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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeDec 16th 2010
   • (edited Dec 16th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexConsider the following simplicial set: $A$ is the simplicial set corepresenting the functor: $F(X)=\{(a,b):a\in X_3,b\in X_1:d_1(a)=s_0(b), d_2(a)=s_1(b)\}$. It looks to me that $A$ is exactly the nerve of the contractible groupoid on two generators. Am I correct? Essentially, maps out from this thing classify edges with two-sided inverses, that is to say, equivalences.

   Consider the following simplicial set:

   $A$ is the simplicial set corepresenting the functor:

   $F(X)=\{(a,b):a\in X_3,b\in X_1:d_1(a)=s_0(b), d_2(a)=s_1(b)\}$.

   It looks to me that $A$ is exactly the nerve of the contractible groupoid on two generators. Am I correct?

   Essentially, maps out from this thing classify edges with two-sided inverses, that is to say, equivalences.

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeDec 16th 2010
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   Author: DavidRoberts
   Format: MarkdownItexI presume that $X$ is a simplicial set?

   I presume that $X$ is a simplicial set?

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeDec 16th 2010
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   Author: Harry Gindi
   Format: MarkdownItexYes.

   Yes.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeDec 16th 2010
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   Author: Mike Shulman
   Format: MarkdownItexI don't see why that simplicial set is even a quasicategory. There's another 3-horn or two that I don't see how to fill.

   I don’t see why that simplicial set is even a quasicategory. There’s another 3-horn or two that I don’t see how to fill.

5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeDec 17th 2010
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   Author: Harry Gindi
   Format: MarkdownItexHmm... Which one?

   Hmm… Which one?

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 17th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexIt seems off to me too. The $k$-skeleton of the nerve of that groupoid should be a $k$-sphere: these $k$-spheres approximate the infinite-dimensional sphere which is the total space of the classifying bundle for $\mathbb{Z}_2$.

   It seems off to me too. The $k$-skeleton of the nerve of that groupoid should be a $k$-sphere: these $k$-spheres approximate the infinite-dimensional sphere which is the total space of the classifying bundle for $\mathbb{Z}_2$.

7. • CommentRowNumber7.
   • CommentAuthorHarry Gindi
   • CommentTimeDec 17th 2010
   • (edited Dec 17th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexHmm... how are you computing this stuff? My picture looks like a filled tetrahedron with the following edges: Three copies of the distinguished edge e ($0\to 1$, $0\to 2$, and $2\to 3$), Two degenerate edges ($0\to 2$ and $1\to 3$), and Some other edge $1\to 2$ That is, this object classifies 3-cells of that form. That is, given an edge $e:\Delta^1\to X$ and the composite $\Delta^1\to \Delta^3\to A$ classifying the edge $\Delta^{\{0,1\}}\to \Delta^3$, the lifts from that thing classify the two-sided inverses of the edge $e$. Since it's a quotient of $\Delta^3$, isn't it safe to say that it has no nondegenerate cells of dimension higher $3$? So I mean, it looks to me like there are: Two vertices of this object Two nondegenerate edges: $e$ and $e^{-1}$ Two nondegenerate 2-simplices: $e \circ e^{-1} = s_0(1)$ and $e^{-1}\circ e = s_0(0)$ One nondegenerate 3-simplex tracking associativity

   Hmm… how are you computing this stuff?

   My picture looks like a filled tetrahedron with the following edges:

   Three copies of the distinguished edge e ($0\to 1$, $0\to 2$, and $2\to 3$),

   Two degenerate edges ($0\to 2$ and $1\to 3$), and

   Some other edge $1\to 2$

   That is, this object classifies 3-cells of that form. That is, given an edge $e:\Delta^1\to X$ and the composite $\Delta^1\to \Delta^3\to A$ classifying the edge $\Delta^{\{0,1\}}\to \Delta^3$, the lifts from that thing classify the two-sided inverses of the edge $e$.

   Since it’s a quotient of $\Delta^3$, isn’t it safe to say that it has no nondegenerate cells of dimension higher $3$?

   So I mean, it looks to me like there are:

   Two vertices of this object

   Two nondegenerate edges: $e$ and $e^{-1}$

   Two nondegenerate 2-simplices: $e \circ e^{-1} = s_0(1)$ and $e^{-1}\circ e = s_0(0)$

   One nondegenerate 3-simplex tracking associativity

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeDec 17th 2010
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   Author: DavidRoberts
   Format: MarkdownItex@ Todd >that groupoid which groupoid is this? And Harry, just checking by "the contractible groupoid on two generators" you mean the codiscrete groupoid with 2 objects? Sorry for all the peanut-gallery comments. Off to a company lunch soon and there's no work being done here.

