Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Variant 1.4.4.7. Let $\operatorname{\mathcal{C}}$ be a category. We will say that a morphism $f: C \rightarrow D$ of $\operatorname{\mathcal{C}}$ is a retract of another morphism $f': C' \rightarrow D'$ if it is a retract of $f'$ when viewed as an object of the functor category $\operatorname{Fun}( [1], \operatorname{\mathcal{C}})$. In other words, we say that $f$ is a retract of $f'$ if there exists a commutative diagram

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

in the category $\operatorname{\mathcal{C}}$, where $r \circ i = \operatorname{id}_{C}$ and $\overline{r} \circ \overline{i} = \operatorname{id}_{D}$.

We say that a collection of morphisms $T$ of $\operatorname{\mathcal{C}}$ is closed under retracts if, for every pair of morphisms $f,f'$ in $\operatorname{\mathcal{C}}$, if $f$ is a retract of $f'$ and $f'$ belongs to $T$, then $f$ also belongs to $T$.