 Solution

An opmonoidal monad on a monoidal category is a monad $(T,\eta ,\mu )$ whose endofunctor is an opmonoidal functor and whose unit and multiplication of a monad are opmonoidal natural transformations. Suppress only the canonical parentheses. Explicitly,The composite opmonoidal functor $T^2$ has tensor comparison $T_2(TX,TY)T(T_2(X,Y))$ and unit comparison $T_{0}T(T_0)$, explaining the last two equations.

For two algebras for a monad $(X,a)$ and $(Y,b)$, defineLet $c=(a\otimes b)T_2(X,Y)$. The unit law for a monad algebra follows at once from the opmonoidality of $\eta$:For the multiplication law, naturality of $T_2$, the algebra laws, and the opmonoidality of $\mu$ giveThe two unit-comparison equations above similarly make $(I,T_0)$ a algebra for a monad. If $f,g$ are morphisms of algebras for a monad, naturality of $T_2$ shows that $f\otimes g$ is an algebra morphism.

For a third algebra $(Z,d)$, the base associator is also an algebra morphism: its intertwining equation is precisely the opmonoidal associativity axiom, followed by $a\otimes b\otimes d$. The two base unitors are algebra morphisms by the opmonoidal unit axioms. Their pentagon and triangle commute because they commute after the faithful forgetful functor, and the lifted maps have exactly the same underlying morphisms.

The Eilenberg-Moore category is therefore monoidal, with these lifted constraints. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ preserves the tensor product, unit object and constraints exactly, so it is a strict monoidal functor.