Kerodon

Proof. We proceed as in the proof of Proposition 5.1.5.14. Assume that, for each object $C \in \operatorname{\mathcal{C}}$, the $\infty $-category $\operatorname{\mathcal{E}}_{C}$ is locally $(n-1)$-truncated; we will show that the functor $U$ is essentially $n$-categorical (the reverse implication follows from Proposition 4.8.6.17, and does not require the assumption that $U$ is locally cartesian). By virtue of Proposition 4.8.5.27, we may assume without loss of generality that $\operatorname{\mathcal{C}}= \Delta ^1$. Fix a pair of objects $X,Z \in \operatorname{\mathcal{E}}$; we wish to show that the map of Kan complexes $\operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Z) )$ is $n$-truncated. We may assume that $U(X) \leq U(Z)$ (otherwise, both Kan complexes are empty and there is nothing to prove); in this case, we wish to show that the morphism space $\operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z)$ is $n$-truncated (Example 3.5.9.4). If $U(X) = U(Z)$, then the desired result follows from our hypothesis on the fibers of $U$. It will therefore suffice to treat the case where $U(X) = 0$ and $U(Z) = 1$. Since $U$ is a locally cartesian fibration, we can choose a $U$-cartesian morphism $f: Y \rightarrow Z$ of $\operatorname{\mathcal{E}}$ satisfying $U(Y) = 0$. In this case, composition with the homotopy calss $[f]$ induces a homotopy equivalence of Kan complexes $\operatorname{Hom}_{\operatorname{\mathcal{E}}}( X,Y) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X, Z)$ (Corollary 5.1.2.3). It will therefore suffice to show that the Kan complex $\operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y)$ is $n$-truncated, which follows from our assumption that the fiber $\operatorname{\mathcal{E}}_{0} = U^{-1} \{ 0\} $ is locally $n$-truncated. $\square$