# Inverse of Braiding in Monoidal Category is Braiding

In Mac Lane's Categories for the Working Mathematician in section $\mathrm{XI}.1$ about symmetric monoidal categories we find that a braided monoidal category $M$ is a monoidal category $(M,\oplus,e,\alpha,\lambda,\rho)$ where in addition to $\alpha, \lambda, \rho$ we have a natural equivalence $$\gamma_{a,b}:a\oplus b\cong b\oplus a$$ which must satisfy the following commutative diagrams:

$\qquad\qquad$

$\gamma$ is then called a braiding.

Now Mac Lane writes that the left hexagon for $\gamma$ implies the right hexagon for $\gamma^{-1}$, concluding that $\gamma^{-1}$ is a braiding as well

I don't really see how this works. Let's write $\beta$ for $\gamma^{-1}$, so that $\beta_{ab}=\gamma_{ba}^{-1}$. To turn the left diagram into (the shape of) the right, I replace the upper $\gamma$ by the inverse $\beta$ while the lower $\gamma$'s are replaced by the inverses of their opposites (instead of $\gamma_{bc}$ we have the inverse $\beta_{bc}$ of $\gamma_{cb}$. Now, the inner hexagon commutes, but I don't see why the outer hexagon should commute.

On the other hand, the outer hexagon commutes if and only if the following diagram commutes:

I don't have my copy of Cats Work at hand to check that those diagrams were correctly transcribed, but those are not the correct hexagon diagrams for a braiding. (They would be okay for a symmetry, where $\gamma_{ab}^{-1} = \gamma_{ba}$ by one of the axioms.) Edit: How do I know they're incorrect? Because they fail on the groupoid of braids!

The easy way to remember the correct diagrams is to pretend for a moment that the underlying monoidal category is strict monoidal (so that in particular the associativity constraints $\alpha$ are identities). On account of the coherence theorem for monoidal categories, this is no real loss of generality. In this case the axioms are (suppressing the $\otimes$ symbols)

$$(abc \stackrel{\gamma_{a,bc}}{\to} bca) = (abc \stackrel{\gamma_{a,b}c}{\to} bac \stackrel{b\gamma_{a,c}}{\to} bca),$$

$$(abc \stackrel{\gamma_{ab,c}}{\to} cab) = (abc \stackrel{a\gamma_{b,c}}{\to} acb \stackrel{\gamma_{a,c}b}{\to} cab),$$

which have very easily digested interpretations as braids. It's pretty clear in this simplified case that by taking inverses on the first condition, one gets the second condition applied to the transformation $\gamma^{-1}$, and vice-versa. And, it's not hard to reinsert associativities to get the corresponding correct hexagon identities, and check the claim there directly.

Now that I've taken a closer look at your post and see that you've posted images, I guess those were correctly "transcribed". This is a genuine error in the text in that case. Assuming that, what I think must have happened is this: the first edition of Cats Work was from the early 70's, well before braided monoidal categories were introduced, but certainly with material there on symmetric monoidal categories. As said before, the displayed images are fine for that case. Now, there are in fact several ways to present the hexagon identities for the symmetric monoidal case; with the "right" presentation, one can define a braiding just by retaining the hexagons and by dropping the axiom $\gamma_{a,b}\gamma_{b,a} = 1_{ba}$. But this won't work if you choose the "wrong" one. Mac Lane, who wrote the second edition when he was well into his 80's, might not have noticed the "wrong" one was used, and all this slipped through the cracks of the editorial process.

• The diagrams are correctly transcribed. – Zhen Lin Oct 24 '13 at 10:32
• I actually thought that all the diagrams that I've seen so far (in the book, in Wikipedia, at nLab) were equal, but now I see that in CWM we have the equality of $γ1∘γ=1γ$ while in the other diagrams we have $γ=γ1∘1γ$ and $γ=1γ∘γ1$. – Stefan Hamcke Oct 24 '13 at 16:37
• And you are right: Replacing $γ_{a,b}$ by $\beta_{b,a}$ whenever possible simply turns the corrected left into the right diagram. Thank you. – Stefan Hamcke Oct 24 '13 at 16:58
• Maybe you can help with another question: In CWM he has $λγ=ρ$ as an axiom, but in nLab I read that this follows from the other axioms. Do you know who is right? – Stefan Hamcke Oct 25 '13 at 22:54
• I don't recall for certain, and it'd take a bit of time for me to work this out from first principles (I should do this and then put it on the nLab), but I expect the nLab is right again. (Mac Lane has at least one other redundancy in his book: he assumes for monoidal categories the axiom $\rho_I = \lambda_I$, but this follows from the others.) A pretty definitive study is G.M. Kelly, On Mac Lane’s conditions for coherence of natural associativities, commutativities, etc., J. Algebra 1 (1964), 397-402. He works with symmetries, not braidings, but it should make no difference for this question. – user43208 Oct 26 '13 at 0:36