Are $\infty$-categories functorially colimits of their simplices?

Question

Let $\mathcal C$ be an $\infty$-category. If $C$ is a quasicategory modeling $\mathcal C$, then we have a coend decomposition

$$\mathcal C = \int^{[n] \in \Delta} \Delta[n] \times C_n.$$

This allows us to write $\mathcal C$ as a colimit in $Cat_\infty$ of a diagram which takes values in coproducts of copies of $\Delta[n]$'s, in at least two (closely related ways):

1. There is a diagram $Tw(\Delta) \to \Delta^{op} \times \Delta \xrightarrow{\Delta[-] \times C_{?}} Cat_\infty$ whose colimit is $\mathcal C$. Here the indexing cateogry is $Tw(\Delta)$, the twisted arrow category of $\Delta$.

2. There is a sort of bar construction $\mathcal C = |\amalg \Delta[n_0] \times \Delta([n_0],[n_1]) \times \cdots \times \Delta([n_{\bullet-1}], [n_\bullet]) \times C_{n_\bullet}|$. Here the indexing category is $\Delta$.

Unfortunately, although these colimit decompositions are functorial in $C$, they don't appear to be functorial in $\mathcal C$ -- functoriality requires us to rigidify from the $\infty$-category $Cat_\infty$ to the 1-category $qCat$.

One way to recover functoriality is to write

$$\mathcal C = \int^{[n] \in \Delta} \Delta[n] \times \iota(\mathcal C^{[n]}),$$

where $\iota : Cat_\infty \to Gpd_\infty$ is the core functor, throwing out noninvertible morphisms. This expresses every category as a colimit of products of $\Delta[n]$'s with spaces, rather than sets, though (either via a $Tw(\Delta)$-indexed diagram or a $\Delta^{op}$-indexed diagram). But this leaves open the following

Question: Is there a functor $\Phi : Cat_\infty \to Fun(J, Cat_\infty)$ for some 1-category $J$ (maybe $Tw(\Delta)$ or $\Delta^{op}$), such that $\Phi(\mathcal C)(j)$ is always a coproduct of $\Delta[n]$'s, and such that $colim \circ \Phi : Cat_\infty \to Cat_\infty$ is equivalent to the identity?

Notes:

• One upshot of the above discussion is that it would suffice to have a functorial coend decomposition.

• One guess would be that maybe we have

$$\mathcal C \overset ? = \int^{[n] \in \Delta} \Delta[n] \times \pi_0 (\iota(\mathcal C^{[n]}))$$

This would be functorial, but I'm not sure whether it's correct.

• Another upshot of the above discussion is that it would suffice to analogously show that spaces are functorially colimits of discrete spaces. Which perhaps makes the idea seem a bit far-fetched...

Answer 1

This is just an expended version of the comment. The answer to the question as asked is no.

The problem is that for any ($\infty$-)category $J$ the category $D_J$ of functors $J \to Cat_\infty$ that are levelwise coproduct of the $\Delta[n]$ is a $1$-category. So if the Identity of $Cat_\infty$ could be factored through $Colim: D_J \to Cat_\infty$ then it would factor through some $1$-category, and hence through the homotopy category of $Cat_\infty$.

This is of course impossible as that would make $Cat_\infty$ equivalent to a $1$-category, or more explicitly, show that any invertivle natural transformation $\alpha:F \to F$ is isomorphic to the identity.

What you can do however if you want such a functorial expressions of categories as colimits of their simplicies, is to replace "coproducts" by $\infty$-groupoid indexed colimit. Then we can simply observe that because complete Segal spaces are a full subcategory of simplicial spaces that any (complete) Segal space can be written as the coend

$$ X = \int^{[n] \in \Delta} X_n \times \Delta[n]$$

and then follow the discussion as in the OP, but this time realize $X$ as a colimits of things of the form $S \times \Delta[n]$ where $S$ is a space (i.e. an $\infty$-groupoid) instead of a set.

I should say this is very much in the spirit of the discussion here: If you don't want to break the equivalence principles, ($\infty$-)categories are not structure on sets, they are structure on ($\infty$-)groupoids. So I've taken the simplex construction in the beginning and replaced it with the $\infty$-groupoid of $n$-simplex.