 Solution

With the right-comodule braiding convention in part (a), the defining formula isConsequently the scalar form on generators is the following table, with row the first argument and column the second:For example, the extra $e_2\otimes e_1$ component of $C(e_2\otimes e_1)$ is $r(b,c)=q^{-1/2}(q-q^{-1})$. The other off-diagonal component gives $r(d,a)=q^{-1/2}$.

For arbitrary elements of the FRT bialgebra, extend this table by bilinearity, the unit rule $r(1,z)=r(z,1)=\epsilon (z)$, andThese are an explicit recursive algorithm for any two words in the generators, so the table determines the entire cobraiding. The braid equation for $C$ ensures the algorithm annihilates the defining coefficient relations in either argument: moving one crossing past two successive crossings gives precisely the two reductions of a quadratic colinearity relation. Thus it descends from the free algebra over a field to the FRT bialgebra. For the convolution inverse, use the inverse of $R=\tau C$, where $\tau$ flips the tensor factors: its coefficients are $\bar{r}(t_{ki},t_{lj})=(R^{-1}{)}_{kl,ij}$. Equivalently, in the flipped coefficient formula use $\tau C^{-1}\tau$, not $C^{-1}$ alone. This gives $\bar{r}(a,a)=\bar{r}(d,d)=q^{-1/2}$, $\bar{r}(a,d)=\bar{r}(d,a)=q^{1/2}$, $\bar{r}(b,c)=-q^{1/2}(q-q^{-1})$, and all other generator values zero. This specifies the coquasitriangular structure completely, not just its values on degree-one elements.