Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 4.6.1.8. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category. For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the morphism space $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is a Kan complex.

Proof. By definition, the morphism space $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ can be identified with the fiber of the restriction map

\[ \theta : \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^1, \operatorname{\mathcal{C}}) \]

over the vertex $(X,Y) \in \operatorname{Fun}( \operatorname{\partial \Delta }^1, \operatorname{\mathcal{C}})$. Corollary 4.4.5.3 guarantees that $\theta $ is an isofibration, so that $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is an $\infty $-category. To show that it is a Kan complex, it will suffice to show that every morphism $u$ in $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is an isomorphism (Proposition 4.4.2.1). By virtue of Corollary 4.4.3.15, it will suffice to show that the image of $u$ in the $\infty $-category $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$ is an isomorphism. This follows from Theorem 4.4.4.4, since evaluation functors

\[ \operatorname{ev}_0: \operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(\{ 0\} , \operatorname{\mathcal{C}}) \simeq \operatorname{\mathcal{C}}\quad \quad \operatorname{ev}_1: \operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(\{ 1\} , \operatorname{\mathcal{C}}) \simeq \operatorname{\mathcal{C}} \]

carry $u$ to the identity morphisms $\operatorname{id}_{X}$ and $\operatorname{id}_{Y}$, respectively. $\square$