# Kerodon

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

Corollary 8.5.2.5. Let $\operatorname{\mathcal{C}}$ be a category and let $e: X \rightarrow X$ be an idempotent endomorphism in $\operatorname{\mathcal{C}}$. The following conditions are equivalent:

$(1)$

The idempotent endomorphism $e$ splits. That is, $e$ admits a factorization $X \xrightarrow {r} Y \xrightarrow {i} X$, where $r \circ i = \operatorname{id}_{Y}$.

$(2)$

The pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$ admits an equalizer in $\operatorname{\mathcal{C}}$.

$(3)$

The pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$ admits a coequalizer in $\operatorname{\mathcal{C}}$.

Proof. The implications $(1) \Rightarrow (2)$ and $(1) \Rightarrow (3)$ follow from Proposition 8.5.2.4. We will show that $(2)$ implies $(1)$; the proof of the implication $(3) \Rightarrow (1)$ is similar. Suppose that there exists a morphism $i: Y \rightarrow X$ which exhibits $Y$ as an equalizer of the pair of morphisms $(e,\operatorname{id}_ X): X \rightrightarrows X$. Since $e$ is idempotent, we have $e \circ e = e = \operatorname{id}_{X} \circ e$. Invoking the universal property of $Y$, we deduce that there is a unique morphism $r: X \rightarrow Y$ satisfying $e = i \circ r$. To complete the proof, it will suffice to show that $r \circ i$ is the identity morphism from $Y$ to itself. Since $i$ is a monomorphism, this follows from the calculation

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