Implication Details

Assumptions: finitary algebraic, pointed

Reason: We have a constant in every algebra, let us denoted it by $0$ . Then the projection $A \times B \to A$ is clearly surjective, hence an epimorphism. To show that $A \sqcup_{A \times B} B$ is trivial, let $R$ be an algebra which admits homomorphisms $f : A \to R$ , $g : B \to R$ such that $f(p_1(a,b)) = g(p_2(a,b))$ for all $(a,b) \times A \times B$ . This means $f(a) = g(b)$ . In particular, $f(a) = g(0) = 0$ . Likewise, $g(b) = 0$ , and we are done.