isomorphic to the right center ${\mathcal{Z}}_r({}_H{\mathcal{M}})$.

Take $(M, \tau_{-,M})\in {\mathcal{Z}}_r({}_H{\mathcal{M}})$. We know from [Proposition \[pr:1.0.1\]]{} that $(M, \sigma_{M,-}=\tau^{-1}_{-,M})\in {\mathcal{Z}}_l({}_H{\mathcal{M}})$. Take the corresponding left-left Yetter-Drinfeld $(M,\lambda)$, as in [Theorem \[th:2.4\]]{}, and define $\rho:\ M\to M\ot H$ by $${\label{eq:2.6.0}}
\rho(m)=m_{[0]}\ot m_{[1]}=m_{[0]}\ot S^{-1}(m_{[-1]}).$$ It follows from [(\[eq:2.3.1\])]{} that $\rho(m)\in M\ot_t H$. The coassociativity of $\rho$ follows immediately from the coassociativity of $\lambda$ and the anti-comultiplicativity of $S^{-1}$. Also $$\varepsilon(m_{[1]})m_{[0]}=\varepsilon(S^{-1}(m_{[-1]}))m_{[0]}
=\varepsilon(m_{[-1]})m_{[0]}=m.$$ From [(\[eq:2.4.2\])]{}, it follows that $${\label{eq:2.6.3}}
\tau_{V,M}(1_{(1)}v\ot 1_{(2)}m)=m_{[0]}\ot m_{[1]}v.$$ In particular, $\tau_{M,H}(1_{(1)}\ot 1_{(2)}m)=\rho(m)$, and the fact that $\tau_{M,H}$ is left $H$-linear implies [(\[eq:2.6.2\])]{}. Hence $(M,\rho)$ is a left-right Yetter-Drinfeld module.\
Conversely, if $(M,\rho)$ is a left-right Yetter-Drinfeld module, then $(M,\tau_{-,M})$, with $\tau$ defined by [(\[eq:2.6.3\])]{} is an object of ${\mathcal{Z}}_r({}_H{\mathcal{M}})$.

[\[co:2.7\]]{} Let $M$ be a $k$-module with a left $H$-action and a right $H$-coaction. Then $M$ is a left-right Yetter-Drinfeld module if and only if $${\label{eq:2.7.1}}
\rho(hm)=h_{(2)}m_{[0]}\ot h_{(3)}m_{[1]}S^{-1}(h_{(1)}).$$

[\[co:2.7a\]]{} Let $M$ be a left-right Yetter-Drinfeld module. For all $y\in H_s$, $z\in H_t$ and $m\in M$, we have that $${\label{eq:2.7a.1}}
\rho(ym)=m_{[0]}\ot ym_{[1]}~~;~~\rho(zm)=m_{[0]}\ot m_{[1]}S^{-1}(z).$$

[\[co:2.7b\]]{} Let $M$ be a left-right Yetter-Drinfeld module. Then $${\label{eq:2.7b.1}}
1_{(2)}m_{[0]}\ot m_{[1]}S^{-1}(1_{(1)})=\rho(m),$$ for all $m\in M$.

Apply $S^{-1}\ot M$ to $\lambda(m)=1_{(1)}S(m_{[1]})\ot 1_{(2)}m_{[0]}$.

[\[co:2.8\]]{} The category ${}_H{\mathcal{YD}}^H$ is a braided monoidal category, 