Theorem 9.1.8.7. Let $\kappa $ and $\lambda $ be regular cardinals satisfying $\kappa \triangleleft \lambda $ and let $\operatorname{\mathcal{C}}$ be a $\lambda $-small, $\kappa $-filtered $\infty $-category. Then there exists a $\lambda $-small, $\kappa $-directed partially ordered set $A$ and a right cofinal functor $F: \operatorname{N}_{\bullet }(A) \rightarrow \operatorname{\mathcal{C}}$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Using Proposition 9.1.8.5 (and Remark 9.1.8.6), we can choose a $\lambda $-small, $\kappa $-directed partially ordered set $(A, \leq )$ and a diagram
\[ \mathscr {F}: \operatorname{N}_{\bullet }(A) \rightarrow \operatorname{\mathcal{QC}}_{< \lambda } \quad \quad (\alpha \in A) \mapsto \operatorname{\mathcal{E}}_{\alpha } \]
having colimit $\operatorname{\mathcal{C}}$, where each $\operatorname{\mathcal{E}}_{\alpha }$ has a final object. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{N}_{\bullet }(A)$ be a cocartesian fibration with covariant transport representation $\mathscr {F}$ (for example, we can take $\operatorname{\mathcal{E}}$ to be the $\infty $-category of elements $\int _{\operatorname{\mathcal{C}}} \mathscr {F}$ of Definition 5.6.2.1) and let $W$ be the collection of all $U$-cocartesian morphisms of $\operatorname{\mathcal{E}}$. Since $\operatorname{\mathcal{C}}$ is a colimit of the diagram $\mathscr {F}$, there is a functor $G: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ which exhibits $\operatorname{\mathcal{C}}$ as a localization of $\operatorname{\mathcal{E}}$ with respect to $W$ (Proposition 7.4.5.20), and is therefore right cofinal (Proposition 7.2.1.10). Since each fiber of $U$ has a final object, the functor $U$ is a reflective localization (Corollary 7.1.5.22): that is, it has a (fully faithful) right adjoint $V: \operatorname{N}_{\bullet }(A) \rightarrow \operatorname{\mathcal{E}}$. The functor $V$ is also right cofinal (Corollary 7.2.3.7), so the composition $(G \circ V): \operatorname{N}_{\bullet }(A) \rightarrow \operatorname{\mathcal{C}}$ is right cofinal. $\square$