$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Corollary Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $u: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. The following conditions are equivalent:


The morphism $u$ is an isomorphism.


There exists a Kan complex $\operatorname{\mathcal{E}}$, a morphism $\overline{u}: \overline{X} \rightarrow \overline{Y}$ in $\operatorname{\mathcal{E}}$, and a functor $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ satisfying $F(\overline{u}) = u$.


There exists a contractible Kan complex $\operatorname{\mathcal{E}}$, a morphism $\overline{u}: \overline{X} \rightarrow \overline{Y}$ in $\operatorname{\mathcal{E}}$, and a functor $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ satisfying $F(\overline{u}) = u$.

Proof. If $u$ is an isomorphism, then it belongs to the image of the inclusion functor $\operatorname{\mathcal{C}}^{\simeq } \hookrightarrow \operatorname{\mathcal{C}}$. Since the core $\operatorname{\mathcal{C}}^{\simeq }$ is a Kan complex, this proves that $(1) \Rightarrow (2)$. Conversely, if we can write $u = F( \overline{u} )$ for some functor $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ where $\operatorname{\mathcal{E}}$ is a Kan complex, then Remark guarantees that $u$ is an isomorphism in $\operatorname{\mathcal{C}}$ (since $\overline{u}$ is automatically an isomorphism in $\operatorname{\mathcal{E}}$, by virtue of Proposition This proves that $(2) \Rightarrow (1)$.

The implication $(3) \Rightarrow (2)$ is immediate. We will complete the proof by showing that $(2)$ implies $(3)$. Let $\operatorname{\mathcal{E}}$ be a Kan complex, let $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor, and let $\overline{u}$ be an edge of $\operatorname{\mathcal{E}}$ satisfying $F( \overline{u} ) = u$. Let us identify $\overline{u}$ with a morphism of simplicial sets $\Delta ^1 \rightarrow \operatorname{\mathcal{E}}$. By virtue of Proposition, this morphism factors as a composition $\Delta ^1 \xrightarrow {v} \operatorname{\mathcal{E}}' \xrightarrow {q} \operatorname{\mathcal{E}}$, where $v$ is anodyne and $q$ is a Kan fibration. Since $\operatorname{\mathcal{E}}$ is a Kan complex and $q$ is a Kan fibration, the simplicial set $\operatorname{\mathcal{E}}'$ is a Kan complex (Remark Because $\Delta ^1$ is weakly contractible and $v$ is a weak homotopy equivalence, the Kan complex $\operatorname{\mathcal{E}}'$ is contractible. We can then write $u = F'(v)$ where $F' = F \circ q$. $\square$