 Monad algebra forgetful functor creates split coequalizers

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 $q,Tq,T^{2}q$ 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.