A very easy question I can't seem to answer: For a universal rform on a coquasitriangular Hopf algebra why is $r(a \otimes 1) = r(1 \otimes a) = \epsilon(a)$?

$\begingroup$ John, To my mind, coquasitriangular Hopf algebras and quasitriangular Hopf algebras are essentially equivalent notions (the only real difference is that comodules are automatically locally finite, whereas modules need to be explicitly declared as such for $R$matrix to converge in many common examples). In particular, any statement about the former is just the adjoint of the latter. So looking at Proposition 2 in Chapter 8, where the analogous statement is proven for $R$matrices, will show you the proof for $r$forms. It looks like you will do some trick with writing $r=r\cdot 1$. $\endgroup$ – David Jordan Aug 18 '10 at 18:32

$\begingroup$ I'll check back in a day or so and see if you want more details. $\endgroup$ – David Jordan Aug 18 '10 at 18:32

$\begingroup$ Sorry Proposition 2, Chapter eight, in Klymik and Schmudgen's text. I seem to have gone from your previous thread to this one, and carried the same standing references =] $\endgroup$ – David Jordan Aug 18 '10 at 23:52
Like David said, the proof is almost identical to the earlier one for $R$matrices:
$r(x\otimes 1) = r\circ (id\otimes\mu)(x\otimes 1\otimes 1) = (r_{13}\ast r_{12})(x\otimes 1\otimes1)= \sum r(x'\otimes 1)r(x''\otimes 1) = (r \ast r)(x\otimes 1).$
Since $r$ is invertible, $r(x\otimes 1)=\epsilon(x)$.