Proposition Let $f: X \rightarrow Y$ be a morphism of simplicial sets. The following conditions are equivalent:


The morphism $f: X \rightarrow Y$ is a categorical equivalence. That is, for every $\infty $-category $\operatorname{\mathcal{C}}$, precomposition with $f$ induces a bijection

\[ \pi _0( \operatorname{Fun}(Y, \operatorname{\mathcal{C}})^{\simeq } ) \rightarrow \pi _0( \operatorname{Fun}(X, \operatorname{\mathcal{C}})^{\simeq } ). \]

For every $\infty $-category $\operatorname{\mathcal{C}}$, precomposition with $f$ induces a homotopy equivalence of Kan complexes $\operatorname{Fun}(Y, \operatorname{\mathcal{C}})^{\simeq } \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{C}})^{\simeq }$.


For every $\infty $-category $\operatorname{\mathcal{C}}$, precompostion with $f$ induces an equivalence of $\infty $-categories $\operatorname{Fun}(Y, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{C}})$.

Proof. The implication $(2) \Rightarrow (1)$ follows from Remark, and the implication $(3) \Rightarrow (2)$ follows from Remark We will complete the proof by showing that $(1)$ implies $(3)$. Assume that $f$ is a categorical equivalence of simplicial sets, let $\operatorname{\mathcal{C}}$ be an $\infty $-category, and let $f^{\ast }: \operatorname{Fun}(Y, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{C}})$ denote the functor given by precomposition with $f$. We wish to show that $[f^{\ast }]$ is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{Cat}_{\infty }}$. For this, it will suffice to show that for any $\infty $-category $\operatorname{\mathcal{D}}$, the induced map

\[ \theta : \pi _0( \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{Fun}(Y, \operatorname{\mathcal{C}}) )^{\simeq } ) \rightarrow \pi _0( \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{Fun}(X, \operatorname{\mathcal{C}}) )^{\simeq } ) \]

is bijective. We conclude by observing that $\theta $ can be identified with the map

\[ \pi _0( \operatorname{Fun}(Y, \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{\mathcal{C}}) )^{\simeq } ) \rightarrow \pi _0( \operatorname{Fun}(X, \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{\mathcal{C}}) )^{\simeq } ) \]

given by precomposition with $f$. $\square$