Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 122 / 2

An opmonoidal functor $F:(\mathcal{C},\otimes ,I)\to (\mathcal{D},\otimes ,J)$ consists of a functor, a natural transformationand a morphism $F_0:FI\to J$. These maps need not be invertible. If $\alpha ,\lambda ,\rho$ are the respective associators and unitors, its axioms areThe first equation has domain $F((X\otimes Y)\otimes Z)$ and codomain $FX\otimes (FY\otimes FZ)$, which fixes the direction of every arrow. This is also called a colax monoidal functor.

An opmonoidal natural transformation $\tau :F\Rightarrow G$ between opmonoidal functors is a natural transformation satisfyingIt respects both the tensor and unit comparison maps. These are commutative diagrams and a unit commutative diagram; the arrow directions are opposite to those for a lax monoidal functor.

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.

We use the left internal-hom convention in which $[M,N]$ is right adjoint to $-{\otimes }_{k}M$, so its evaluation morphism has domain $[M,N]{\otimes }_{k}M$. For a Hopf algebra over the commutative ring $k$, take the full $k$-moduleand give it the internal hom for modules over a Hopf algebraThe antipode is an antihomomorphism, $S(hg)=S(g)S(h)$, soThe unit acts identically. Thus this is a left $H$-module.

The evaluation morphism is $H$-linear. Using the diagonal bialgebra action and the antipode identity,Now let $q:L{\otimes }_{k}M\to N$ be $H$-linear and define its ordinary curried map $\hat{q}(\ell )(u)=q(\ell \otimes u)$. To see that it too is $H$-linear, computeConversely, any $H$-linear map $L\to [M,N]$ uncurries to an $H$-linear map by the already established linearity of evaluation. The ordinary tensor–hom adjunction therefore restricts to a natural bijectionPostcomposition by an $H$-linear map makes $[M,-]$ a functor. This proves that the category of left $H$-modules is a left closed monoidal category. All modules are allowed; no inverse antipode or finite-dimensional dual is required. Stating the tensor–hom convention explicitly avoids confusing this construction with the closure on the opposite side.