Corollary 9.2.9.11. Let $\kappa \trianglelefteq \lambda \trianglelefteq \mu $ be regular cardinals. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category which admits $\lambda $-small $\kappa $-filtered colimits, let $\operatorname{\mathcal{D}}$ be an $\infty $-category which admits $\mu $-small $\kappa $-filtered colimits, and let $h: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor with $\operatorname{Ind}_{\lambda }^{\mu }$-extension $H: \operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{D}}$. The following conditions are equivalent:
- $(1)$
The functor $H$ is $(\kappa ,\mu )$-cocontinuous.
- $(2)$
The functor $H$ is $(\kappa ,\lambda )$-cocontinuous.
- $(3)$
The functor $h$ is $(\kappa ,\lambda )$-cocontinuous.
Proof.
The implication $(1) \Rightarrow (2)$ is trivial (Remark 9.1.9.18) and the implication $(2) \Rightarrow (3)$ follows from the observation that the functor $\operatorname{\mathcal{C}}\rightarrow \operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}})$ is $(\kappa ,\lambda )$-cocontinuous (Proposition 9.2.3.16). We will complete the proof by showing that $(3)$ implies $(1)$ (the implication $(2) \Rightarrow (1)$ is a special case of Proposition 9.1.9.19, but we will not need this). As in the proof of Corollary 9.2.9.10, fix 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 $\operatorname{\mathcal{C}}$ and set $G = \operatorname{Ind}_{\lambda }^{\mu }(g)$, so that $g$ and $G$ admit left adjoints $f$ and $F = \operatorname{Ind}_{\lambda }^{\mu }(f)$, respectively.
Suppose we are given a diagram $Q: \operatorname{\mathcal{K}}\rightarrow \operatorname{Ind}_{\lambda }^{\mu }(\operatorname{\mathcal{C}})$, where $\operatorname{\mathcal{K}}$ is a $\mu $-small $\kappa $-filtered $\infty $-category; we wish to show that the colimit of $Q$ is preserved by the functor $H$. Since $G$ is fully faithful, $Q$ is isomorphic to the composite $F \circ \widehat{Q}$, where $\widehat{Q} = G \circ Q$. The functor $F$ is a left adjoint, and therefore preserves all colimits (Corollary 7.1.4.22). It will therefore suffice to show that the colimit of $\widehat{Q}$ is preserved by the functor $(H \circ F): \operatorname{Ind}_{\lambda }^{\mu }(\widehat{\operatorname{\mathcal{C}}} ) \rightarrow \operatorname{\mathcal{D}}$.
Using Theorem 9.2.9.4, we can identify $\operatorname{Ind}_{\lambda }^{\mu }(\widehat{\operatorname{\mathcal{C}}})$ with an $\operatorname{Ind}_{\kappa }^{\mu }$-completion of $\operatorname{\mathcal{C}}$. Let $H': \operatorname{Ind}_{\lambda }^{\mu }(\widehat{\operatorname{\mathcal{C}}} ) \rightarrow \operatorname{\mathcal{D}}$ be the $\operatorname{Ind}_{\kappa }^{\mu }$-extension of the functor $h$. We will complete the proof by showing that $H'$ is isomorphic to $(H \circ F)$. Since both $H'$ and $H \circ F$ are $(\lambda ,\mu )$-cocontinuous, it will suffice to show that they have the same restriction to $\widehat{\operatorname{\mathcal{C}}}$: that is, that $H'|_{ \widehat{\operatorname{\mathcal{C}}} }$ is isomorphic to the composition $\widehat{\operatorname{\mathcal{C}}} \xrightarrow {f} \operatorname{\mathcal{C}}\xrightarrow {h} \operatorname{\mathcal{D}}$. By construction, they become isomorphic after precomposition with the functor $g$. It will therefore suffice to show that $h \circ f$ is $(\kappa ,\lambda )$-cocontinuous, which follows from assumption $(3)$.
$\square$