How to show that Yetter-Drinfeld condition is equialent to $\Psi$ is a braiding Let $H$ be a Hopf algebra and $V$ a right $H$-module and right $H$-comodule. The module $V$ is a Yetter-Drinfeld module over $H$ if and only if
\begin{align}
( v \triangleleft h_{(2)} )_{(0)}  \otimes h_{(1)} ( v \triangleleft h_{(2)} )_{(1)} = ( v_{(0)} \triangleleft h_{(1)} )  \otimes  v_{(1)} h_{(2)}.  \quad (1)
\end{align}
Let $\Psi: U \otimes W \to W \otimes U$ be a braiding given by
\begin{align}
\Psi(u \otimes w) = w_{(0)} \otimes ( u \triangleleft w_{(1)} ).
\end{align}
Then
\begin{align}
\Psi_{12} \Psi_{23} \Psi_{12} = \Psi_{23} \Psi_{12} \Psi_{23}.
\end{align}
The algebra $H$ is in the right-right Yetter-Drinfeld category $YD^H_H$. Suppose that $V \in  YD^H_H$. Let $\Psi: V \otimes H \to H \otimes V$ be a braiding. Let $u \in H, v \in V, h \in H$. Then
\begin{align}
& \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes h) \\
& = h_{(0)} \otimes (v_{(0)} \triangleleft h_{(1)}) \otimes ( u \triangleleft v_{(1)} h_{(1)} )  \quad (2) \\
& \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes h) \\
& = h_{(0)} \otimes (v \triangleleft h_{(2)})_{(0)} \otimes ( u \triangleleft h_{(1)} ( v \triangleleft h_{(2)} )_{(1)} ). \quad (3)
\end{align}
How to show that $\Psi$ is a braiding is equavalent to the compatibility condition (1)? My problem is: how to remove $u$ in (2) and (3)? Thank you very much.
 A: The YD-condition is not equivalent to $\Psi$ being a braiding. The condition (1) is related to $\Psi$ being a morphism of $H$-modules (see the more detailed answer to your other question Reference request: compatibility conditions of four versions of Yetter-Drinfeld modules). 
The Quantum Yang-Baxter equation is a consequence of the action and the coaction conditions (and the YD-condition), which are of course part of the YD-structure. These give the two hexagon axioms of a braiding (see standard references on braided monoidal categories for these conditions).
$$\Psi_{V,W\otimes Z}=(id_W\otimes \Psi_{V,Z})(\Psi_{V,W}\otimes id_Z),$$
$$\Psi_{V\otimes W, Z}=(\Psi_{V,Z}\otimes id_W)(id_V\otimes \Psi_{W,Z}).$$
However, the two hexagon axioms are stronger than the quantum Yang-Baxter equation 
$$\Psi_{12}\Psi_{23}\Psi_{12}=\Psi_{23}\Psi_{12}\Psi_{23}.$$
This is proved by observing that
$\Psi_{12}\Psi_{23}=\Psi_{V,V\otimes V}.$ This allows us to rewrite the right right hand side as $\Psi_{V,V\otimes V}(id_V\otimes \Psi_{V,V})$, but by naturality of the braiding applied to $\Psi_{V,V}$ (which is a morphism of $H$-modules by the YD-condition (1) and hence naturality applies) this equals $( \Psi_{V,V}\otimes id_V)\Psi_{V,V\otimes V}$ which can be translated into the left hand side by the same reasoning.
A: We can also prove that Yetter-Drinfeld condition implies that $\Psi$ is a braiding as follows.
Let $u, v, w \in V $. Then
\begin{align}
& \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes w) \\
& =  \Psi_{12} \Psi_{23}(  (u_{(-1)}.v) \otimes u_{(0)} \otimes w ) \\
& =  \Psi_{12}  (  (u_{(-1)}.v) \otimes (u_{(0)})_{(-1)}.w \otimes (u_{(0)})_{(0)} ) \\
& =  ( u_{(-1)}.v )_{(-1)}.( ( (u_{(0)})_{(-1)} ).w ) \otimes (u_{(-1)}.v)_{(0)} \otimes ( u_{(0)} )_{(0)}  \\
& =  ( (u_{(-1)})_{(1)}.v )_{(-1)}.(  (u_{(-1)})_{(2)}.w ) \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} \\
& =  ( ((u_{(-1)})_{(1)}.v )_{(-1)} (u_{(-1)})_{(2)} ).w \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)}, 
\end{align}
\begin{align}
& \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes w) \\
& = \Psi_{23} \Psi_{12} (u \otimes v_{(-1)}.w \otimes v_{(0)}) \\
& = \Psi_{23} ((u_{(-1)}v_{(-1)}).w \otimes u_{(0)} \otimes v_{(0)}) \\
& = (u_{(-1)} v_{(-1)}).w \otimes (u_{(0)})_{(-1)}.v_{(0)} \otimes ( u_{(0)} )_{(0)} \\
& = ( (u_{(-1)})_{(1)} v_{(-1)}).w \otimes (u_{(-1)})_{(2)}.v_{(0)} \otimes u_{(0)}.      
\end{align}
Therefore the Yetter-Drinfeld condition 
\begin{align}
& (h_{(1)}.v )_{(-1)}   h_{(2)} \otimes (h_{(1)}.v)_{(0)}\\
& = h_{(1)} v_{(-1)} \otimes h_{(2)}.v_{(0)}.
\end{align}
implies that (every $u_{(-1)}$ is some $h \in H$)
\begin{align}
& ((u_{(-1)})_{(1)}.v )_{(-1)} (u_{(-1)})_{(2)} \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} \\
& = (u_{(-1)})_{(1)} v_{(-1)} \otimes (u_{(-1)})_{(2)}.v_{(0)} \otimes u_{(0)}.
\end{align}
Therefore
\begin{align}
& \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes w) = \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes w).
\end{align}
