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A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying the unit and multiplication of a monad lawsAn algebra for a monad is an object $A$ equipped with $a:TA\to A$ such thatA morphism of algebras for a monad $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $fa=bT(f)$. Identities and composition obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^T$ and its forgetful functor $U^T(A,a)=A$.

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with induced monad $T=UF$, the Eilenberg-Moore comparison functor isThe functor $U$ is monadic when it has such a left adjoint and this comparison is an equivalence of categories; a stricter convention requires an isomorphism of categories over $\mathcal{C}$. We will identify which limit-lifting conclusion each convention supports in part (iii).

A monad on $\mathcal{C}$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfying, at every object $A$,These are the unit and associativity laws for the unit and multiplication of a monad.

An algebra for a monad is a pair $(A,a)$ with $a:TA\to A$ satisfyingA monad algebra morphism $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ with $fa=bTf$. Identities satisfy this condition, and composing two such maps preserves it by functoriality of $T$. The resulting Eilenberg-Moore category is denoted ${\mathcal{C}}^T$. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ sends $(A,a)$ to $A$ and leaves the underlying arrows unchanged.

Define the forgetful functor $G^{\mathbb{T}}(A,a)=A$, acting as the identity on underlying morphisms. Define the free algebra functor byThe monad laws make ${\mu }_X$ an algebra for a monad structure, and naturality of $\mu$ makes $T(h)$ an monad algebra morphism.

For $(A,a)$, define the transpose mapsThe map $\Psi (k)$ is an monad algebra morphism, because naturality of $\mu$ and the algebra associativity law giveFor $k:X\to A$, naturality of $\eta$ gives $\Phi \Psi (k)=aT(k){\eta }_X=a{\eta }_{A}k=k$. Conversely, if $h$ is an monad algebra morphism, thenPrecomposing by a map into $X$ or postcomposing by an monad algebra morphism respects both formulas. Thus the bijection is natural and $F^{\mathbb{T}}⊣G^{\mathbb{T}}$. This is the free-forgetful Eilenberg-Moore adjunction.

Its adjunction unit is the given ${\eta }_X$; its adjunction counit at $(A,a)$ is the monad algebra morphism $a:(TA,{\mu }_A)\to (A,a)$. Hence $G^{\mathbb{T}}{F}^{\mathbb{T}}=T$, and the induced multiplication is the underlying counit at $(TX,{\mu }_X)$, namely ${\mu }_X$. Therefore this adjunction induces exactly the original monad, including its unit and multiplication of a monad.

A monad consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1\Rightarrow T$ and $\mu :T^2\Rightarrow T$, the unit and multiplication of a monad, satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aTa$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ satisfies $ha=bTh$. These objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$; its composition works because $T$ is a functor.

For the list monad, $TX={\coprod }_{n\ge 0}{X}^n$ is the set of finite ordered lists, including the empty list. The map $Tf$ applies $f$ to each entry; ${\eta }_X(x)=[x]$ and ${\mu }_X$ concatenates a list of lists. The unit laws say that adding singleton brackets and then flattening changes nothing. Associativity says that flattening a list of lists of lists in either order produces the same ordered sequence. These descriptions also prove naturality.

If $a:TX\to X$ is an algebra for a monad, defineThe singleton law gives $a([x])=x$. Apply the algebra associativity law to $[[],[x]]$ and $[[x],[]]$ to obtain $e\ast x=x=x\ast e$. Applying it to $[[x,y],[z]]$ and $[[x],[y,z]]$ showsThus $(X,\ast ,e)$ is a monoid. Applying the same law to $[[x_1,\dots ,x_{n-1}],[x_n]]$ shows inductively that $a$ is necessarily ordered multiplication of its entries, with the empty product $e$.

Conversely, any monoid defines such a list-fold map $a$. The monoid unit proves $a\eta =1$, and associativity and the unit prove that multiplying flattened lists equals multiplying their individual products, including empty sublists. Hence $a\mu =aTa$. An algebra morphism preserves the empty-list value and two-entry-list values, so it is a monoid homomorphism; conversely a monoid homomorphism preserves every ordered product and is an algebra morphism. ThereforeThus list-monad algebras are monoids, with the identification also matching every morphism.

If the multiplication in the unit and multiplication of a monad $\mu$ is invertible, the two unit identities of the monadmake both $T\eta$ and $\eta T$ its inverse. Hence idempotence implies equality of the two unit insertions:This is the first implication in the characterization of an idempotent monad.

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.

On the monoidal category of modules over the commutative ring $k$, consider the monad $T=H{\otimes }_k-$ coming from the unit and multiplication of the bialgebra $H$. Its opmonoidal functor structure has comparison mapsHere and below Sweedler notation abbreviates the comultiplication $\Delta h=\sum h_{(1)}\otimes h_{(2)}$. The opmonoidal associativity and unit axioms are the coassociativity and counit laws of the coalgebra. The unit and multiplication of a monad are opmonoidal natural transformations because the bialgebra axioms sayThe Eilenberg-Moore category of this opmonoidal monad is the category of left $H$-modules: a monad-algebra map $H{\otimes }_{k}M\to M$ is exactly a unital associative action.

Applying the preceding construction gives the diagonal bialgebra action and the unit actionThe usual associators and unitors for the tensor product of modules are $H$-linear, and the underlying tensor product is exactly ${\otimes }_k$. The forgetful functor into $k$-modules is strict monoidal. The base need not be a field; the modules need be neither flat modules nor finitely generated modules.