Proposition 9.3.3.7. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $n$ be an integer, and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. Then the morphism $f$ is $n$-truncated (in the sense of Definition 9.3.3.1) if and only if it is $n$-truncated when regarded as an object of the slice $\infty $-category $\operatorname{\mathcal{C}}_{/Y}$ (in the sense of Definition 9.3.1.1).
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. By definition, $f$ is $n$-truncated as an object of $\operatorname{\mathcal{C}}_{/Y}$ if and only if, for every morphism $g: C \rightarrow Y$ of $\operatorname{\mathcal{C}}$, the morphism space $K = \operatorname{Hom}_{ \operatorname{\mathcal{C}}_{/Y} }( g, f )$ is $n$-truncated. Using Corollary 4.6.9.18, we can identify $K$ with the homotopy fiber of the composition map $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,X) \xrightarrow { [f] \circ } \operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,Y)$ over the vertex $g \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(C,Y)$. The desired result now follows from Corollary 3.5.9.12. $\square$