Kerodon

Proof. Let $J$ be an infinite set, and let $\operatorname{\mathcal{J}}$ be the corresponding indiscrete category (that is, the category having object set $\operatorname{Ob}(\operatorname{\mathcal{J}}) = J$ and $\operatorname{Hom}_{\operatorname{\mathcal{J}}}(j,j') = \ast $ for every pair of elements $j,j' \in J$). Then the nerve $\operatorname{N}_{\bullet }(\operatorname{\mathcal{J}})$ is a contractible Kan complex. Setting $\widetilde{\operatorname{\mathcal{C}}} = \operatorname{N}_{\bullet }(\operatorname{\mathcal{J}}) \times \operatorname{\mathcal{C}}$, it follows that the projection map $\pi : \widetilde{\operatorname{\mathcal{C}}} \rightarrow \operatorname{\mathcal{C}}$ is a trivial Kan fibration. We will complete the proof by showing that $\widetilde{\operatorname{\mathcal{C}}}$ satisfies condition $(\ast )$. Let $K$ be a finite simplicial subset of $\widetilde{\operatorname{\mathcal{C}}}$, so that the inclusion map $K \hookrightarrow \widetilde{\operatorname{\mathcal{C}}}$ can be identified with a pair of diagrams

Since $J$ is infinite, we can choose an element $j \in J$ which is not of the form $f(x)$ for any vertex $x \in K$. It follows that $f$ admits a unique extension $\overline{f}: K^{\triangleright } \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{J}})$ which carries the cone point of $K^{\triangleright }$ to the element $j \in J$. Our assumption that $\operatorname{\mathcal{C}}$ is filtered guarantees that $g$ admits an extension $\overline{g}: K^{\triangleright } \rightarrow \operatorname{\mathcal{C}}$. We complete the proof by observing that the pair $(\overline{f}, \overline{g} )$ determines a monomorphism of simplicial sets $K^{\triangleright } \rightarrow \widetilde{\operatorname{\mathcal{C}}}$. $\square$