Corollary 9.2.7.17. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $g: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. If another morphism $f: A \rightarrow B$ is left orthogonal to $g$, then any retract of $f$ (in the $\infty $-category $\operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}})$) is also left orthogonal to $g$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. We proceed as in the proof of Proposition 9.2.5.14. Let $f': A' \rightarrow B'$ be a retract of $f$ (in the $\infty $-category $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$); we will show that $f'$ is left orthogonal to $g$. Let $\pi : \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ be the projection map, and let us identify $g$ with an object $\widetilde{X} \in \operatorname{\mathcal{C}}_{/Y}$ satisfying $\pi ( \widetilde{X} ) = X$. By virtue of Corollary 9.2.7.13, it will suffice to show that for any morphism $\widetilde{f}'$ of $\operatorname{\mathcal{C}}_{/Y}$ satisfying $\pi ( \widetilde{f}' ) = f'$, the object $\widetilde{X}$ is $\widetilde{f}'$-local. It follows from Corollary 4.2.5.2 that $\pi $ induces a right fibration $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}_{/Y} ) \rightarrow \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$. Applying Remark 8.5.1.23, we deduce that $\widetilde{f}'$ is a retract of a morphism $\widetilde{f}$ of $\operatorname{\mathcal{C}}_{/Y}$ satisfying $U( \widetilde{f} ) = f$. By virtue of Variant 9.2.1.7, it will suffice to show that the object $\widetilde{X}$ is $\widetilde{f}$-local, which follows from our assumption that $f$ is left orthogonal to $g$ (Corollary 9.2.7.13). $\square$