3
$\begingroup$

Let $\mathcal{C}$ be an $(\infty,1)$-category (meaning quasi-category) admitting (finite) limits and $X\colon N(\Delta)^\text{op}\rightarrow\mathcal{C}$ a simplicial object of $\mathcal{C}$. For a (finite) simplicial set $K$, let $X[K]$ denote the composite functor $N(\Delta_{/K})^\text{op}\rightarrow N(\Delta)^\text{op}\rightarrow\mathcal{C}$ and $X(K)$ its limit.

The simplicial object $X$ is called a category object of $\mathcal{C}$ if the spine inclusions $\mathrm{Sp}^n\rightarrow\Delta^n$ induce equivalences $X(\Delta^n)\rightarrow X(\mathrm{Sp}^n)$ for all $n\ge2$, equivalently the inclusions $\Delta^{\{0,\dotsc,i\}}\cup\Delta^{\{i,\dotsc,n\}}\rightarrow\Delta^n$ induce equivalences $X(\Delta^n)\rightarrow X(\Delta^{\{0,\dotsc,i\}}\cup\Delta^{\{i,\dotsc,n\}})$ for all $n\ge2$ and $0\le i\le n$. If, more generally, the inclusions $\Delta^S\cup\Delta^{S^{\prime}}\rightarrow\Delta^n$ induce equivalences $X(\Delta^n)\rightarrow X(\Delta^S\cup\Delta^{S^{\prime}})$ for any $n\ge2$ and any partition $[n]=S\cup S^{\prime}$ s.t. $S\cap S^{\prime}$ consists of a single vertex, then $X$ is called a groupoid object of $\mathcal{C}$. In the proof of Proposition 1.1.8. in $(\infty,2)$-categories and the Goodwillie Calculus, Jacob Lurie seems to implicitly use the following:

(Claim): A simplicial object $X$ is a groupoid object if and only if it is a category object and the horn inclusion $\Lambda_0^2\rightarrow\Delta^2$ induces an equivalence $X(\Delta^2)\rightarrow X(\Lambda_0^2)$.

This claim has also been explicitly reproduced at the nlab or in Proposition A.5 of this paper, either referring back to the above Proposition 1.1.8., which itself however only refers to the proof of Proposition 6.1.2.6. in HTT. The relevant portion is the implication $(4') \Rightarrow (3)$ in that proof, which starts with a w.l.o.g. step, namely "Replacing $i$ by $n-i$ if necessary", so all the argument there demonstrates, unless I'm missing something crucial yet implicit, is that $X$ is a groupoid object if and only if it is a category object and both horn inclusions $\Lambda_0^2\rightarrow\Delta^2$ and $\Lambda_2^2\rightarrow\Delta^2$ induce equivalences $X(\Delta^2)\rightarrow X(\Lambda_0^2)$ resp. $X(\Delta^2)\rightarrow X(\Lambda_2^2)$. It's not clear to me, even in the simpler case where $\mathcal{C}$ is only a $1$-category, why the conditions on the two horns should be equivalent to one another.

Question: Is (Claim) true? If so, what additional insight is missing?

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ is this answered by mathoverflow.net/questions/436321 ? $\endgroup$
    – Tim
    Commented Oct 22 at 16:52
  • 1
    $\begingroup$ @Tim I don't see how, that's more akin to a special case. I understand that lifting against one outer $2$-horn implies the existence of inverses in some sense, but it doesn't transfer completely to the internal context, as far as I can tell. E.g. it's possible to construct a morphism $X_1\rightarrow X_1$ that's supposed to function as "inversion morphism" under the assumption that either outer horn inclusion induces an equivalence, but I cannot show this morphism is involutive (up to equivalence) without assuming that the other horn inclusion induces an equivalence, too. $\endgroup$
    – Thorgott
    Commented Oct 22 at 18:23
  • 2
    $\begingroup$ @Thorgott Can you promote the external statement to an internal one via Yoneda? $\endgroup$
    – Phil Tosteson
    Commented Oct 22 at 18:28
  • $\begingroup$ @PhilTosteson Thanks. The proof of Prop. 1.1.8. should show post hoc that $X(\Delta^2)\rightarrow X(\Lambda_0^2)$ is an equivalence iff $X(\Delta^2)\rightarrow X(\Lambda_2^2)$ is an equivalence for a category object in $\mathrm{Spc}$. For a category object in $\mathcal{C}$, we can post-compose with functors $\mathrm{map}_{\mathcal{C}}(t,-)\colon\mathcal{C}\rightarrow\mathrm{Spc}$ to obtain category objects in $\mathrm{Spc}$ and since these functors are limit-preserving, the Yoneda lemma should imply the result. Still, I was hoping it's something more straightforward I was missing... $\endgroup$
    – Thorgott
    Commented Oct 23 at 9:13

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .