Coequalizer of an idempotent and the identity implies the idempotent splits
Since $e\circ e=e\circ 1_C$, there is, by the universal property of the coequalizer $(R,r)$, a unique map $i:R\to C$ such that $i\circ r=e$.
Since $e\circ e=e\circ 1_C$, there is, by the universal property of the coequalizer $(R,r)$, a unique map $i:R\to C$ such that $i\circ r=e$.