Is the tensor product of two Yetter-Drinfeld modules a Yetter-Drinfeld module? Let $U,V$ be two Yetter-Drinfeld modules over a bialgebra $H$. Is $U \otimes V$ a Yetter-Drinfeld modules over $H$? Thank you very much.
 A: Yes. Yetter-Drinfeld modules over a bialgebra form a tensor category.
You need to check that the following holds

$\left(h_{(1)}\rhd(v\otimes w)\right)^{(-1)}h_{(2)}\otimes \left(h_{(1)}\rhd(v\otimes w)^{(0)}\right)=\\=h_{(1)}\left(v\otimes w\right)^{(-1)}\otimes h_{(2)}\rhd\left(v\otimes w\right)^{(0)}$

where $\Delta h=h_{(1)}\otimes h_{(2)}$ and $\delta(v\otimes w)=(v\otimes w)^{(-1)}\otimes (v\otimes w)^{(0)}$ is a coaction on tensor product (You should know that the simultaneous modules and comodules form tensor category, hence it is well-defined).
A: We have 
\begin{align}
& h.(v \otimes w) = h_{(1)}.v \otimes h_{(2)}.w, \\
& \delta(v \otimes w) = v_{(-1)} w_{(-1)} \otimes v_{(0)} \otimes w_{(0)}.
\end{align}
Therefore
\begin{align}
& ( h_{(1)}.(v \otimes w) )_{(-1)} h_{(2)} \otimes ( h_{(1)}.(v \otimes w) )_{(0)} \\
& = ( h_{(1)}.v \otimes h_{(2)}.w )_{(-1)} h_{(3)} \otimes ( h_{(1)}.v \otimes h_{(2)}.w )_{(0)} \\
& = ( h_{(1)}.v )_{(-1)} ( h_{(2)}.w )_{(-1)} h_{(3)} \otimes ( h_{(1)}.v)_{(0)} \otimes (h_{(2)}.w )_{(0)} \\
& = ( h_{(1)}.v )_{(-1)}  h_{(2)} w_{(-1)} \otimes ( h_{(1)}.v)_{(0)} \otimes h_{(3)}.w_{(0)} \\
& =  h_{(1)} v_{(-1)} w_{(-1)} \otimes  h_{(2)}.v_{(0)} \otimes h_{(3)}.w_{(0)} \\
& =  h_{(1)} (v \otimes w)_{(-1)} \otimes  h_{(2)}.(v \otimes w)_{(0)}.
\end{align}
