# Subfunctor of internal Hom

Let $$\mathcal{H}$$ be a Hopf algebra over $$\mathbb{C}$$. Let $$\textrm{mod}_\mathcal{H}$$ be the monoidal abelian category of finite-dimensional modules over $$\mathcal{H}$$. Fix $$X\in\textrm{Obj}(\textrm{mod}_\mathcal{H})$$. We know that the functor $$(-\otimes X):\textrm{mod}_\mathcal{H}\rightarrow\textrm{mod}_\mathcal{H}$$ is left adjoint to functor $$\textrm{Hom}_{\mathbb{C}}(X,-):\textrm{mod}_\mathcal{H}\rightarrow\textrm{mod}_\mathcal{H}.$$

Moreover, $$\textrm{Hom}_{\mathcal{H}}(X,Y)$$ is a submodule of $$\textrm{Hom}_{\mathbb{C}}(X,Y)$$, for all $$Y\in\textrm{Obj}(\textrm{mod}_\mathcal{H})$$. It seems that $$\textrm{Hom}_{\mathcal{H}}(X,-)$$ defines a subfunctor of $$\textrm{Hom}_{\mathbb{C}}(X,-)$$, since composition of morphisms of modules is a morphism of modules, as well. Am I right about that? If this is the case, does $$\textrm{Hom}_{\mathcal{H}}(X,-)$$ have any (left or right) adjoints? Does anyone know of any references that deal with this? Thanks in advance for answers.

• I don't a precise reference apart elementary module theory, but I think $-\otimes_{\cal H} X \dashv \hom_{\cal H}(X,-)$... (where the left adjoint is a suitable coequalizer obtained from $-\otimes_{\mathbb C}X$). – Fosco Nov 25 '19 at 8:38
• @Fosco Doc, you cannot tensor two right $H$-modules unless you turn left at Albuquerque, first... – Bugs Bunny Dec 17 '19 at 6:57

Doc, let us discern what catechism you are coddling. We know that $$hom_H(X,Y)=hom_{\mathbb C}(X,Y)^H.$$ Thus, your basic functor does not take values in the category you write. Instead, $$hom_H(X,-)=mod_H \rightarrow mod_{\mathbb C}.$$ Its left adjoint functor you know and cherish already: $$-\otimes X= mod_{\mathbb C} \rightarrow mod_{H}.$$ If you impertinently persist with a functor to $$mod_H$$, you are taking a composition $$mod_H \xrightarrow{hom_H(X,-)} mod_{\mathbb C}\xrightarrow{T} mod_{H}$$ where the functor $$T$$ treats a vector space as a trivial $$H$$-module. The left adjoint to $$T$$ is the coinvariant functor $$(-)_H: mod_H \rightarrow mod_{\mathbb C}.$$ Blending these two judicious remarks together, we conclude that the left adjoint to $$hom_H(X,-)=mod_H \rightarrow mod_{H}$$ is $$(-)_H\otimes X: mod_H \rightarrow mod_{H}.$$ Gee, ain’t I a stinker?