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Corollary 4.8.8.9. Let $f: X \rightarrow Z$ be a morphism of Kan complexes and let $n$ be an integer. Then $f$ factors as a composition $X \xrightarrow {f'} Y \xrightarrow {f''} Z$, where $f''$ is $n$-truncated and $f'$ is $(n+1)$-connective.

Proof. Using Theorem 4.8.8.3, we can factor $f$ as a composition $f'' \circ f'$, where $f'': \operatorname{\mathcal{C}}\rightarrow Z$ is an essentially $n$-categorical functor of $\infty$-categories and $f': X \rightarrow \operatorname{\mathcal{C}}$ is categorically $(n+1)$-connective. If $n \leq -1$, then $f''$ induces an equivalence from $\operatorname{\mathcal{C}}$ to a summand of $Z$, so that $\operatorname{\mathcal{C}}$ is a Kan complex. If $n \geq 0$, Remark 4.8.7.10 guarantees that $\operatorname{\mathcal{C}}$ is a Kan complex. Setting $Y = \operatorname{\mathcal{C}}$, we observe that $f''$ is $n$-truncated (Example 4.8.6.3) and $f'$ is $(n+1)$-connective (Example 4.8.7.3). $\square$