Timeline for Zigzags and contractibility of categories
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2014 at 16:43 | answer | added | Ilias A. | timeline score: 6 | |
| Aug 14, 2014 at 15:07 | comment | added | Vidit Nanda | @OmarAntolín-Camarena Yes, sorry about that. And thanks for the helpful comments which brought the question to its present form. | |
| Aug 14, 2014 at 15:06 | comment | added | Omar Antolín-Camarena | For future readers: if all the above comments seem to make no sense, it's because the question was substantially edited. :) | |
| Aug 14, 2014 at 12:20 | vote | accept | Vidit Nanda | ||
| Aug 14, 2014 at 7:40 | answer | added | Fernando Muro | timeline score: 7 | |
| Aug 14, 2014 at 1:41 | history | edited | Vidit Nanda | CC BY-SA 3.0 |
added 136 characters in body
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| Aug 14, 2014 at 1:29 | history | edited | Vidit Nanda | CC BY-SA 3.0 |
modified in light of comments
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| Aug 14, 2014 at 1:22 | comment | added | Omar Antolín-Camarena | I still don't get what construction you mean. For a group $G$ thought of as a one object category I think what you wrote can be interpreted to produce (1) $G$ again, (2) the free product $G \ast G^{\mathrm{op}} \cong G \ast G$, or (3) the coproduct of monoids $G \ast_{\mathrm{mon}} G$. | |
| Aug 13, 2014 at 20:43 | comment | added | Vidit Nanda | Dear @Fedotov, I understand. So if the coherent nerve of the hammock localization has an initial element, can we then conclude that the original category is contractible as well? And does the localization functor induce a homotopy equivalence on classifying spaces in this case? | |
| Aug 13, 2014 at 17:37 | comment | added | Ilias A. | So I would say, that you can not compare the nerves of $Z\mathsf{C}$ and $\mathsf{C}$ but you you can compare the nerve of $\mathsf{C}$ and the coherent nerve of $L_{\mathsf{C}}\mathsf{C}$ (which is a simplicially enriched category). | |
| Aug 13, 2014 at 17:36 | comment | added | Ilias A. | @ViditNanda I think that up the equivalence relation that you made for id morphisms (which I'm not sure to understand), the category $Z\mathsf{C}$ looks like the underling category of the Dwyer-Kan Hammock localization $L_{\mathsf{C}}\mathsf{C}$ at all morphisms of $\mathsf{C}$. More precisely $Z\mathsf{C}=L_{\mathsf{C}}\mathsf{C}_{0}$ i.e. you take only the 0 cells of $L_{\mathsf{C}}\mathsf{C}$ which is now an ordinary category. | |
| Aug 13, 2014 at 15:20 | comment | added | Vidit Nanda | @ZhenLin certainly I don't mean to use this for anything like $\mathbf{Set}$. But tons of examples can be generated by taking various copies of $x \to x_0 \gets x_1 \cdots$ and gluing them all at $x$ for instance. The category so obtained does not have an initial object, but the zigzag category contracts to $x$. | |
| Aug 13, 2014 at 14:33 | comment | added | Fernando Muro | I think Omar is right. You seem to be describing the Gabriel-Zisman localization of $\bf C$ at all arrows, i.e. the groupoidification of $\bf C$. Hence $Z\bf C$ is the fundamental groupoid of the nerve of $\bf C$, and it has an initial object iff it is trivial, but this says nothing about the higher homotopy groups of $\bf C$, unfortunately. | |
| Aug 13, 2014 at 14:16 | comment | added | Omar Antolín-Camarena | Do you mean for x→y←x to be the identity on x in ZC when the right- and left-pointing arrows are the same? If so, then your ZC is a groupoid and only has an initial object when it has a single object. If not, then powers of $x \to y \leftarrow x$ are probably all distinct, and I would expect $Z\mathbf{C}$ to not have an initial object unless $\mathbf{C}$ is a single object groupoid. (I haven't thought this through very carefully.) | |
| Aug 13, 2014 at 13:28 | comment | added | Zhen Lin | Do you have an interesting example where $Z \mathcal{C}$ has an initial object? The only ones I can think of are categories $\mathcal{C}$ with a unique strict initial object such as $\mathbf{Set}$, and those are contractible simply by virtue of having an initial object. | |
| Aug 13, 2014 at 12:59 | history | asked | Vidit Nanda | CC BY-SA 3.0 |