= Solution
We use right <comodules> and write their <coactions> as $u\mapsto\sum u_{(0)}\otimes u_{(1)}$. A <coquasitriangular structure> gives the <braiding> on these <comodules>
$$
b_{M,N}(u\otimes v)=\sum v_{(0)}\otimes u_{(0)}\,\gamma(u_{(1)},v_{(1)}).
$$
In an arbitrary <symmetric monoidal category> this notation abbreviates a composite of the two coactions, the ambient symmetry, and $\gamma$; it does not assume that the objects have elements. The coquasitriangular axioms ensure that this composite is a comodule morphism, is invertible using the convolution inverse of $\gamma$, and satisfies the two hexagon laws. More explicitly, in scalar notation those laws come from
$$
\gamma(ab,c)=\sum\gamma(a,c_{(1)})\gamma(b,c_{(2)}),\qquad\gamma(a,bc)=\sum\gamma(a_{(1)},c)\gamma(a_{(2)},b),
$$
while the comodule-morphism condition is
$$
\sum\gamma(a_{(1)},b_{(1)})a_{(2)}b_{(2)}=\sum b_{(1)}a_{(1)}\gamma(a_{(2)},b_{(2)}).
$$
Unit normalizations give the unit constraints. These descriptions are identities of morphisms in the ambient symmetric category, with the displayed reordering carried by its symmetry.
Apply the preceding self-<braiding> result to the regular right <comodule> $(H,\Delta)$. Its <Yang–Baxter operator> is
$$
R(a\otimes b)=\sum b_{(1)}\otimes a_{(1)}\gamma(a_{(2)},b_{(2)}).
$$
Compose its braid equation with the three <counits>. Expanding $R_{23}R_{12}R_{23}$ and cancelling the leading counit factors gives
$$
\sum\gamma(a_{(1)},b_{(1)})\gamma(a_{(2)},c_{(1)})\gamma(b_{(2)},c_{(2)}).
$$
Expanding $R_{12}R_{23}R_{12}$ instead gives
$$
\sum\gamma(a_{(2)},b_{(2)})\gamma(b_{(1)},c_{(1)})\gamma(a_{(1)},c_{(2)}).
$$
They are equal by the <Yang–Baxter operator> equation. \b[This is the required scalar identity:]
$$
\boxed{\sum\gamma(a_{(1)},b_{(1)})\gamma(a_{(2)},c_{(1)})\gamma(b_{(2)},c_{(2)})=\sum\gamma(a_{(2)},b_{(2)})\gamma(b_{(1)},c_{(1)})\gamma(a_{(1)},c_{(2)}).}
$$
Indeed, after $\Delta^{\otimes3}$ the ordered factors are $a_{(1)},a_{(2)},b_{(1)},b_{(2)},c_{(1)},c_{(2)}$. The two ambient symmetries on the left produce the three pairings $(a_{(1)},b_{(1)})$, $(a_{(2)},c_{(1)})$, $(b_{(2)},c_{(2)})$. On the right, the central symmetry produces $a_{(1)},a_{(2)},b_{(2)},b_{(1)},c_{(1)},c_{(2)}$, yielding exactly the other three pairings. Hence the calculation identifies the actual morphisms requested, also when the category is not a category of vector spaces.
Back to article page
