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For a left module over a Hopf algebra, a vector $v$ is invariant when $hv=\epsilon (h)v$. In the internal hom for modules over a Hopf algebra, invariant maps are exactly the R-module homomorphisms for the acting algebra.

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.

The induced internal hom for modules over a Hopf algebra on ${\mathrm{Hom}}_k(H,H)$ iswhere both occurrences of $H$ have their left regular action. If $T$ is a morphism of left $H$-modules, thenSo every such morphism is an invariant vector of a module over a Hopf algebra.

For the converse, a calculation valid for every $k$-linear $T$ isIf $T$ is invariant, its left-hand side is instead $\sum \epsilon (h_{(1)})T(h_{(2)}x)=T(hx)$. Hence $T(hx)=hT(x)$ for all $h,x$, which is precisely the R-module homomorphism condition.

In fact an endomorphism of the left regular module is determined by $a=T(1)$ and has $T(x)=xa$. Conversely each right multiplication map is left $H$-linear. Thus the invariant space has the explicit descriptionAs a vector space it is isomorphic to $H$, and under composition its algebra is the opposite algebra $H^{\mathrm{op}}$. No dimension or semisimplicity hypothesis is needed.