Proposition 11.5.0.12. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ be the full subcategory spanned by the subterminal objects of $\operatorname{\mathcal{C}}$. Then the construction $X \mapsto [X]$ induces a trivial Kan fibration $U: \operatorname{\mathcal{C}}' \rightarrow \operatorname{N}_{\bullet }( \operatorname{Sub}(\operatorname{\mathcal{C}}) )$.
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Proof. Let $X$ and $Y$ be objects of $\operatorname{\mathcal{C}}'$. Since $Y$ is subterminal, the morphism space $\operatorname{Hom}_{\operatorname{\mathcal{C}}'}(X,Y) = \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is either empty or contractible. It follows that the induced map $\operatorname{Hom}_{\operatorname{\mathcal{C}}'}(X,Y) \rightarrow \operatorname{Hom}_{\operatorname{N}_{\bullet }( \operatorname{Sub}(\operatorname{\mathcal{C}}) )}( [X], [Y] )$ is a homotopy equivalence. Allowing $X$ and $Y$ to vary, we deduce that the functor $U$ is fully faithful. By construction, $U$ is surjective on objects, and therefore essentially surjective. Applying Theorem 4.6.2.21, we conclude that $U$ is an equivalence of $\infty $-categories. Since $\operatorname{N}_{\bullet }( \operatorname{Sub}(\operatorname{\mathcal{C}}) )$ is the nerve of a category, $U$ is automatically an inner fibration (Proposition 4.1.1.10). Moreover, every isomorphism in $\operatorname{N}_{\bullet }( \operatorname{Sub}(\operatorname{\mathcal{C}}) )$ is an identity morphism, so $U$ is an isofibration. Applying Proposition 4.5.5.20, we conclude that $U$ is a trivial Kan fibration. $\square$