 Idempotent splitting through a coequalizer

If $e=ir$ and $ri=1$, then $r$ is the coequalizer of $(e,1)$: an arrow $h$ with $he=h$ factors uniquely as $(hi)r$. Conversely a coequalizer $q$ makes $e$ factor as $iq=e$; epimorphic cancellation gives $qi=1$. Thus the splitting of an idempotent morphism is equivalent to this particular coequalizer, without assuming arbitrary coequalizers exist.