Zoran Skoda 1(x)1-Delta(1)

For any element $h$ in a weak bialgebra $H$, denote

$$I(h) = \epsilon(1_{(2)}h)1_{(1)}\otimes 1 - 1\otimes \epsilon(1_{(1)}h)1_{(2)} = \bar\Pi^L(h)\otimes 1 - 1\otimes\Pi^L(h).$$

We claim that $\Delta(1)\cdot I(h) = 0$. Indeed,

$$\array{ 1_{(1)}\epsilon(1_{(2')}h)1_{(1')}\otimes 1_{(2)} &=& \epsilon((1_{(1)} 1_{(1')})_{(2)}) (1_{(1)} 1_{(1')})_{(1)}\epsilon(1_{(2')}h)1_{(1')}\otimes 1_{(2)}\\ &=& \epsilon(1_{(2)}1_{(2')})\epsilon(1_{(3')}h)1_{(1)}1_{(1')}\otimes 1_{(3)}\\ &=& \epsilon(1_{(2)}1_{(2')}h)1_{(1)}1_{(1)'}\otimes 1_{(3)}\\ &=& \epsilon(1_{(2)} h) 1_{(1)} \otimes 1_{(3)}\\ &=& 1_{(1)}\otimes\epsilon(1_{(2)}h)1_{(3)}\\ &=& 1_{(1)}\otimes\epsilon(1_{(2)}1_{(1')}h) 1_{(3)}1_{(2')}\\ &=& 1_{(1)}\otimes\epsilon(1_{(2)}1_{(2')})\epsilon(1_{(1')}h)1_{(3)}1_{(3')}\\ &=& 1_{(1)}\otimes 1_{(2)}\epsilon(1_{(1')}h)1_{(2')} }$$

It follows that $(1\otimes 1 - \Delta(1)) I(h) = I(h) - 0 = I(h)$. In particular, for every $h$, the element $I(h)$ belongs to the right principal ideal generated by $1\otimes 1 - \Delta(1)$.