$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Theorem 4.8.8.3. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories and let $n$ be an integer. Then $F$ admits a factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$ with the following properties:
The functor $G$ is essentially $n$-categorical: that is, it is $m$-full for $m \geq n+2$.
The functor $F'$ is categorically $(n+1)$-connective: that is, it is $m$-full for $m \leq n+1$.
Proof of Theorem 4.8.8.3.
Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories and let $n$ be an integer. We wish to show that $F$ factors as a composition $G \circ F'$, where $G$ is essentially $n$-categorical and $F'$ is categorically $(n+1)$-connective. Using Proposition 4.1.3.2, we can reduce to the case where $F$ is an inner fibration. In this case, the factorization
\[ \operatorname{\mathcal{C}}\xrightarrow {F'} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \xrightarrow {G} \operatorname{\mathcal{C}} \]
of Remark 4.8.8.15 has the desired properties: Proposition 4.8.8.14 guarantees that $G$ is an $n$-categorical inner fibration (and is therefore essentially $n$-categorical, by virtue of Proposition 4.8.6.35), and Corollary 4.8.8.19 guarantees that $F'$ is categorically $(n+1)$-connective.
$\square$