Kerodon

Proof. Let $f: A \rightarrow B$ be a transfinite composition of morphisms of $S$; we wish to show that $f$ is weakly 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 Remark 9.1.5.12, it will suffice to show that for every morphism $\widetilde{f}: \widetilde{A} \rightarrow \widetilde{B}$ of $\operatorname{\mathcal{C}}_{/Y}$ satisfying $\pi ( \widetilde{f} ) = f$, the object $\widetilde{X}$ is weakly $\widetilde{f}$-local.

Choose an ordinal $\alpha $ and a diagram $F: \operatorname{N}_{\bullet }( \mathrm{Ord}_{\leq \alpha } ) \rightarrow \operatorname{\mathcal{C}}$ which exhibits $f$ as a transfinite composition of morphisms of $S$ (see Definition 9.1.2.1). We will assume that $\alpha > 0$ (otherwise, $f$ is an identity morphism and the desired result follows from Example 9.1.5.13). In this case, Lemma 4.3.7.8 guarantees that the inclusion map $\operatorname{N}_{\bullet }( \{ 0 < \alpha \} ) \hookrightarrow \operatorname{N}_{\bullet }( \mathrm{Ord}_{\leq \alpha } )$ is right anodyne. Since $\pi $ is a right fibration (Proposition 4.3.6.1), we can lift $F$ to a diagram $\widetilde{F}: \operatorname{N}_{\bullet }( \mathrm{Ord}_{ \leq \alpha } ) \rightarrow \operatorname{\mathcal{C}}$ for which the associated morphism $\widetilde{F}(0) \rightarrow \widetilde{F}(\alpha )$ coincides with $\widetilde{f}$. For every nonzero limit ordinal $\lambda \leq \alpha $, Proposition 7.1.3.19 guarantees that the restriction $\widetilde{F}|_{ \operatorname{N}_{\bullet }( \mathrm{Ord}_{\leq \lambda } ) }$ is a colimit diagram in the $\infty $-category $\operatorname{\mathcal{C}}_{/Y}$. Using Remark 9.1.5.12, we see that $\widetilde{F}$ exhibits $\widetilde{f}$ as a transfinite composition of morphisms of $\operatorname{\mathcal{C}}_{/Y}$ with respect to which $\widetilde{X}$ is weakly local. Applying Proposition 9.1.3.17, we conclude that $\widetilde{X}$ is weakly $\widetilde{f}$-local, as desired. $\square$