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Proposition Let $\operatorname{\mathcal{C}}$ be a category, let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$, and let $S$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ which are weakly left orthogonal to $S$. Then $S$ is closed under retracts.

Proof. Let $f'$ be a morphism of $\operatorname{\mathcal{C}}$ which belongs to $S$ and let $f$ be a retract of $f'$, so that there exists a commutative diagram

\[ \xymatrix@C =40pt@R=40pt{ C \ar [r]^-{i} \ar [d]^{f} & C' \ar [d]^{f'} \ar [r]^-{r} & C \ar [d]^{f} \\ D \ar [r]^-{\overline{i}} & D' \ar [r]^-{\overline{r}} & D } \]

with $r \circ i = \operatorname{id}_{C}$ and $\overline{r} \circ \overline{i} = \operatorname{id}_{D}$. We wish to show that $f$ also belongs to $S$. Consider a lifting problem $\sigma :$

\[ \xymatrix@C =40pt@R=40pt{ C \ar [d]^{f} \ar [r]^-{u} & X \ar [d]^{g} \\ D \ar [r]^-{v} \ar@ {-->}[ur]^{h} & Y, } \]

where $g$ belongs to $T$. Our assumption $f' \in S$ ensures that the associated lifting problem

\[ \xymatrix@C =40pt@R=40pt{ C' \ar [d]^{f'} \ar [r]^-{u \circ r} & X \ar [d]^{g} \\ D' \ar [r]^-{v \circ \overline{r}} \ar@ {-->}[ur] & Y } \]

admits a solution: that is, we can choose a morphism $h': D' \rightarrow X$ satisfying $g \circ h' = v \circ \overline{r}$ and $h' \circ f' = u \circ r$. Then the morphism $h = h' \circ \overline{i}$ is a solution to the lifting problem $\sigma $, by virtue of the calculations

\[ g \circ h = g \circ h' \circ \overline{i} = v \circ \overline{r} \circ \overline{i} = v \]

\[ h \circ f = h' \circ \overline{i} \circ f = h' \circ f' \circ i = u \circ r \circ i = u. \]