A quasi-bialgebra need not be quasi-Hopf

The monoid laws make multiplication associative and unital. On basis elements, $\mathrm{\Delta }(mn)=mn\otimes mn=(m\otimes m)(n\otimes n)$, $\mathrm{\Delta }(1)=1\otimes 1$, $\epsilon (mn)=1=\epsilon (m)\epsilon (n)$, and coassociativity and the counit equations hold. Thus $H$ is a bialgebra and, with $\mathrm{\Phi }=1$, a quasi-bialgebra.

Suppose $(S,\alpha ,\beta )$ is a quasi-antipode. With $\mathrm{\Phi }=1$, its normalization equations give $\beta \alpha =1=\alpha \beta$, so ${\alpha }^{-1}=\beta$. Define $S^{\mathrm{\prime }}(h)={\alpha }^{-1}S(h)\alpha$. The quasi-antipode equations imply $\sum S^{\mathrm{\prime }}(h_{(1)})h_{(2)}=\epsilon (h)1$ and $\sum h_{(1)}{S}^{\mathrm{\prime }}(h_{(2)})=\epsilon (h)1$; thus $S^{\mathrm{\prime }}$ is an ordinary Hopf antipode of the bialgebra in step 1.1.

The basis element $e$ is group-like: $\mathrm{\Delta }(e)=e\otimes e$ and $\epsilon (e)=1$. The antipode equations in [F3] would make $e$ invertible, since $S^{\mathrm{\prime }}(e)e=1=e{S}^{\mathrm{\prime }}(e)$. But an invertible idempotent is $1$ (multiply $e^2=e$ by $e^{-1}$), contradicting $e\mathrm{\ne }1$. Hence no quasi-antipode exists, refuting the statement.