Kerodon

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

Proposition 4.5.2.14. A commutative diagram of $\infty $-categories

4.24
\begin{equation} \begin{gathered}\label{equation:characterize-categorical-pullback2} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}_{01} \ar [r] \ar [d] & \operatorname{\mathcal{C}}_0 \ar [d] \\ \operatorname{\mathcal{C}}_1 \ar [r] & \operatorname{\mathcal{C}}} \end{gathered} \end{equation}

is a categorical pullback square if and only if, for every simplicial set $X$, the diagram of Kan complexes

4.25
\begin{equation} \begin{gathered}\label{equation:characterize-categorical-pullback22} \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(X, \operatorname{\mathcal{C}}_{01} )^{\simeq } \ar [r] \ar [d] & \operatorname{Fun}( X, \operatorname{\mathcal{C}}_0)^{\simeq } \ar [d] \\ \operatorname{Fun}(X, \operatorname{\mathcal{C}}_1)^{\simeq } \ar [r] & \operatorname{Fun}(X, \operatorname{\mathcal{C}})^{\simeq } } \end{gathered} \end{equation}

is a homotopy pullback square.

Proof. By definition, the diagram (4.24) is a categorical pullback square if and only if the induced map $\theta : \operatorname{\mathcal{C}}_{01} \rightarrow \operatorname{\mathcal{C}}_{0} \times ^{\mathrm{h}}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1$ is an equivalence of $\infty $-categories. Using the criterion of Proposition 4.5.1.22, we see that this is equivalent to the requirement that $\theta $ induces a homotopy equivalence $\theta _{X}: \operatorname{Fun}( X, \operatorname{\mathcal{C}}_{01} )^{\simeq } \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{C}}_{0} \times ^{\mathrm{h}}_{\operatorname{\mathcal{C}}} \operatorname{\mathcal{C}}_1)^{\simeq }$ for every simplicial set $X$. Using Remarks 4.5.2.6 and 4.5.2.7, we can identify $\theta _{X}$ with the map

\[ \operatorname{Fun}( X, \operatorname{\mathcal{C}}_{01} )^{\simeq } \rightarrow \operatorname{Fun}(X, \operatorname{\mathcal{C}}_0)^{\simeq } \times ^{\mathrm{h}}_{\operatorname{Fun}(X, \operatorname{\mathcal{C}})^{\simeq } } \operatorname{Fun}(X, \operatorname{\mathcal{C}}_1)^{\simeq } \]

determined by the commutative diagram (4.25). The desired result now follows from the criterion of Corollary 3.4.1.6. $\square$