The left dual of a finite-dimensional Hopf module

Let $H$ be a bialgebra over a field $k$, let $S:H\to H$ be an antipode satisfying

$\sum S(h_{(1)})h_{(2)}=\epsilon (h)1_H=\sum h_{(1)}S(h_{(2)}),$

and let $V$ be a finite-dimensional left $H$-module. The algebraic dual $V^{\ast }$ is the vector space of linear functionals $V\to k$ from Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$. Give it the candidate left $H$-action

$(h\cdot f)(v):=f(S(h)\cdot v)\mathrm{.}$

Set $\mathrm{ev}:V^{\ast }\otimes V\to k$, $f\otimes v\mapsto f(v)$, and set $\mathrm{coev}:k\to V\otimes V^{\ast }$, $1\mapsto {\sum }_i{e}_i\otimes e^i$ for any finite basis $(e_i)$ and its dual basis $(e^i)$. The tensor in this formula is basis-independent by Finite tensor duality and basis-independent coevaluation. In this page's convention, “left dual” means these displayed orientations $V^{\ast }\otimes V\to k$ and $k\to V\otimes V^{\ast }$. Some sources use “right dual” for the same orientations; the formulas here fix the convention. The next theorem proves the candidate action and maps give a left dual, without assuming $S$ bijective.