Kerodon

Proof. Let $\overline{v}: F(Y) \rightarrow F(X)$ be a homotopy inverse to $F(u)$. Since $F$ is fully faithful, the natural map $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y, X) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(Y), F(X))$ is a homotopy equivalence. We may therefore assume without loss of generality that $\overline{v} = F(v)$, for some morphism $v: Y \rightarrow X$ in the $\infty $-category $\operatorname{\mathcal{C}}$. Let $v \circ u$ be a composition of $u$ and $v$ in the $\infty $-category $\operatorname{\mathcal{C}}$. Since $F(u)$ is homotopy inverse to $F(v)$, the morphism $F( v \circ u)$ is homotopic to $\operatorname{id}_{ F(C) } = F( \operatorname{id}_{C} )$. Since the map $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X, X) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(X))$ is a homotopy equivalence, it follows that $v \circ u$ is homotopic to $\operatorname{id}_{C}$: that is, $v$ is a left homotopy inverse to $u$. A similar argument (with the roles of $u$ and $v$ reversed) shows that $v$ is also a right homotopy inverse to $u$. It follows that $u$ is an isomorphism. $\square$