Connectedness of truncated version of cosimplicial indexing category

Question

Let $F:\mathbf{\Delta}\to\mathcal{S}_{\leqslant n-1}$ be a cosimplicial object in the $\infty$-category of $(n-1)$-truncated spaces. Is it always a right Kan extension of its restriction along $\mathbf{\Delta}^{\leqslant n}\hookrightarrow\mathbf{\Delta}$? (Then one might call $F$ to be $n$-coskeletal.)

I want to use this to show that $\mathbf{\Delta}^{\leqslant n}_{/[m]}:=\mathbf{\Delta}^{\leqslant n}\times_{\mathbf{\Delta}}\mathbf{\Delta}_{/[m]}$ is $n$-connected ($\pi_{<n}\simeq 0$), for every $[m]\in\mathbf{\Delta}$ (which I view as a finite version of Quillen’s theorem A). Other ways to confirm is also fine (without solving the above question).

One might replace $\mathbf{\Delta}$ with $\mathbf{\Delta}_s$ everywhere, e.g. by (proof of) [HTT, Lemma 6.5.3.8]; in the case $m\leqslant n$, it is easy to see that it has a final object hence is in fact contractible.

Answer 1

I don't believe the first claim is true, but I can give a somewhat formal argument for the connectivity of $\Delta_{\le n} \times_{\Delta} \Delta_{/m}$. Let us write $u$ for the inclusion $\Delta_{\le n} \to \Delta$. The crucial result follows the following special case of Appendix A of https://arxiv.org/abs/2207.09256:

Theorem 1. Given a diagram $F : \Delta \to \mathcal{S}_{\le n-1}$ the canonical map $\lim_{\Delta} F \to \lim_{\Delta_{\le n}} u^*F$ is invertible.

(This may already be the result you were after, but we can carry this through to actually obtain the connectivity result you requested.)

Fix an arbitrary $F : \Delta \to \mathcal{S}_{\le n-1}$. By Theorem 1, we know that the following is an equivalence:

\begin{align*} &\lim F \\ &\cong \hom_{\mathbf{Fun}(\Delta,\mathcal{S})}(\mathbf{1}, F) \\ &\to \hom_{\mathbf{Fun}(\Delta_{\le n},\mathcal{S})}(u^*\mathbf{1}, u^*F) \\ &\cong \hom_{\mathbf{Fun}(\Delta_{\le n},\mathcal{S})}(\mathbf{1}, u^*F) \\ &\cong \lim u^*F \end{align*}

Transposing, we conclude that $\hom_{\mathbf{Fun}(\Delta,\mathcal{S})}(\mathbf{1}, F) \to \hom_{\mathbf{Fun}(\Delta,\mathcal{S})}(u_!u^*\mathbf{1}, F)$ is an equivalence and, consequently, that $u_!u^* \mathbf{1} \to \mathbf{1}$ is orthogonal to any $(n-1)$-truncated object in $\mathbf{Fun}(\Delta, \mathcal{S})$. Accordingly, $u_!u^*\mathbf{1} = u_!\mathbf{1}$ is $n$-connected. We therefore conclude that $(u_!\mathbf{1})(m) : \mathcal{S}$ is $n$-connected for any $m : \Delta$. Let us unfold this using the formula for left Kan extensions: $$ (u_!\mathbf{1})(m) = \mathrm{colim}_{ \Delta_{\le n} \times_{\Delta} \Delta_{/m} } \mathbf{1} = L(\Delta_{\le n} \times_{\Delta} \Delta_{/m}) $$

The conclusion then follows.

Incidentally, this is just slight alteration of the usual proof of one direction of Quillen's theorem A. The other direction (that suitably connective fibers implies $n$-cofinality) also holds.