define a homotopy between functors as a zig-zag of natural transformations, then clearly by your proof we retract $Q(F)$ to a point by deformation. But why is this the right notion of homotopy which agrees with the definition of contractibility that uses the nerve? | |
| Aug 31, 2020 at 0:54 | comment | added | display llvll | OK. But how do you show that a zig-zag of natural transformations connecting the identity functor and a constant functor is a sufficient condition for contractibility? It seems that is where all the meat of the proof is. | |
| Aug 27, 2020 at 3:47 | comment | added | Zhen Lin | I'm surprised you say that you "don't understand anything of the proof", then – this is a much easier proof than what you proposed. It doesn't even rely on Quillen's Theorem A, which in my view is a non-trivial result concerning homotopy colimits. | |
| Aug 27, 2020 at 3:36 | review | Low quality posts | |||
| Aug 27, 2020 at 7:52 | |||||
| Aug 27, 2020 at 3:12 | comment | added | display llvll | Thanks but I am not interested in another proof, there's a proof of this result on any book and they are all similar to this one. I wanted to see if my proof could work and I realized there is a point that does not work (thank you for making me think more about the last bullet). I never asked for a different proof so I'm sorry but I don't see how this is an answer to my question. | |
| Aug 27, 2020 at 1:58 | history | answered | Zhen Lin | CC BY-SA 4.0 |