Proposition 1.5.4.11. Let $\operatorname{\mathcal{C}}$ be a category, let $T$ be a collection of morphisms in $\operatorname{\mathcal{C}}$, and let $S$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ which are weakly left orthogonal to $T$. Then $S$ is closed under transfinite composition.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Let $\alpha $ be an ordinal and suppose we are given a functor $ \mathrm{Ord}_{\leq \alpha } \rightarrow \operatorname{\mathcal{C}}$, given by a pair
\[ ( \{ C_{\beta } \} _{\beta \leq \alpha }, \{ f_{\gamma ,\beta } \} _{\beta \leq \gamma \leq \alpha } ) \]
which satisfies condition $(a)$ of Definition 1.5.4.10. Assume that each of the morphisms $f_{\beta +1,\beta }$ belongs to $S$. We wish to show that the morphism $f_{\alpha ,0}$ also belongs to $S$. For this, we must show that every lifting problem $\sigma :$
\[ \xymatrix@C =40pt@R=40pt{ C_0 \ar [d]_{ f_{\alpha ,0} } \ar [r]^-{u} & X \ar [d]^{g} \\ C_{\alpha } \ar@ {-->}[ur] \ar [r]^-{v} & Y } \]
admits a solution, provided that $g$ belongs to $T$. We construct a collection of morphisms $\{ u_{\beta }: C_{\beta } \rightarrow X \} _{\beta \leq \alpha }$, satisfying the requirements $g \circ u_{\beta } = v \circ f_{\alpha ,\beta }$ and $u_{\beta } = u_{\gamma } \circ f_{\gamma ,\beta }$ for $\beta \leq \gamma $, using transfinite recursion. Fix an ordinal $\gamma \leq \alpha $, and assume that the morphisms $\{ u_{\beta } \} _{ \beta < \gamma }$ have been constructed. We consider three cases:
If $\gamma = 0$, we set $u_{\gamma } = u$.
If $\gamma $ is a nonzero limit ordinal, then our hypothesis that $C_{\gamma }$ is the colimit of the diagram $\{ C_{\beta } \} _{\beta < \gamma }$ guarantees that there is a unique morphism $u_{\gamma }: C_{\gamma } \rightarrow X$ satisfying $u_{\beta } = u_{\gamma } \circ f_{\gamma ,\beta }$ for $\beta < \gamma $. Moreover, our assumption that the equality $g \circ u_{\beta } = v \circ f_{\alpha ,\beta }$ holds for $\beta < \gamma $ guarantees that it also holds for $\beta = \gamma $.
Suppose that $\gamma = \beta +1$ is a successor ordinal. In this case, we take $u_{\gamma }$ to be any solution to the lifting problem
\[ \xymatrix@C =70pt@R=70pt{ C_{\beta } \ar [d]_{ f_{\beta +1,\beta } } \ar [r]^-{u_{\beta }} & X \ar [d]^{g} \\ C_{\beta +1} \ar@ {-->}[ur] \ar [r]^-{v \circ f_{\alpha , \beta +1}} & Y, } \]which exists by virtue of our assumption that $f_{\beta +1,\beta }$ belongs to $S$.
We now complete the proof by observing that $u_{\alpha }$ is a solution to the lifting problem $\sigma $. $\square$