= Solution
A <split coequalizer> consists of $f,g:A\rightrightarrows B$, $q:B\to Q$, $s:Q\to B$ and $t:B\to A$ with $qf=qg$, $qs=1_Q$, $ft=1_B$ and $gt=sq$. If $hf=hg$, then $h=hft=hgt=hsq$. Thus $hs$ factors $h$ through $q$, and the factor is unique because $q$ has the right inverse $s$. This proves the <coequalizer> property directly. Every <functor> preserves the diagram, since all these equations are preserved.
For <idempotent splitting through a coequalizer>, first suppose $e=ir$ with $ri=1_Q$. Then $re=r$. Any $h:E\to Z$ satisfying $he=h$ factors as $h=(hi)r$, uniquely since $r$ is a <split epimorphism>. Hence $r$ coequalizes $(e,1_E)$.
Conversely, let $q:E\to Q$ coequalize $(e,1_E)$. Since $e^2=e$, the arrow $e:E\to E$ itself equalizes the pair, so there is a unique $i:Q\to E$ with $iq=e$. Then $qi q=qe=q$, and every <coequalizer> is an <epimorphism>, so $qi=1_Q$. Thus $e=iq$ is a <splitting of an idempotent morphism>.
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