quantum groups - Establishing the Co-Quasi- Triangular Structure of FRT Algebras

Let me try to rephrase everything in more modern terms.

First of all $Rinmathrm{End}(Votimes V)$. Then let me denote by $U$ the $mathbb{C}langle u^i_j rangle$-valued matrix with entries being $u^i_j$'s ($Uin{rm End}(V)langle u^i_j rangle$).

Now the two-sided ideal $mathcal J(R)$ is generated by the entries of the matrix
$$
M(U):=U_1U_2R-RU_2U_1.
$$
This expression lies in ${rm End}(Votimes V)langle u^i_j ranglecong{rm End}(V)otimes{rm End}(V)otimes mathbb{C}langle u^i_j rangle$. Viewing elements of this space as tensors with $3$ components some people rewrite it as follows:
$$
U_{1,3}U_{2,3}R_{1,2}-R_{1,2}U_{2,3}U_{1,3}.
$$

REMARK: it seems that Klimyk and Schmudgen chose a similar but different $M(U)$, but it is not very important.

Now we want to define $r:mathbb{C}langle u^i_j rangleotimes mathbb{C}langle u^i_j rangleto mathbb{C}$. It is sufficient to define it on generators, and the best way to organize the corresponding coefficients is to give an expression for
$$
r(Uotimes U)in{rm End}(V)otimes{rm End}(V)cong{rm End}(Votimes V).
$$
We define naively $r(Uotimes U):=R$.

We then need to check that the elements $r(M(U)otimes U)$ and $r(Uotimes M(U))$, lying in ${rm End}(V^{otimes3})$, vanish. Let me try with the second one.

First part:
$$
r(Uotimes U_1U_2R)=r(Uotimes U_1U_2)R_{2,3}=r(Uotimes U_1)r(Uotimes U_2)R_{2,3}=R_{1,2}R_{1,3}R_{2,3}
$$

Second part:
$$
r(Uotimes RU_2U_1)=R_{2,3}r(Uotimes U_2U_1)=R_{2,3}r(Uotimes U_2)r(Uotimes U_1)=R_{2,3}R_{1,3}R_{1,2}
$$

So there is no problem here since we find Yang-Baxter.

Let me try now with the first one.

First part:
$$
r(U_1U_2Rotimes U)=r(U_1U_2otimes U)R_{1,2}=r(U_1otimes U)r(U_2otimes U)R_{1,2}=R_{1,3}R_{2,3}R_{1,2}
$$

Second part:
$$
r(RU_2U_1otimes U)=R_{1,2}r(U_2U_1otimes U)=R_{1,2}r(U_2otimes U)r(U_1otimes U)=R_{1,2}R_{2,3}R_{1,3}
$$

There seems to be a problem here since we find an expression which is not Yang-Baxter:
$$
R_{1,3}R_{2,3}R_{1,2}-R_{1,2}R_{2,3}R_{1,3}
$$
But there is no: applying the flip $tau_{1,2}$ we get
$$
R_{2,3}R_{1,3}R_{2,1}-R_{2,1}R_{1,3}R_{2,3}
$$
which is an avatar of Yang-Baxter (as far as $R^{op}=R^{-1}$).