Kerodon

Proof. The implication $(1) \Rightarrow (2)$ is trivial and the implication $(2) \Rightarrow (3)$ follows from Corollary 9.4.7.18. If condition $(3)$ is satisfied, then we can choose a small regular cardinal $\lambda $ for which the $\infty $-category $\operatorname{\mathcal{C}}$ is $\lambda $-accessible and $(\iota \circ L)$ is $\lambda $-finitary (Remark 9.4.7.2). It then follows that $\operatorname{\mathcal{C}}'$ is closed under small $\lambda $-filtered colimits, which proves that $(3) \Rightarrow (4)$. It will therefore suffice to show that $(4)$ implies $(1)$.

Fix a small regular cardinal $\kappa $ such that $\operatorname{\mathcal{C}}$ is $\kappa $-accessible and the full subcategory $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ is closed under small $\kappa $-filtered colimits. Then $\operatorname{\mathcal{C}}'$ admits small $\kappa $-filtered colimits which are preserved by the inclusion functor $\iota $. Consequently, to prove $(1)$, it will suffice to show that the $\infty $-category $\operatorname{\mathcal{C}}'$ is accessible. In fact, we will show that $\operatorname{\mathcal{C}}'$ is $\kappa $-accessible. Let $\operatorname{\mathcal{C}}_{< \kappa }$ be the full subcategory of $\operatorname{\mathcal{C}}$ spanned by the $\kappa $-compact objects, and let $\operatorname{\mathcal{C}}'_0 \subseteq \operatorname{\mathcal{C}}'$ be the essential image of the functor $L|_{ \operatorname{\mathcal{C}}_{< \kappa } }$. Since the $\infty $-category $\operatorname{\mathcal{C}}_{< \kappa }$ is essentially small (Proposition 9.4.6.2) and $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ is locally small (Remark 9.4.6.14), the full subcategory $\operatorname{\mathcal{C}}'_0 \subseteq \operatorname{\mathcal{C}}'$ is essentially small. Our assumption that $\iota $ is $\kappa $-finitary, guarantees that the functor $L$ is $\kappa $-compact (Example 9.4.2.17), so that each object of $\operatorname{\mathcal{C}}'_0$ is $\kappa $-compact when viewed as an object of $\operatorname{\mathcal{C}}'$. Since $\operatorname{\mathcal{C}}$ is $\kappa $-accessible, every object $X \in \operatorname{\mathcal{C}}$ can be realized as the colimit of a small $\kappa $-filtered diagram in $\operatorname{\mathcal{C}}_{< \kappa }$, so that $L(X) \in \operatorname{\mathcal{C}}'$ can be realized as the colimit of a small $\kappa $-filtered diagram in $\operatorname{\mathcal{C}}'_0$. It follows that $\operatorname{\mathcal{C}}'$ is an $\operatorname{Ind}_{\kappa }$-completion of the essentially small subcategory $\operatorname{\mathcal{C}}'_0$, and therefore $\kappa $-accessible. $\square$

Proof. Assume that $F$ is accessible; we will show that condition $(2)$ is satisfied (the converse is immediate from the definitions). Using Remark 9.4.7.2, we can choose a small regular cardinal $\kappa $ for which $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are $\kappa $-accessible and the functor $F$ is $\kappa $-finitary. Let $\operatorname{\mathcal{C}}_{< \kappa }$ be the full subcategory of $\operatorname{\mathcal{C}}$ spanned by the $\kappa $-compact objects. It follows from Proposition 9.4.6.2 that $\operatorname{\mathcal{C}}_{< \kappa }$ is essentially small. Using Remark 9.4.6.8, we can choose a small regular cardinal $\lambda \geq \kappa $ such that $F( \operatorname{\mathcal{C}}_{< \kappa } ) \subseteq \operatorname{\mathcal{D}}_{< \lambda }$: that is, $F$ carries $\kappa $-compact objects of $\operatorname{\mathcal{C}}$ to $\lambda $-compact objects of $\operatorname{\mathcal{D}}$. Enlarging $\lambda $ if necessary, we may assume that $\kappa \trianglelefteq \lambda $ (see Proposition 9.1.7.8). We will complete the proof by showing that the functor $F$ is $\lambda $-compact. Since the functor $F$ is $\kappa $-finitary, it is also $\lambda $-finitary. It will therefore suffice to show that for every $\lambda $-compact object $C \in \operatorname{\mathcal{C}}$, the image $F(C) \in \operatorname{\mathcal{D}}$ is also $\lambda $-compact. Using Corollary 9.3.6.7, we can realized $C$ as the colimit of a $\lambda $-small $\kappa $-filtered diagram $G: \operatorname{\mathcal{K}}\rightarrow \operatorname{\mathcal{C}}_{< \kappa } \subseteq \operatorname{\mathcal{C}}$. Since the functor $F$ is $\kappa $-finitary, the object $F(C)$ is a colimit of the composite functor

and is therefore $\lambda $-compact by virtue of Corollary 9.2.5.25. $\square$

Proof. The implication $(1) \Rightarrow (2)$ follows from Example 9.4.7.16 and Remark 9.4.7.7. Conversely, suppose that condition $(2)$ is satisfied. Since $\operatorname{\mathcal{D}}$ is accessible, it admits a dense full subcategory $\operatorname{\mathcal{D}}_0 \subseteq \operatorname{\mathcal{D}}$ which is essentially small (for example, if $\operatorname{\mathcal{D}}$ is $\kappa $-accessible, we can take $\operatorname{\mathcal{D}}_0$ to be the full subcategory spanned by the $\kappa $-compact objects). Applying Proposition 8.4.1.8, we see that the restricted Yoneda embedding

is fully faithful. It will therefore suffice to show that there exists a small regular cardinal $\lambda $ for which the composite functor $h \circ F$ preserves small $\lambda $-filtered colimits. By virtue of Proposition 7.1.8.2, this is equivalent to the requirement that the composite functor $\operatorname{\mathcal{C}}\xrightarrow {F} \operatorname{\mathcal{D}}\xrightarrow { \operatorname{Hom}_{\operatorname{\mathcal{D}}}(D, \bullet ) } \operatorname{\mathcal{S}}$ preserves small filtered colimits for each $D \in \operatorname{\mathcal{D}}_0$. The existence of $\lambda $ follows by combining $(2)$ with our assumption that $\operatorname{\mathcal{D}}_0$ is essentially small. $\square$

Proof. The functor $F$ preserves all colimits which exist in $\operatorname{\mathcal{C}}$ (Corollary 7.1.4.28) and is therefore accessible. To show that $G$ is accessible, it will suffice to show that for each object $C \in \operatorname{\mathcal{C}}$, the composite functor

is accessible (Proposition 9.4.7.17). Since $G$ is right adjoint to $F$, this composite functor is corepresented by the object $F(C) \in \operatorname{\mathcal{D}}$, so the desired result follows from Example 9.4.7.16. $\square$