Direct comparison from the Rezk hom to the hom of a simplicial category along the coherent nerve?

Question

Consider the following construction: Define $G_n$ to be the contractible groupoid on $n+1$ objects. Choosing a linear order on the objects of each $G_n$ turns $G_*$ into a cosimplicial object. Define the cosimplicial simplicial set $J_*=N(G_*)$.

This cosimplicial object defines a Quillen pair between the Quillen model structure on sSet and the Joyal model structure. The right adjoint is the "core Kan complex" functor.

Given a pair of simplicial sets $X,Y$, define $\operatorname{Map}(X,Y)_n=\operatorname{Hom}(J_n \times X,Y)$.

Then when $X$ is a quasicategory and $x,x'\in X_0$ are vertices, define the Rezk hom $$X(x,x')=\operatorname{Map}(\Delta[1],X)\times_{\operatorname{Map}(\partial\Delta[1],X)} \Delta[0]$$ to be the fibre over $(x,x')$.

This definition is actually just the definition of the Hom of the associated complete Segal space.

Question: When $X$ is the coherent nerve of a fibrant simplicial category, $C$, has anyone computed a direct comparison between the Rezk Hom and the Hom of $C$.

Lurie compares a different presentation of this hom object (the right hom) using straightening/unstraightening over a point.

What we can do is define $$K_n=\partial\Delta[1] \coprod_{J_n\times\partial\Delta[1]} J_n \times \Delta[1]$$ and define $Q_n=\operatorname {Hom}_{\mathfrak{C}[K_n]}(0,1)$, but the combinatorics looks really crazy to show that $Q_n$ is naturally weakly equivalent to the $n$-simplex.

Has anyone worked this out without using straightening/unstraightening? Can we simplify the $J_i$ by replacing them all with finite fragments the way people usually do with $J_1$?

Answer 1

Edit: The proof in the original answer below the line is correct, but it doesn't prove everything we need to establish to prove the equivalence between qCat and sCat. It turna out this was all worked out in great detail in the paper Mapping Spaces in Quasicategories by Dugger and Spivak.

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I think I have a proof of a comparison by a zigzag of weak equivalences (and it becomes sort of clear after doing the proof why it's unlikely that there is an actual direct comparison map).

First, notice that the cosimplicial simplicial set $J_*\times \Delta[1]$ is a Reedy-cofibrant resolution of $\Delta[1]$ with respect to the Joyal model structure. It is also the case that $J_*\times \partial\Delta[1]\to J_*\times \Delta[1]$ is a Reedy cofibration, and since the Reedy model structure is left-proper, it follows that $J_*\times \Delta[1] \to K_*$ is a homotopy pushout of a Reedy weak equivalence. Then it follows that $K_*$ is a Reedy-cofibrant resolution of $\Delta[1]$ as well.

Then since $\mathfrak{C}$ is a left-Quillen functor, it follows that $\mathfrak{C}[K_*]$ is a Reedy-cofibrant resolution of $\mathfrak{C}[\Delta[1]]=[1]=[1](\Delta[0])$.

Note that $[1](J_*)$ is Reedy-fibrant replacement of $[1]$ because each $[1](J_n)$ is Dwyer-Kan equivalent to $[1]$ and fibrant, so since $\mathfrak{C}[K_*]\cong [1](J_*)$ in the homotopy category of $\operatorname{sCat}^\Delta$, we can lift this isomorphism to a weak equivalence of cosimplicial objects $\mathfrak{C}[K_*]\to [1](J_*)$.

We can also give the obvious componentwise natural Dwyer-Kan equivalence $[1](\Delta[n])\to [1](J_n)$.

Now, passing to $\operatorname{Hom}(0,1)$, we obtain a pair of natural weak equivalences of cosimplicial simplicial sets $Q_*\to J_*\leftarrow \Delta[*]$.

In order to prove the claim, it suffices to show that the associated $Q_*$-latching maps are all cofibrations, that is to say, that $|\partial\Delta[n]|_Q\to |\Delta[n]|_Q=Q_n$ is a cofibration, of simplicial sets for $n\geq 0$, which is an exercise.