= Monad algebra forgetful functor creates split coequalizers
{title2=$cTq=qb$}
For algebra maps $f,g:(A,a)\rightrightarrows(B,b)$ whose underlying maps have a <split coequalizer> $q:B\to Q$, every functor preserves the split diagram, including $T$ and $T^2$. Since $qb$ equalizes $Tf,Tg$, it factors uniquely to the displayed action $c:TQ\to Q$. Cancellation of the epimorphisms $q,Tq,T^2q$ proves the unit and multiplication laws. Any algebra-valued map equalizing the pair factors through $q$, and cancelling $Tq$ proves that its factor preserves the action. This establishes creation of that coequalizer.
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