split epimorphism,equalizer,universal property

Question

Let $g:B\to A$ be a split epimorphism with $f:A\to B$, $g\circ f=\operatorname{id}_A$. Why is $g$ a coequalizer of $f\circ g$ and $\operatorname{id}_B$?
Commutativity is clear,but the universal property is not.

Answer 1

Suppose $h : B \to C$ satisfies $hfg = h$. Suppose $u : A \to C$ satisfies $ug = h$, then $u = ugf = hf$, and reciprocally if $u = hf$ then $ug = hfg = h$. This proves the universal property, ie. that there exists a unique morphism $A \to C$ making the obvious diagram commute.