# Kerodon

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Proposition 7.6.3.15. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category 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 an isomorphism, then $\sigma$ is a pullback square if and only if $f'$ is also an isomorphism.

$(2)$

If $f'$ is an isomorphism, then $\sigma$ is a pushout square if and only if $f$ is also an isomorphism.

Proof. We will prove $(1)$; the proof of $(2)$ is similar. Note that $\sigma$ restricts to a diagram

$\sigma _0: \operatorname{N}_{\bullet }(  \times  \setminus \{ (0,0) \} ) \rightarrow \operatorname{\mathcal{C}}$

satisfying $\sigma _0( 0,1) = X$, $\sigma _0( 1,1) = Y$, and $\sigma _0( 1,0 ) = Y'$. The assumption that $f$ is an isomorphism guarantees that $\sigma _0$ is right Kan extended from the full subcategory

$\{ 1\} \times \Delta ^1 \subseteq \operatorname{N}_{\bullet }( \times  \setminus \{ (0,0) \} ).$

It follows that $\sigma$ is a pullback diagram if and only if the restriction $\sigma |_{ \operatorname{N}_{\bullet }( \{ (0,0) < (1,0) < (1,1) \} )}$ is a limit diagram (Corollary 7.3.7.2) By virtue of Corollary 7.2.2.6, this is equivalent to the requirement that $f'$ is an isomorphism. $\square$