It is well known that many constructions involving bialgebras extends to monoidal categories, and often becomes more natural in that framework.

If one cares about the category of finite dimensional modules, then the existence of an antipode on a given bialgebra is equivalent to the rigidity of its category of finite dimensional modules.

I wonder what happens if one cares instead about the category of all modules on some Hopf algebra $H$. At first I thought existence of an antipode was equivalent to existence of internal Hom's in $H-mod$, which clearly generalizes being rigid: if $V,W$ are $H$-modules, then the internal Hom from $V$ to $W$ is the space of linear maps between those, equipped with the adjoint action (which does of course involve the antipode). But it turns out that modules over a bialgebra also have internal Hom's for abstract non-sense reasons: the functors "tensoring with $V$" on either side are cocontinuous so they have right adjoints. WHat is true is that a bialgebra is an Hopf algebra iff the fiber functor to Vect is compatible with internal Hom's, and in fact the same holds for Hopf monads.

But this characterization does involve the fiber functor, which I don't want.

Note that if one looks at categories of comodules instead, then it is generated under filtered colimits by the finite dimensional ones, and those are dualizable iff we start with something Hopf. This is the "correct" generalization of being rigid in the large setting.

What is special about the category of all modules over an Hopf algebra, which doesn't involve the fiber functor ?