Proposition 7.6.3.23. Let $U: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty $-categories and let $\sigma : \Delta ^1 \times \Delta ^1 \rightarrow \operatorname{\mathcal{C}}$ be a commutative square, represented informally by the diagram
\[ \xymatrix@R =50pt@C=50pt{ X' \ar [r]^-{f'} \ar [d] & Y' \ar [d] \\ X \ar [r]^-{f} & Y. } \]
Then:
- $(1)$
If $f$ is $U$-cartesian, then $\sigma $ is a $U$-pullback square if and only if $f'$ is also $U$-cartesian.
- $(2)$
If $f'$ is $U$-cocartesian, then $\sigma $ is a $U$-pushout square if and only if $f$ is also $U$-cocartesian.
Proof.
We will prove $(1)$; the proof of $(2)$ is similar. Note that $\sigma $ restricts to a diagram
\[ \sigma _0: \operatorname{N}_{\bullet }( \{ (0,1) < (1,1) > (1,0) \} ) \rightarrow \operatorname{\mathcal{C}} \]
satisfying $\sigma _0( 0,1) = X$, $\sigma _0( 1,1) = Y$, and $\sigma _0( 1,0 ) = Y'$. The assumption that $f$ is $U$-cartesian guarantees that $\sigma _0$ is $U$-right Kan extended from the full subcategory
\[ \{ 1\} \times \Delta ^1 \subseteq \operatorname{N}_{\bullet }( \{ (0,1) < (1,1) > (1,0) \} ). \]
It follows that $\sigma $ is a $U$-pullback diagram if and only if the restriction $\sigma |_{ \operatorname{N}_{\bullet }( \{ (0,0) < (1,0) < (1,1) \} )}$ is a $U$-limit diagram (Proposition 7.3.8.1) By virtue of Corollary 7.2.2.5, this is equivalent to the requirement that $f'$ is $U$-cartesian.
$\square$