Kerodon

Proof. Choose a functor $g: \operatorname{\mathcal{C}}\rightarrow \widehat{\operatorname{\mathcal{C}}}$ which exhibits $\widehat{\operatorname{\mathcal{C}}}$ as an $\operatorname{Ind}_{\kappa }^{\lambda }$-completion of $\operatorname{\mathcal{C}}$. Assume first that $\operatorname{\mathcal{C}}$ is $(\kappa ,\lambda )$-cocomplete, so that $g$ admits a left adjoint $f: \widehat{\operatorname{\mathcal{C}}} \rightarrow \operatorname{\mathcal{C}}$. Then the functor $F = \operatorname{Ind}_{\lambda }^{\mu }(f)$ is left adjoint to $G = \operatorname{Ind}_{\lambda }^{\mu }(g)$ (Exercise 9.3.3.11). Since $g$ is fully faithful, the functor $G$ is also fully faithful (Corollary 9.3.2.2), and therefore induces an equivalence from $\operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}})$ to a reflective subcategory of $\operatorname{Ind}_{\lambda }^{\mu }( \widehat{\operatorname{\mathcal{C}}} )$ (Remark 6.3.3.5). By virtue of Corollary 7.1.4.29, it will suffice to show that the $\infty $-category $\operatorname{Ind}_{\lambda }^{\mu }( \widehat{\operatorname{\mathcal{C}}} )$ is $(\kappa ,\mu )$-cocomplete. This follows from the equivalence $\operatorname{Ind}_{\lambda }^{\mu }( \widehat{\operatorname{\mathcal{C}}}) \simeq \operatorname{Ind}_{\kappa }^{\mu }(\operatorname{\mathcal{C}})$ supplied by Theorem 9.3.6.4.

We now prove the converse. Assume that $\operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}})$ is $(\kappa ,\mu )$-cocomplete. If $\operatorname{\mathcal{C}}$ is idempotent-complete, then it can be identified with the full subcategory of $\operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}})$ spanned by the $(\lambda ,\mu )$-compact objects (Corollary 9.4.1.21). This full subcategory is closed under all $\lambda $-small colimits which exist in $\operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}})$ (Proposition 9.2.5.24), and therefore admits $\lambda $-small $\kappa $-filtered colimits. $\square$