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Proposition 5.1.5.6. Suppose we are given a commutative diagram of $\infty $-categories

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d]^-{U} & \operatorname{\mathcal{C}}' \ar [d]^-{U'} \\ \operatorname{\mathcal{D}}\ar [r]^-{ \overline{F} } & \operatorname{\mathcal{D}}', } \]

where the functors $U$ and $U'$ are isofibrations and the functors $F$ and $\overline{F}$ are equivalences of $\infty $-categories. Let $g: Y \rightarrow Z$ be a morphism in $\operatorname{\mathcal{C}}$. Then $g$ is $U$-cartesian if and only if $F(g)$ is $U'$-cartesian.

Proof. It follows from Lemma 5.1.5.5 that if $F(g)$ is $U'$-cartesian, then $g$ is $U$-cartesian. For the converse, suppose that $g$ is $U$-cartesian. Arguing as in the proof of Corollary 5.1.5.2, we can construct a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [r]^-{G} \ar [d]^-{U'} & \operatorname{\mathcal{C}}\ar [d]^-{U} \\ \operatorname{\mathcal{D}}' \ar [r]^-{\overline{G}} & \operatorname{\mathcal{D}}, } \]

where $G$ and $\overline{G}$ are homotopy inverses of the equivalences $F$ and $\overline{F}$, respectively. Then $G( F(g) )$ is isomorphic to $g$ as an object of the arrow $\infty $-category $\operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}})$. Invoking Corollary 5.1.2.5, we conclude that $G( F(g) )$ is $U$-cartesian, so that $F(g)$ is $U'$-cartesian by virtue of Lemma 5.1.5.5. $\square$