I have a dumb question.

Given a Hopf algebra $H$, take an invertible element $J\in H\otimes H$ and define $\Delta^J=J^{-1} \Delta J$. This becomes a new coproduct when $J$ satisfies a certain condition. Such $J$ are called twists, and let's denote the new Hopf algbera obtained this way by $H^J$. Two twists are called gauge equivalent when there is an invertible element $u \in H$ such that $J_2=\Delta(v) J_1 (v^{-1}\otimes v^{-1})$. Then $H^{J_1}$ and $H^{J_2}$ are isomorphic.

Now, let's take $H=k^G$ (i.e. the function algebra on $G$ with point wise multiplication). I understand that the twists up to gauge equivalent are given by $H^2(G,k^\times)$.

What I don't understand is the following. Given $J\in H^2(G,k^\times)$, $(k^G)^J$ is still commutative. So it should still be a function algebra on a group, and the group in question should be $G$ itself, by considering the representations of the Hopf algebra. So I believe $(k^G)^J$ is isomorphic to $k^G$. (And there is a statement to this effect in Proposition 1.36.5 and Remark 1.36.6 in this pdf .)

But I can't construct an explicit isomorphism, mapping $\Delta^J$ to $\Delta$.

Could someone enlighten me?