# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

Proposition 1.4.4.9. Let $\operatorname{\mathcal{C}}$ be a category, let $S$ be a collection of morphisms of $\operatorname{\mathcal{C}}$, and let $T$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ having the left lifting property with respect to $S$. Then $T$ is closed under retracts.

Proof. Let $f'$ be a morphism of $\operatorname{\mathcal{C}}$ which belongs to $T$ 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 $T$. Consider a lifting problem $\sigma :$

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

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

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

admits a solution: that is, we can choose a morphism $h': D' \rightarrow X$ satisfying $p \circ h' = v \circ \overline{r}$ 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

$p \circ h = p \circ h' \circ \overline{i} = v \circ \overline{r} \circ \overline{i} = v$

$h \circ f = h' \circ \overline{i} \circ f = h' \circ f' \circ i = u \circ r \circ i = u.$
$\square$