Given a simplicial category $\mathcal{C}_{\ast}$ (if necessary, you may assume it's fibrant), denote as $\mathcal{C}$ its underlying ordinary category, and as $\mathcal{W}$ the class of all equivalences in $\mathcal{C}_{\ast}$ (in the simplicially enriched sense, i.e. $f: A \to B$ is an equivalence if there exists $g: B \to A$ and 1-simplices $gf \to id_A$ and $fg \to id_B$ in Map(A,A) and Map(B,B) respectively). Regard $\mathcal{W}$ and $\mathcal{C}$ as discrete simplicial categories, and form the following homotopy pushout in the Joyal model structure on simplicial sets:

$\require{AMScd}$ \begin{CD} N(\mathcal{W}) @>>> N(\mathcal{C})\\ @VVV @VVV\\ Ex^{\infty}N(\mathcal{W}) @>>> \mathcal{C}^{\infty}_{\mathcal{W}} \end{CD}

which induces a functor of $\infty$-categories $q: \mathcal{C}^{\infty}_{\mathcal{W}} \to N_{\Delta}(\mathcal{C}_{\ast})$. Now, my intuition tells me that $N_{\Delta}(\mathcal{C}_{\ast})$ should be the $\infty$-category obtained by localizing at the equivalences, in other words, that $q$ should be an equivalences of $\infty$-categories.

Is there a self-contained, slick way to show that this is the case? If it can help, I'd like to point out that this is the same as showing that the corresponding adjoint map $\tilde{q}: \mathfrak{C}[\mathcal{C}^{\infty}_{\mathcal{W}}] \to \mathcal{C}_{\ast}$ is a weak equivalence of simplicial categories. Furthermore, we know that $\tilde{q}$ is induced by the homotopy pushout (in the Bergner model structure on simplicial categories) given by the image of the above square along the functor $\mathfrak{C}$.


1 Answer 1


This is not true. Here is a counter example. We let $\mathcal{C}_*$ be the following simplicial category. It has two objects 0 and 1. Their only endomorphisms are the identity. There are no morphisms from 1 to 0. The simplicial set of morphisms from 0 to 1 is the simplicial circle $\Delta[1] / \partial \Delta[1]$. if you want a fibrant version, you could just as well use any reduced model of the simplicial circle such as the Bar construction on $\mathbb{Z}$.

Now $\mathcal{C}= [1]$, the free walking 1-cell. And its nerve is the one-simplex. There are no non-trivial equivalences, so $W \cong \partial [1] = 0 \sqcup 1$. Then $Ex^\infty N(W) = W$ and so your $\mathcal{C}^\infty_W = N(\mathcal{C}) = \Delta[1]$. But this is definitely not equivalent to your original.

$$\mathfrak{C}(\Delta[1]) = [1] \to \mathcal{C}_*$$

is NOT a DK-equivalence.


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