Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 4.8.2.14. Let $n$ be an integer and let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories which exhibits $\operatorname{\mathcal{D}}$ as a local $n$-truncation of $\operatorname{\mathcal{C}}$. Then $\operatorname{\mathcal{C}}$ is locally $n$-truncated if and only if $F$ is an equivalence of $\infty $-categories.

Proof. By assumption, $F$ is essentially surjective. It follows from Theorem 4.6.2.21 that $F$ is an equivalence of $\infty $-categories if and only if, for every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the induced map of Kan complexes

\[ F_{X,Y}: \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X, Y) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) ) \]

is a homotopy equivalence. Since $F_{X,Y}$ exhibits $\operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$ as an $n$-truncation of $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$, this is equivalent to the requirement that $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is $n$-truncated. $\square$