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Corollary 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:


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 $\operatorname{\mathcal{D}}$.


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 Proposition, 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 We will complete the proof by showing that $\eta $ is also an isomorphism. Fix an object $C \in \operatorname{\mathcal{C}}$, we wish to show that the map $\eta _{C}: C \rightarrow (G \circ F)(C)$ is an isomorphism in the $\infty $-category $\operatorname{\mathcal{C}}$ (Theorem By virtue of assumption $(1)$, it will suffice to show that $F( \eta _{C} )$ is an isomorphism in the $\infty $-category $\operatorname{\mathcal{D}}$. The compatibility of $\eta $ and $\epsilon $ guarantees that the diagram

\[ \xymatrix { & (F \circ G \circ F)(C) \ar [dr]^{ \epsilon _{F(C)} } & \\ F(C) \ar [ur]^{ F( \eta _ C ) } \ar [rr]^{\operatorname{id}} & & F(C) } \]

commutes in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}$. Since $\epsilon _{F(C)}$ is an isomorphism, it follows that $F( \eta _ C )$ is also an isomorphism (Remark $\square$