# Does one of the hexagon identities imply the other one?

Suppose we have a monoidal category equipped with additional data that almost makes it a braided monoidal category except that only one of the hexagon identities is satisfied. Can we then prove the other hexagon identity? If not, is there an explicit counterexample and can we prove the other identity under the additional condition that the braiding is symmetric?

Consider the category of $A-$graded vector spaces (here $A$ is an abelian group) with obvious tensor product and trivial associator. Then each isomorphism $a\otimes b\to b\otimes a$ can be specified as a nonzero complex number $B(a,b)$. Now the first hexagon axiom says that the function $B(a,b)$ is linear in the first variable and the second axiom says that $B(a,b)$ is linear in the second variable. Clearly these conditions are independent. The symmetric braiding would correspond to skew-symmetric function $B(a,b)$ and linearity in one variable would imply bilinearity.
Recall first that the free braided monoidal category on one generator is the category of braids. Recall also that to every braid you can associate a "crossing number": every overcrossing counts as +1, and every undercrossing as -1. Given a braid $B$, let me write $\chi(B)$ for this crossing number, since I spell is $\chi$rossing. Anyway, there's another braided category where I multiply every braid $B$ by $q^{\chi(B)}$, for $q \neq 0$ some "variable". What I mean by this is that I'm taking the same monoidal category, but a different braiding, where the basic crossing is rescaled by $q$. Then the crossing of $n$ strands over $m$ strands is rescaled by $q^{nm}$. I guess implicitly I'm taking my category over $k[q,q^{-1}]$, for $k$ a field. An undercrossing $n$ under $m$ is rescaled by $q^{-nm}$.
Now here's what I'd like to do. I'd like to build a "prebraided category", which is the same monoidal category, but where the braiding of $n$ strands over $m$ strands is rescaled by $q^n$ (except when $m=0$, in which case don't rescale). I think you'll find that this satisfies one of the hexagons but not the other.