Let $H$ be a bialgebra and ${}_H^H YD$ the category of Yetter-Drinfeld modules over $H$. It is said that Yetter-Drinfeld condition is equivalent to the condition of $H$-action commutes with braiding.
Suppose that $$ \Psi: v\otimes w \mapsto v_{(-1)}.w \otimes v_{(0)} $$ is a morphism of $H$-modules. Then for $h \in H$, we have $$ \Psi(h.(v \otimes w)) = h.(\Psi(v \otimes w)). $$ The right hand side is $$ h.( (v_{(-1)}.w )\otimes v_{(0)} ) = h_{(1)}.( (v_{(-1)}.w ) ) \otimes h_{(2)}.( v_{(0)} ). $$ The left hand side is $$ \Psi( h_{(1)}.v \otimes h_{(2)}.w ) = ( h_{(1)}.v )_{(-1)}.( h_{(2)}.w ) \otimes h_{(1)}.v )_{(0)}. $$ But the equation $$ h_{(1)}.( (v_{(-1)}.w ) ) \otimes h_{(2)}.( v_{(0)} ) = ( h_{(1)}.v )_{(-1)}.( h_{(2)}.w ) \otimes h_{(1)}.v )_{(0)} \quad (1) $$ is slightly different from the Yetter-Drinfeld condition $$ h_{(1)} v_{(-1)} \otimes h_{(2)}.v_{(0)} = ( h_{(1)}.v )_{(-1)} h_{(2)} \otimes (h_{(1)}.v)_{(0)}. $$ Since $V$ is an $H$-module, (1) is the same as $$ ( h_{(1)} v_{(-1)} ).w \otimes h_{(2)}.( v_{(0)} ) = ( ( h_{(1)}.v )_{(-1)} h_{(2)} ).w \otimes ( h_{(1)}.v )_{(0)}. $$ But how to remove $w$ in the above equation?
Thank you very much.