Kerodon

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$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Corollary 9.2.7.15. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category containing a morphism $g: X \rightarrow Y$ and a $2$-simplex

9.17
\begin{equation} \begin{gathered}\label{equation:two-out-of-three-for-orthogonal} \xymatrix@R =50pt@C=50pt{ & B \ar [dr]^{ f'' } & \\ A \ar [ur]^{f'} \ar [rr]^{f} & & C. } \end{gathered} \end{equation}

Assume that $f'$ is left orthogonal to $g$. Then $f$ is left orthogonal to $g$ if and only if $f''$ is left orthogonal to $g$.

Proof. Assume that $f''$ is left orthogonal to $g$; we will show that $f$ is left orthogonal to $g$ (the proof of the converse is similar). Let $\pi : \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ be the projection map, so that we can 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 the object $\widetilde{X}$ is $\widetilde{f}$-local, for every morphism $\widetilde{f}$ of $\operatorname{\mathcal{C}}_{/Y}$ satisfying $\pi ( \widetilde{f} ) = f$. Since $\pi $ is a right fibration (Proposition 4.3.6.1), we can lift (9.17) to a diagram

\[ \xymatrix@R =50pt@C=50pt{ & \widetilde{B} \ar [dr]^{ \widetilde{f}'' } & \\ \widetilde{A} \ar [ur]^{ \widetilde{f}'} \ar [rr]^{ \widetilde{f}} & & \widetilde{C} } \]

in the $\infty $-category $\operatorname{\mathcal{C}}_{/Y}$. Corollary 9.2.7.13 guarantees that the object $\widetilde{X}$ is both $\widetilde{f}'$-local and $\widetilde{f}''$-local, so the desired result follows from Remark 9.2.1.11. $\square$