 Solution

For an algebra for a monad $(A,a)$, consider the fork in the Eilenberg-Moore categoryHere ${\mu }_A:F(TA)\to F(A)$ is the counit at $F(A)$, and $Ta=F(a)$. The arrow $a$ is an algebra morphism by $a{\mu }_A=aTa$, which also says that it coequalizes the two arrows.

Let $h:F(A)\to (B,b)$ be an algebra morphism with $h{\mu }_A=hTa$. Define $g=h{\eta }_A:A\to B$. Naturality of $\eta$ at $a$ givesSince $h$ is an algebra morphism,Thus $g:(A,a)\to (B,b)$ is an algebra morphism with $ga=h$. Any other such factorization satisfies $g^{\prime }=g^{\prime }a{\eta }_A=h{\eta }_A=g$. This proves the full coequalizer universal property inside the algebra category.

The pair is moreover a reflexive pair: its common section is $F({\eta }_A)$, with underlying map $T{\eta }_A$, because ${\mu }_{A}T{\eta }_A=1_{TA}$ and $TaT{\eta }_A=T(a{\eta }_A)=1_{TA}$. Hence every monad algebra is a reflexive coequalizer of free algebras. This reflexive free-algebra presentation of a monad algebra needs no general existence theorem for arbitrary algebra-category colimits.