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Proposition 9.5.5.9. Let $\kappa \trianglelefteq \lambda $ be regular cardinals and let $f: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories. Suppose that $\operatorname{\mathcal{D}}$ is $(\kappa ,\lambda )$-cocomplete, so that $f$ admits an $\operatorname{Ind}_{\kappa }^{\lambda }$-extension $F: \operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{D}}$ (Definition 9.3.1.12). The following conditions are equivalent:
- $(1)$
The functor $f$ is $\kappa $-right exact.
- $(2)$
The functor $F$ is $\kappa $-right exact.
- $(3)$
The functor $F$ is $\lambda $-right exact.
Proof.
Fix an object $D \in \operatorname{\mathcal{D}}$. Using the criterion of Theorem 9.3.5.15, it will suffice to prove the equivalence of the following conditions:
- $(1_ D)$
The $\infty $-category $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/D}$ is $\kappa $-filtered.
- $(2_ D)$
The $\infty $-category $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}}) \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/D}$ is $\kappa $-filtered.
- $(3_ D)$
The $\infty $-category $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}}) \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/D}$ is $\lambda $-filtered.
Note that the $\infty $-category $\operatorname{\mathcal{E}}= \operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}}) \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/D}$ admits $\lambda $-small $\kappa $-filtered colimits which are preserved by the right fibration $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}})$ (Proposition 7.1.9.8). Applying Proposition 9.3.1.16, we can identify $\operatorname{\mathcal{E}}$ with the $\operatorname{Ind}_{\kappa }^{\lambda }$-completion of the $\infty $-category $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/D}$. The equivalence of $(1_ D)$, $(2_ D)$, and $(3_ D)$ now follows from Corollary 9.3.7.6.
$\square$