Kerodon

Here I get into some trouble when trying to write down a proof for this exercise. By some common reduction, one can reduce the question to extending a simplicial map $\tau_0:\Lambda^n_i\to\mathrm{holim}(X_0\to\cdots\to X_n)$ over $\Delta^n$ to an $n$-simplex $\tau:\Delta^n\to\mathrm{holim}(X_0\to\cdots\to X_n)$ lying over $\mathrm{id}_{\Delta^n}$, where $n\geq 1$ and $0\leq i<n$; $X_j$'s are Kan complexes and $f_j:X_{j-1}\to X_j$ are Kan fibrations. The case $n=1$ is trivial hence let's assume $n\geq 2$ below. Making use of Proposition 050F, $\tau_0$ is represented by a tuple $(\sigma_0,\ldots,\sigma_{i-1},\bullet,\sigma_{i+1},\ldots,\sigma_n)$, where $\sigma_0$ is an $(n-1)$-simplex of $X_1$ and $\sigma_1,\ldots,\sigma_n$ are $(n-1)$-simplices of $X_0$, satisfying that $d^{n-1}_j(\sigma_k)=d^{n-1}_{k-1}(\sigma_j)$ for $0<j<k\leq n$; $f_1(d^{n-1}_0(\sigma_k))=d^{n-1}_{k-1}(\sigma_0)$ for $1<k\leq n$; and $(f_2\circ f_1)(d^{n-1}_0(\sigma_1))=f_2(d^{n-1}_0(\sigma_0))$. And the expected $\tau$ is represented by an $n$-simplex of $X_0$ such that $\sigma_j=d^n_j(\sigma)$ for all $j>0$, and $\sigma_0=f_1(d^n_0(\sigma))$. Evidently if the lifting problem is solvable, $(\sigma_0,f_1(\sigma_1),\ldots,f_1(\sigma_{i-1}),\bullet,f_1(\sigma_{i+1}),\ldots,f_1(\sigma_n))$ defines a simplicial map $\Lambda^n_i\to X_1$; when $i>1$, this implies that $f_1(d^{n-1}_0(\sigma_1))=d^{n-1}_0(\sigma_0)$, seemingly beyond the compatibility conditions obtained above. Many thanks if one could provide a bit hint how to overcome the difficulty mentioned here (or maybe some elementary error is made in the argument).

You are right; the exercise does not look right. Thanks!