# Kerodon

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

Proposition 8.5.2.4. Let $\operatorname{\mathcal{C}}$ be a category containing a retraction diagram

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{r} & \\ Y \ar [ur]^{i} \ar [rr]^{ \operatorname{id}_{Y} } & & Y, }$

and let $e = i \circ r$ be the idempotent endomorphism Example 8.5.2.3. Then:

$(1)$

The morphism $i$ exhibits $Y$ as an equalizer of the pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$.

$(2)$

The morphism $r$ exhibits $Y$ as a coequalizer of the pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$.

Proof. We will prove $(1)$; the proof of $(2)$ is similar. Fix an object $Z \in \operatorname{\mathcal{C}}$ and a morphism $f: Z \rightarrow X$ satisfying $e \circ f = \operatorname{id}_{X} \circ f$; we wish to show that there is a unique morphism $g: Z \rightarrow Y$ satisfying $i \circ g = f$. To prove uniqueness, we note that $g$ is determined by the identity

$g = \operatorname{id}_{Y} \circ g = (r \circ i) \circ g = r \circ (i \circ g) = r \circ f.$

To establish existence, we observe that the composition $g = r \circ f$ satisfies the identity

$i \circ g = i \circ (r \circ f) = (i \circ r) \circ f = e \circ f = \operatorname{id}_{X} \circ f = f.$
$\square$