The dual of a finite-dimensional Hopf algebra is a Hopf algebra

Since $H$ is finite-dimensional, [F3] identifies $(H\otimes H{)}^{\ast }$ with $H^{\ast }\otimes H^{\ast }$; thus for each $\varphi \in H^{\ast }$, the transpose $m^{\ast }(\varphi )$ is a tensor ${\sum }_i{\varphi }_i\otimes {\psi }_i$ characterized by $\varphi (ab)={\sum }_i{\varphi }_i(a){\psi }_i(b)$. The same finite tensor duality for $H\otimes H$ permits transposing every structural identity in [F1] and [F2].

The convolution product on $H^{\ast }$ is associative: evaluating $(\varphi \psi )\theta$ and $\varphi (\psi \theta )$ on $h$ gives respectively $\sum \varphi (h_1)\psi (h_2)\theta (h_3)$ through the two iterated coproducts of $H$, which agree by coassociativity; the counit identities make $\epsilon$ its left and right unit.

The coproduct $m^{\ast }$ on $H^{\ast }$ is coassociative because evaluating $(m^{\ast }\otimes \mathrm{i}\mathrm{d})m^{\ast }(\varphi )$ and $(\mathrm{i}\mathrm{d}\otimes m^{\ast })m^{\ast }(\varphi )$ on $a\otimes b\otimes c$ gives $\varphi ((ab)c)$ and $\varphi (a(bc))$, equal by associativity of $H$; evaluation at $1_H$ is a two-sided counit by $1_{H}a=a=a{1}_H$.

The coproduct on $H^{\ast }$ preserves convolution products: for $\varphi ,\psi \in H^{\ast }$ and $a,b\in H$, ${\mathrm{\Delta }}_{H^{\ast }}(\varphi \psi )(a\otimes b)=(\varphi \psi )(ab)=\sum \varphi (a_1{b}_1)\psi (a_2{b}_2)=({\mathrm{\Delta }}_{H^{\ast }}(\varphi ){\mathrm{\Delta }}_{H^{\ast }}(\psi ))(a\otimes b),$ where the second equality uses that ${\mathrm{\Delta }}_H$ is an algebra map. Finite tensor duality makes equality on all $a\otimes b$ equality of tensors.

The unit ${\epsilon }_H$ is group-like for ${\mathrm{\Delta }}_{H^{\ast }}$, since ${\epsilon }_H(ab)={\epsilon }_H(a){\epsilon }_H(b)$; the counit $\varphi \mapsto \varphi (1_H)$ preserves convolution products because ${\mathrm{\Delta }}_H(1_H)=1_H\otimes 1_H$. Thus the maps in the Statement make $H^{\ast }$ a bialgebra.

Define $S^{\ast }(\varphi )=\varphi \circ S$. If ${\mathrm{\Delta }}_{H^{\ast }}(\varphi )={\sum }_i{\varphi }_i\otimes {\psi }_i$, then for $h\in H$, $(S^{\ast }({\varphi }_{(1)}){\varphi }_{(2)})(h)={\sum }_{(h),i}{\varphi }_i(S(h_1)){\psi }_i(h_2)={\sum }_{(h)}\varphi (S(h_1)h_2)=\epsilon (h)\varphi (1_H)\mathrm{.}$ The last equality is the left antipode equation in [F1]; this is the left antipode equation for $S^{\ast }$ in $H^{\ast }$.

Likewise, $({\varphi }_{(1)}{S}^{\ast }({\varphi }_{(2)}))(h)={\sum }_{(h),i}{\varphi }_i(h_1){\psi }_i(S(h_2))={\sum }_{(h)}\varphi (h_{1}S(h_2))=\epsilon (h)\varphi (1_H),$ using the right antipode equation in [F1]. Both convolution equations hold, so $S^{\ast }$ is an antipode on $H^{\ast }$, and no step uses $S^{-1}$.