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.



How to show that Yetter-Drinfeld condition is equialent to $\Psi$ is a braiding
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](https://mathoverflow.net/questions/242715/reference-request-compatibility-conditions-of-four-versions-of-yetter-drinfeld)). 


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.


