7
$\begingroup$

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 $\mathfrak{C}$.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.