Timeline for Reducing the simplices in the nerve of a category with an object with trivial endomorphism monoid
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| S Mar 31, 2014 at 15:25 | history | bounty ended | CommunityBot | ||
| S Mar 31, 2014 at 15:25 | history | notice removed | CommunityBot | ||
| Mar 30, 2014 at 14:46 | vote | accept | Werner Thumann | ||
| Mar 24, 2014 at 19:32 | vote | accept | Werner Thumann | ||
| Mar 30, 2014 at 14:46 | |||||
| Mar 24, 2014 at 19:29 | vote | accept | Werner Thumann | ||
| Mar 24, 2014 at 19:29 | |||||
| Mar 24, 2014 at 17:44 | answer | added | Werner Thumann | timeline score: 1 | |
| Mar 24, 2014 at 17:29 | comment | added | user43326 | @James Griffin You are right, we can't really define the collapsing as simplicial map since we don't have enough vertices. I guess we will need a barycentric subdivision. On the other hand, collapsing successively should be easier, since in the example of the category with two objects with single map in both direction, we would only need to construct homotopy equivalence between $D^n$ and $D^{n+1}$ instead of a weak equivalence between $D^2$ and $S^{\infty }$. | |
| Mar 23, 2014 at 19:38 | comment | added | Dylan Wilson | Ah! Thank you, that answers my question. | |
| Mar 23, 2014 at 18:00 | comment | added | Werner Thumann | Sorry, I was not aware that the definition could cause so much confusion, but the idea is really simple: Take a simplex of the form $A_1\rightarrow A_2\rightarrow A_3\rightarrow ...\rightarrow A_k$ and look at it as a linear graph. Then delete all objects $A_i$ which are not equal to $X$. Then the number of connected components you get is the number of $X$-components. | |
| Mar 23, 2014 at 16:57 | comment | added | Dylan Wilson | Probably I'm confused by `maximal substring'. Shouldn't your example $X \rightarrow A \rightarrow X$ have only one $X$-component because every string of just $X$s is contained in $X \rightarrow X$ (the composite)? How are you ordering strings? | |
| S Mar 23, 2014 at 14:18 | history | bounty started | Werner Thumann | ||
| S Mar 23, 2014 at 14:18 | history | notice added | Werner Thumann | Draw attention | |
| Mar 23, 2014 at 14:12 | history | edited | Werner Thumann | CC BY-SA 3.0 |
added 183 characters in body
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| Mar 22, 2014 at 18:23 | comment | added | James Griffin | User43326, I don't see how that helps, as instead of collapsing all cells between the first and second occurence of X, one could always collapse all cells between the first and last occurence of X, yielding a direct proof. The content we require is that such a 'collapse' is a homotopy equivalence, but I can't see that the 'collapse' is even a map of simplicial sets, assuming the version of collapse I'm using is the same as yours. | |
| Mar 20, 2014 at 22:05 | history | edited | Werner Thumann | CC BY-SA 3.0 |
deleted 335 characters in body
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| Mar 20, 2014 at 20:45 | history | edited | Ricardo Andrade |
replaced new tag 'simplicial-sets' with existing tag 'simplicial-stuff'
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| Mar 20, 2014 at 14:39 | comment | added | user43326 | I guess a proof would go like this: define $N^r_k(C)$ to be the simplicial subset of $N^r(C)$ with at most $k$ $X$-component. Then the inclusion $N^r_k(C)\subset N^r_{k+1}(C) $ would be a homotopy equivalence, because collapsing all cells between, let's say, the first and second occurence of X would be homotopic to the identity. Since $N^r(C)$ is the colimit of $N^r_k(C)$'s, one would get desired weak equivalence by passing to the limit. | |
| Mar 20, 2014 at 13:46 | history | edited | Werner Thumann | CC BY-SA 3.0 |
added 336 characters in body
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| Mar 19, 2014 at 9:00 | comment | added | Werner Thumann | A single $X$ is also a $X$-component, so your example is $X\rightarrow X\rightarrow Y\rightarrow X\rightarrow Y\rightarrow X$ and has thus three $X$-components. | |
| Mar 19, 2014 at 8:21 | comment | added | Roman Bruckner | Is $N^r(C)$ even a simplicial set? Let $C$ be the groupoid with 2 objects $x$, $y$, and two nontrivial morphisms $f\colon x\to y$, $f^{-1}\colon y\to x$. Let $\sigma = (\operatorname{id}_x, f, f^{-1}, f, f^{-1})\in N(C)$. Then $\sigma\in N^r(C)$, but $d_4(\sigma) = (\operatorname{id}_x,f,f^{-1},\operatorname{id}_x)\not\in N^r(C)$ | |
| Mar 18, 2014 at 15:57 | history | asked | Werner Thumann | CC BY-SA 3.0 |