Theorem 4.6.2.21. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be functor of $\infty $-categories. Then $F$ is an equivalence of $\infty $-categories if and only if it is fully faithful and essentially surjective.
Proof of Theorem 4.6.2.21. Assume first that $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is an equivalence of $\infty $-categories. Then $F$ induces a homotopy equivalence of Kan complexes $F^{\simeq }: \operatorname{\mathcal{C}}^{\simeq } \rightarrow \operatorname{\mathcal{D}}^{\simeq }$ (Remark 4.5.1.19). Passing to connected components, we conclude that the induced map $\pi _0( \operatorname{\mathcal{C}}^{\simeq } ) \rightarrow \pi _0( \operatorname{\mathcal{D}}^{\simeq } )$ is bijective. In particular, $F$ is essentially surjective. We have a commutative diagram of Kan complexes
where the horizontal maps are homotopy equivalences (Theorem 4.5.7.1) and the vertical maps are Kan fibrations (Corollary 4.4.5.4). Applying Proposition 3.2.8.1, we conclude that for every vertex $(X,Y) \in \operatorname{Fun}( \operatorname{\partial \Delta }^1, \operatorname{\mathcal{C}})^{\simeq }$, the induced map of fibers
is a homotopy equivalence. It follows that $F$ is fully faithful.
Now suppose that $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is a functor of $\infty $-categories which is fully faithful and essentially surjective. Using Corollary 4.6.2.11 and Remark 4.6.2.15, we see that the induced map $F^{\simeq }: \operatorname{\mathcal{C}}^{\simeq } \rightarrow \operatorname{\mathcal{D}}^{\simeq }$ is also fully faithful and essentially surjective, and is therefore a homotopy equivalence of Kan complexes (Lemma 4.6.2.22). It follows that the morphism $\theta _0$ in (4.55) is a homotopy equivalence of Kan complexes. Combining our assumption that $F$ is fully faithful with Proposition 3.2.8.1, we conclude that $\theta $ is also a homotopy equivalence. Applying Theorem 4.5.7.1, we conclude that $F$ is an equivalence of $\infty $-categories. $\square$