# Is the simplicial nerve a localization?

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 $$\mathfrak{C}$$.

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