Proposition 4.7.6.13 (Uniqueness). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an equivalence of $\infty $-categories. If $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are minimal, then $F$ is an isomorphism of simplicial sets.
Proof. Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a homotopy inverse to $F$. It follows from Lemma 4.7.6.11 that $F$ and $G$ are monomorphisms of simplicial sets. We will complete the proof by showing that the composite map $(F \circ G): \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}$ is an epimorphism of simplicial sets (so that, in particular, $F$ is an epimorphism). Let $\sigma $ be an $n$-simplex of $\operatorname{\mathcal{D}}$; we wish to show that $\sigma $ belongs to the image of $F \circ G$. The proof proceeds by induction on $n$. Set $\sigma _0 = \sigma |_{ \operatorname{\partial \Delta }^ n }$; our inductive hypothesis then guarantees that we can write $\sigma _0 = (F \circ G)( \tau _0 )$ for some morphism $\tau _0: \operatorname{\partial \Delta }^ n \rightarrow \operatorname{\mathcal{D}}$.
Choose a $2$-simplex
in the $\infty $-category $\operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{\mathcal{D}})$, where $\alpha $ and $\beta $ are isomorphisms. Precomposing with $\tau _0: \operatorname{\partial \Delta }^ n \rightarrow \operatorname{\mathcal{D}}$, we obtain a $2$-simplex
in the $\infty $-category $\operatorname{Fun}( \operatorname{\partial \Delta }^ n, \operatorname{\mathcal{D}})$. Using Corollary 4.4.5.9, we can lift $\alpha (\tau _0)$ to an isomorphism $\widetilde{\alpha }: \sigma \rightarrow \tau $ in the $\infty $-category $\operatorname{Fun}( \Delta ^ n, \operatorname{\mathcal{D}})$. Since $\operatorname{\mathcal{D}}$ is an $\infty $-category, Theorem 1.5.6.1 guarantees that we can lift (4.77) to a $2$-simplex
in the $\infty $-category $\operatorname{Fun}( \Delta ^ n, \operatorname{\mathcal{D}})$. By construction, $\gamma $ is an isomorphism relative to $\operatorname{\partial \Delta }^ n$. Our assumption that $\operatorname{\mathcal{D}}$ is minimal then guarantees that $\sigma = (F \circ G)(\tau )$ belongs to the image of $F \circ G$. $\square$