$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Corollary 6.2.2.21. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories. Then $F$ is an equivalence if and only if it satisfies the following pair of conditions:
- $(1)$
The functor $F$ is conservative. That is, a morphism $u$ of $\operatorname{\mathcal{C}}$ is an isomorphism if and only if $F(u)$ is an isomorphism in $\operatorname{\mathcal{D}}$.
- $(2)$
The functor $F$ admits a fully faithful right adjoint $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$.
Proof.
Suppose that conditions $(1)$ and $(2)$ are satisfied; we will show that $F$ is an equivalence of $\infty $-categories (the converse is immediate from the definitions). Combining assumption $(2)$ with Corollary 6.2.2.19, we can choose a functor $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ 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 $G$. Let $\eta : \operatorname{id}_{\operatorname{\mathcal{C}}} \rightarrow G \circ F$ be a natural transformation which is compatible up to homotopy with $\epsilon $, in the sense of Definition 6.2.1.1. For each object $C \in \operatorname{\mathcal{C}}$, the morphism $\eta _{C}$ fits into a commutative diagram
\[ \xymatrix { & (F \circ G \circ F)(C) \ar [dr]^{ \epsilon _{F(C)} } & \\ F(C) \ar [ur]^{ F( \eta _ C ) } \ar [rr]^{\operatorname{id}_{F(C)}} & & F(C) } \]
in the $\infty $-category $\operatorname{\mathcal{D}}$, where $\epsilon _{ F(C) }$ and $\operatorname{id}_{ F(C) }$ are isomorphisms. It follows that $F( \eta _ C )$ is also an isomorphism in $\operatorname{\mathcal{D}}$. Applying assumption $(1)$, we deduce that $\eta _{C}$ is an isomorphism in $\operatorname{\mathcal{C}}$. Allowing the object $C$ to vary (and invoking the criterion of Theorem 4.4.4.4), we deduce that $\eta $ is also a natural isomorphism, so that $F$ and $G$ are homotopy inverse to one another.
$\square$