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

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.