# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Corollary 6.2.2.13. Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a functor of $\infty$-categories. The following conditions are equivalent:

$(1)$

The functor $G$ is fully faithful and the essential image of $G$ is a reflective subcategory of $\operatorname{\mathcal{C}}$.

$(2)$

The functor $G$ is fully faithful and admits a left adjoint $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$.

$(3)$

There exist a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ and a natural isomorphism $\epsilon : F \circ G \xrightarrow {\sim } \operatorname{id}_{\operatorname{\mathcal{D}}}$ which is the counit of an adjunction between $F$ and $G$.

$(4)$

The functor $G$ admits a left adjoint $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ for which the composition $(F \circ G): \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}$ is an equivalence of $\infty$-categories.

Proof. Let $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ be the essential image of $G$. If $G$ is fully faithful, then it induces an equivalence $\operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}'$ (Corollary 4.6.2.19). The equivalence $(1) \Leftrightarrow (2)$ follows by applying Proposition 6.2.2.7 to the subcategory $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$, and the implication $(2) \Rightarrow (3)$ follows by applying Proposition 6.2.2.11 to the subcategory $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$. To show that $(3) \Rightarrow (2)$, we observe that if a natural isomorphism $\epsilon : F \circ G \xrightarrow {\sim } \operatorname{id}_{\operatorname{\mathcal{D}}}$ is the counit of an adjunction, then $G$ restricts to an equivalence of $\operatorname{\mathcal{D}}$ with a full subcategory of $\operatorname{\mathcal{C}}$ (Proposition 6.2.1.13), and is therefore fully faithful. The equivalence $(3) \Leftrightarrow (4)$ is a special case of Proposition 6.1.4.7. $\square$