Coequalizers in category of algebras Given a monad $(T, \eta, \mu) : Sets \to Sets$, $T$-algebras $(A, f : TA \to A)$ and $(B, g : TB \to B)$, and $T$-algebra maps $j, k : (A, f) \to (B, g)$, is it known how to compute the coequalizer of $j$ and $k$, as a $T$-algebra? I am particularly interested in the case where $T$ is the monad corresponding to an algebraic theory. 
 A: This not a complete answer, but rather a collection of, hopefully easy, exercises which when completed answer the question. 
In a similar way to usual universal algebra let us call a $T$-subalgebra of $(B,g)\times (B,g)$ a congruence, when its underlying set is an equivalence relation.
Exercise 1 Show that arbitrary meets of congruences are congruences.
Let $(R,i) \subseteq (B,g)\times (B,g)$ be the smallest congruence containing $\{(j(a),k(a))\,|\, a \in A\}$.
Exercise 2 Use Exercise 1 to show that $(R,i)$ exists.
Let $q:B\to B/R$ be the coequalizer (in the category of sets) of the two projections $r_1,r_2$ of the congruence $R$.
Exercise 3 Show that $r_1,r_2 : R\to B$ is the kernel pair of $q$.
Exercise 4 Use the axiom of choice to show that there are maps $t: B/R\to B$ such that $qt=1_{B/R}$, and then use Exercise 3 to show that there is a map $s: B\to R$ such that $r_1 s = 1_B$ and $r_2 s = tq$.
Exercise 5 Use Exercise 4 to show that for any functor $F:Sets \to \mathscr{C}$, $F(q)$ is the coequalizer of $F(r_1)$ and $F(r_2)$ (this is a standard fact about split forks).
Exercise 6 Use Exercise 5 to equip $B/R$ with a $T$-algebra structure and show that $q$ is the coequalizer of $r_1,r_2 : (R,i) \to (B,g)$.
Exercise 7 Let $e : (B,g) \to (C,h)$ be a morphism such that $ej=ek$. Explain why the kernel pair of $e$ is a congruence containing $\{(j(a),k(a))\,|\, a \in A\}$.
Exercise 8 Use Exercise 7 and the definition of $(R,i)$ to show that $q$ is the coequalizer of $j,k : (A,f)\to (B,g)$.
A: For a finitary monad, the coequalizer of $j,k$ is given by the canonical algebra structure on the coequalizer among sets of the two canonical functions $x,y:f\coprod f\to f\coprod g$, where you need to already know how to construct coproducts of algebras. The latter is reasonably complex business, but in particular cases is no problem. To read about the latter, it seems better to think in terms of algebraic theories and consult the first fifty pages or so of Adamek et l' said book here.
