Kerodon

Example 4.8.8.6. When $n = 0$, Theorem 4.8.8.3 asserts that every functor of $\infty $-categories $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ admits a factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$, where the functor $G$ is faithful and the functor $F'$ is both full and essentially surjective. When $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are (nerves of) ordinary categories, this follows from Proposition 4.8.8.1. To handle the general case, we can use (the proof of) Proposition 4.8.8.1 to factor the functor $\mathrm{h} \mathit{F}$ as a composition $\mathrm{h} \mathit{\operatorname{\mathcal{C}}} \xrightarrow {F'_0} \operatorname{\mathcal{D}}'_0 \xrightarrow {G_0} \mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ where $G_0$ is a faithful functor and $F'_0$ is a full functor which is essentially surjective (or even bijective on objects). To prove Theorem 4.8.8.3, we can take $\operatorname{\mathcal{D}}'$ to be the fiber product $\operatorname{N}_{\bullet }( \operatorname{\mathcal{D}}'_0 ) \times _{ \operatorname{N}_{\bullet }(\mathrm{h} \mathit{\operatorname{\mathcal{D}}}) } \operatorname{\mathcal{D}}$, and $G: \operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$ to be the functor given by projection onto the second factor (which is faithful by virtue of Proposition 4.8.5.8).