# Constructing the inverse of a braiding in a braided pivotal category

Assume we have a braided pivotal monoidal category. This means we assume the braiding $$c$$ to be a natural isomorphism. But looking at the corresponding string diagram, it seems to me as if we could obtain $$c^{-1}$$ as the composition of the following morphisms (treating the monoidal structure as strict): $$A \otimes B \xrightarrow{\eta_{B^{*}} \otimes A \otimes B} B^{**} \otimes B^{*} \otimes A \otimes B \xrightarrow{B^{**} \otimes c_{A, B^{*}} \otimes B} B^{**} \otimes A \otimes B^{*} \otimes B \xrightarrow{B^{**} \otimes A \otimes \varepsilon_{B}} B^{**} \otimes A = B \otimes A$$

As a string diagram, this construction would look like this:

This begs the questions:

• Is it enough to assume the braiding as a morphism instead of assuming it to be an iso?

• When working 2-categorically, we now have a non-trivial 2-cell at $$c^{-1}$$, should this be filled to make the notion well-behaved?

• Braided pivotal categories are discussed in Section 2.3 of my paper arxiv.org/pdf/1509.02937.pdf – André Henriques Mar 8 at 12:49
• This is true, as André says below, and in fact I don't think you need pivotality, just rigidity (use the other duality $1\rightarrow B \otimes B^*$). This is an analog of the fact that in a quasi-triangular Hopf algebra the R-matrix satisfies $(S\otimes id)(R)=R^{-1}$. – Adrien Mar 8 at 14:05
• @TimCampion There is a notion of Lax braiding. I don't know much about it, but OP's question is answered in Prop. 1.3 of Day, Panchadcharam, and Street's "Lax Braidings and the Lax Centre". The braiding is invertible as long as the category is right rigid, for a correct choice of left versus right. – Adrien Mar 8 at 14:11
• @Adrien Nice! I think that reference constitutes an answer. I wonder if this statement is an instance of the fact that any monoidal natural transformation between monoidal functors between categories with duals is an isomorphism... – Tim Campion Mar 8 at 14:17
• @TimCampion rather it has to do with the fact that a lax monoidal functor between rigid categories is automatically strong, I think. The braiding is a monoidal structure on the identity functor from $C$ to $C$ with opposite multiplication. – Adrien Mar 8 at 14:38

As Tim says this should perhaps be an answer: yes this is true, and doesn't require pivotal. As long as the category is right rigid, the braiding is automatically invertible. See e.g. Prop 1.3 in Day, Panchadcharam, and Street's "Lax Braidings and the Lax Centre" (http://science.mq.edu.au/~street/laxcentre.pdf).

• I marked this as correct as this was what I was looking for and answers the first of my questions. The second one is a bit fuzzy, I guess. I take a away from this that not assuming the inverse of the braiding might actually might have unwanted homotopical effects at the level of 2-cells... – javra Mar 8 at 14:40

The equality between these two diagrams is automatic.
Starting from the RHS:

(1) Push the rightmost vertical strand under its adjancent strand by a Reidemeister II move (thus creating three crossings).

(2) slide the cup up away from under the strand, by using naturality of the braiding (thus bringing back the number of crossings to one).

(3) undo the zig-zag

• Given that they are both inverses of the original braiding we do not need a string diagram proof to show that the equation holds, that's just true in any category. My question was about why we can't define its left-hand side to be the right-hand side. – javra Mar 8 at 13:03
• Of course you can use this to prove that the braiding is automatically invertible. (Providing an explicit inverse is definitely a valid proof that something is invertible.) – André Henriques Mar 8 at 13:50
• @AndréHenriques I am a little confused. Suppose that the braiding is just assumed to be a natural transformation, and not assumed to be invertible (so you only have, say, right-handed crossings for your string diagrams). If the category is rigid or pivotal you can define a morphism with the formula from the RHS as in the OP. Is it automatically an inverse to the braiding? Your proof uses a Reidemeister II move, which presupposes that the braiding has an inverse. What if you don't a priori assume the braiding to be invertible? – Chris Schommer-Pries Mar 8 at 14:13
• @ChrisSchommer-Pries I agree it's not obvious -- when removing the hypothesis that the braiding be invertible, one must go back to the definition of a braided monoidal category, which doesn't explicitly include a Reidemeister II move (the Reidemeister II move involves duals, after all). But I think the Reidemeister II move can be deduced, when a dual exists, from a hexagon identity plus naturality of of the braiding (and some kind of unit law), so it turns out to be okay. – Tim Campion Mar 8 at 14:22
• @ChrisSchommer-Pries Oops! Of course you're completely right. I was confused myself. – André Henriques Mar 8 at 14:50