Let $A$ be a Hopf algebra dually paired with a quasi-triangular Hopf algebra $B$. If $x$ is some fixed element of $A$, then we can define a linear map $$ P_x: A \to \mathbb{C} $$ by setting $$ P_x:a \mapsto \langle R,x \otimes a \rangle. $$
Let us take the case $A = SL_q(2)$, $B = U_q({\mathfrak sl}_2)$, and let $R$ be the standard universal $R$-matrix for $U_q({\mathfrak sl}_2)$, for which $$ \langle R, u^i_m \otimes u^j_n \rangle = R^{ij}_{mn} = q^{-\frac{1}{2}}.(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta (i-j)\delta_{in}\delta_{jm}), $$ where $\theta$ is the Heaviside symbol. If we take $x=u^k_l$, then $$ P_{u^k_l}(a) = \langle R, u^k_l \otimes a \rangle. $$ Now since $ab = qba$, we should have $$ P_{u^k_l}(u^1_1u^1_2) = q P_{u^k_l}(u^1_2u^1_1), \qquad \qquad \text{ for all } \quad k,l = 1,2. $$ However, $$ P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12}. $$ From the formula for $R^{ij}_{mn}$, we get that $$ P(u^1_1u^1_2) = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12} = R^{21}_{11}R^{11}_{12} + R^{21}_{21}R^{21}_{12} = 0.0 + q^{-\frac{1}{2}}.1.q^{-\frac{1}{2}}.(q-q^{-1}) = q^{-1}(q-q^{-1}). $$
On the other hand, we have $$ qP_{u^2_1}(u^1_2u^1_1) = \langle R,u^2_1 \otimes u^1_2u^1_1 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_2 \rangle \langle R,u^z_1 \otimes u^1_1 \rangle =\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11}. $$ From the formula for $R^{ij}_{mn}$, we now get that $$ qP_{u^2_1}(u^1_2u^1_1) = q\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11} = qR^{21}_{12}R^{11}_{11} + qR^{21}_{22}R^{21}_{11} = q.q^{-\frac{1}{2}}.(q-q^{-1}).q^{-\frac{1}{2}}.q + q.0.0 = q(q-q^{-1}). $$
Thus, the two results are not equal, but instead differ by a factor of $q^2$. A similar problem arises for the action of $P_{u^2_1}$ on $bd - qdb$. We get $$ P_{u^2_1}(u^1_2u^2_2) = q^{-1}(q-q^{-1}), $$ whereas $$ qP_{u^2_1}(u^2_2u^1_2) = q(q-q^{-1}). $$
I've checked and rechecked everything very carefully but can't seem to spot my error. Can anyone see what is going wrong here?