Kerodon

Proof. Suppose first that $\operatorname{\mathcal{C}}$ is $(\kappa ,\mu )$-compactly generated: that is, it can be identified with the $\operatorname{Ind}_{\kappa }^{\mu }$-completion of some $\infty $-category $\operatorname{\mathcal{D}}$. Applying Theorem 9.3.6.4, we conclude that $\operatorname{\mathcal{C}}$ can also be identified with the $\operatorname{Ind}_{\lambda }^{\mu }$-completion of the $\infty $-category $\operatorname{Ind}_{\kappa }^{\lambda }(\operatorname{\mathcal{D}})$, and is therefore also $(\lambda ,\mu )$-compactly generated.

Conversely, suppose that $\operatorname{\mathcal{C}}$ is $(\lambda ,\mu )$-compactly generated: that is, it can be identified with the $\operatorname{Ind}_{\lambda }^{\mu }$-completion of the full subcategory $\operatorname{\mathcal{C}}_{< \lambda } \subseteq \operatorname{\mathcal{C}}$. In this case, we wish to show that $\operatorname{\mathcal{C}}$ is $(\kappa ,\mu )$-compactly generated if and only if $\operatorname{\mathcal{C}}_{< \lambda }$ is $(\kappa ,\lambda )$-compactly generated. Without loss of generality, we may assume that $\mu $ is uncountable (otherwise, both conditions are vacuous). In this case, the $\infty $-category $\operatorname{\mathcal{C}}$ is idempotent-complete. Since the full subcategory $\operatorname{\mathcal{C}}_{< \lambda }$ is closed under retracts (Remark 9.2.5.18), it is also also idempotent-complete (Proposition 8.5.4.9). The desired equivalence now follows by applying Proposition 9.4.1.25 to the $\infty $-category $\operatorname{\mathcal{C}}_{< \lambda }$. $\square$