   @ Todd

   that groupoid

   which groupoid is this? And Harry, just checking by “the contractible groupoid on two generators” you mean the codiscrete groupoid with 2 objects?

   Sorry for all the peanut-gallery comments. Off to a company lunch soon and there’s no work being done here.

9. • CommentRowNumber9.
   • CommentAuthorHarry Gindi
   • CommentTimeDec 17th 2010
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   Author: Harry Gindi
   Format: MarkdownItexI mean the contractible groupoid with two objects.

   I mean the contractible groupoid with two objects.

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeDec 17th 2010
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    Author: Mike Shulman
    Format: MarkdownItexI'm saying there should be more nondegenerate 3-simplices tracking associativity. The one you've got does associativity for $e e^{-1} e$; what about $e^{-1} e e^{-1}$? And you need some 4-simplices, and so on. The nerve of the contractible groupoid on 2 objects has nondegenerate k-simplices for every k.

    I’m saying there should be more nondegenerate 3-simplices tracking associativity. The one you’ve got does associativity for $e e^{-1} e$; what about $e^{-1} e e^{-1}$? And you need some 4-simplices, and so on. The nerve of the contractible groupoid on 2 objects has nondegenerate k-simplices for every k.

11. • CommentRowNumber11.
    • CommentAuthorHarry Gindi
    • CommentTimeDec 17th 2010
    • (edited Dec 17th 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexHmm... Good point! How should I think of that object $A$ then? Is it weakly equivalent (for the Joyal model structure) to the nerve of that groupoid? At least mapping into a fixed quasicategory, it seems like they're identical (since we can lift everything on the target).

    Hmm…

    Good point! How should I think of that object $A$ then? Is it weakly equivalent (for the Joyal model structure) to the nerve of that groupoid?

    At least mapping into a fixed quasicategory, it seems like they’re identical (since we can lift everything on the target).

12. • CommentRowNumber12.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 17th 2010
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    Author: Todd_Trimble
    Format: MarkdownItex@David: as Harry said, the category $E\mathbb{Z}_2$ is the chaotic or codiscrete groupoid with two objects. There is a $\mathbb{Z}_2$-action on this category which permutes the two objects, and the quotient by this action $p: E\mathbb{Z}_2 \to B\mathbb{Z}_2$ maps down to the group $\mathbb{Z}_2$ considered as the one-object groupoid. The (realization of the) nerve of the functor $p$ is the classifying bundle. Intuitively, if the two objects of $E\mathbb{Z}_2$ are $0$ and $1$, the chain with $k$ edges which runs as $0 \to 1 \to \ldots \to 0/1$ is a non-degenerate $k$-simplex which represents the northern hemisphere of a $k$-sphere, and the chain which runs as $1 \to 0 \to \ldots 1/0$ represents the southern hemisphere. Hence the $k$-skeleton if a $k$-sphere. The union of the $k$-spheres is an infinite-dimensional sphere which is the total space of the classifying bundle for the group $\mathbb{Z}_2$.

    @David: as Harry said, the category $E\mathbb{Z}_2$ is the chaotic or codiscrete groupoid with two objects. There is a $\mathbb{Z}_2$-action on this category which permutes the two objects, and the quotient by this action $p: E\mathbb{Z}_2 \to B\mathbb{Z}_2$ maps down to the group $\mathbb{Z}_2$ considered as the one-object groupoid. The (realization of the) nerve of the functor $p$ is the classifying bundle.

    Intuitively, if the two objects of $E\mathbb{Z}_2$ are $0$ and $1$, the chain with $k$ edges which runs as $0 \to 1 \to \ldots \to 0/1$ is a non-degenerate $k$-simplex which represents the northern hemisphere of a $k$-sphere, and the chain which runs as $1 \to 0 \to \ldots 1/0$ represents the southern hemisphere. Hence the $k$-skeleton if a $k$-sphere. The union of the $k$-spheres is an infinite-dimensional sphere which is the total space of the classifying bundle for the group $\mathbb{Z}_2$.

13. • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeDec 17th 2010
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    Author: Mike Shulman
    Format: MarkdownItexI think you're right that they're Joyal-equivalent; I think I've even seen that written down somewhere, but I can't find it right now. Mapping into a fixed quasicategory would be good enough for that, since a map A→B of simplicial sets is a Joyal-equivalence iff [B,X]→[A,X] is an equivalence for all quasicategories X.

    I think you’re right that they’re Joyal-equivalent; I think I’ve even seen that written down somewhere, but I can’t find it right now. Mapping into a fixed quasicategory would be good enough for that, since a map A→B of simplicial sets is a Joyal-equivalence iff [B,X]→[A,X] is an equivalence for all quasicategories X.