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Corollary 9.3.7.6. Let $\kappa \trianglelefteq \lambda $ be regular cardinals. For every $\infty $-category $\operatorname{\mathcal{C}}$, the following conditions are equivalent:

$(1)$

The $\infty $-category $\operatorname{\mathcal{C}}$ is $\kappa $-filtered.

$(2)$

The $\infty $-category $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}})$ is $\kappa $-filtered.

$(3)$

The $\infty $-category $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}})$ is $\lambda $-filtered.

Proof. The implications $(3) \Rightarrow (2) \Rightarrow (1)$ follow from Proposition 9.3.7.4 (and do not require the assumption that $\kappa \trianglelefteq \lambda $). We will show that $(1)$ implies $(3)$. Choose a regular cardinal $\mu $ satisfying $\lambda \trianglelefteq \mu $ such that $\operatorname{\mathcal{C}}$ is essentially $\mu $-small. If $\operatorname{\mathcal{C}}$ is $\kappa $-filtered, then Proposition 9.3.7.4 implies that the $\infty $-category $\operatorname{Ind}_{\kappa }^{\mu }(\operatorname{\mathcal{C}})$ has a final object. Theorem 9.3.6.4 guarantees that $\operatorname{Ind}_{\kappa }^{\mu }(\operatorname{\mathcal{C}})$ can be viewed as an $\operatorname{Ind}_{\lambda }^{\mu }$-completion of $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}})$. Applying Proposition 9.3.7.4 again, we conclude that $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{C}})$ is $\lambda $-filtered. $\square$