Let $\mathcal{C}$ be a simplicially enriched category whose Homobjects are all Kan complexes. Denote by $N\mathcal{C}$ the homotopycoherent nerve of $\mathcal{C}$, which is a quasicategory. Suppose I have a simplicial object $X: N\Delta \to N\mathcal{C}$. Can I always find a simplicial object $\tilde{X}: \Delta \to \mathcal{C}$ such that $X$ is weakly equivalent to $N\tilde{X}$?

$\begingroup$ Recall that in any model category $\mathcal M$, if $X$ is cofibrant and $Y$ is fibrant, then any morphism $X \to Y$ in the homotopy category of $\mathcal M$ lifts to a morphism in $X \to Y$ in $\mathcal M$. The morphism $X$ descends to a morphism in the homotopy category of quasicategories, which is the same as the homotopy category of simplicial categories. Since $\mathcal C$ is Bergnerfibrant, this lifts to a simpliciallyenriched functor from a Bergnercofibrant replacement of $\Delta$ to $\mathcal C$. But since $\Delta$ is not itself Bergnercofibrant, it's not clear one can do better. $\endgroup$– Tim Campion ♦Commented Oct 4, 2023 at 14:45

$\begingroup$ It's possible that there is another model structure on simplicial categories in which $\Delta$ is cofibrant  see e.g. discussion here. If $\mathcal C$ is fibrant in that model structure, then we'd have a positive answer to your question. $\endgroup$– Tim Campion ♦Commented Oct 4, 2023 at 14:50

$\begingroup$ @TimCampion even if $\Delta$ were cofibrant, would that argument give the result I want? I was thinking that that would imply that there was a natural transformation between $X$ and $N\tilde{X}$, but not necessarily a weak equivalence. $\endgroup$– K. StrongCommented Oct 4, 2023 at 15:50
1 Answer
It is easy to construct counterexamples already for the full subcategory $\{[0],[1]\}⊂Δ$. The category $\cal C$ can be constructed by applying the nerve functor to homobjects of a category $D$ enriched in groupoids. All groupoids have a finite set of objects and at most one morphism between any pair of objects.
The category $\def\id{{\rm id}}D$ has two objects $0$, $1$ and nonidentity morphisms $$s_0:0→1, \quad d_0,d_1:1→0, \quad s_0d_0=s_0d_1=\id_1:1→1, \quad d_0s_0≅d_1s_0≅\id_0:0→0.$$ The latter three 1morphisms are isomorphic to each other. However, $d_0$ and $d_1$ are not isomorphic to each other.
Now we have an obvious functor $N Δ_{≤1} → N D$ of quasicategories. Observe that the definition of the homotopy coherent nerve functor allows for an isomorphism $d_0s_0≅d_1s_0$. On the other hand, a simplicial object $Δ→D$ does not allow for such an isomorphism. Since any object weakly equivalent to the one constructed above must have $d_0$ and $d_1$ as the images of the corresponding face maps, we see that it is impossible to rectify the homotopy coherent simplicial object to a strict one.

$\begingroup$ I'm sorry, I'm having trouble deducing all of the argument  I assume that the obvious functor you are talking about is the one that takes the $2$simplex $(d_0,s_0)$ to the $2$simplex $(d_0, s_0)$ but with the additional data of $d_0s_0 \cong \text{id}_0$ (and similarly for $(d_1,s_1)$). Why is it obvious that this is not weakly equivalent to the nerve applied to the obvious functor $\Delta_{\le 1} \to D$? $\endgroup$ Commented Oct 5, 2023 at 18:33

$\begingroup$ Ah wait, I see now  there is no ``obvious functor'' $\Delta_{\le 1} \to D$; in particular you cannot send $d_0$ to $d_0$ because $d_0$ has no right inverse in $D$. The only simplicial objects on the simpliciallyenriched category level are constant. $\endgroup$ Commented Oct 5, 2023 at 23:02