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Throughout let $(\mathscr{C}, \otimes, \mathbf{1})$ be a monoidal category (I suppressed unitors and associators for simplicity).

The usual definition a rigid monoidal category is done in two steps: 1) Defining what is means right and left dual objects for an object $X$. 2) All objects have both right and left duals. Also we demand the rigidity axioms as a coherence condition. My first question is as follows:

The rigidity puts an extra structure, dualizable objects, on $\mathscr{C}$. New structures should respect old structures. So the question is the right dual functor, $*$, a monoidal functor? It should be!

So I investigated and found that $*$ can be defined as a monoidal functor $*:(\mathscr{C}, \otimes)\to (\mathscr{C}^\text{op}, \otimes^\text{op})$, where $\otimes^\text{op}$ is a bifunctor like $\otimes$ such that $X\otimes^\text{op} Y=Y\otimes X$. This make perfect sense to me, but I want to make sure that is in fact true.

Now we move onto the second question: Suppose $\mathscr{C}$ has a braiding structure (remove rigidity). Braiding $\sigma$ is itself a natural transformation between the bifunctors $\otimes$ and $\otimes^\text{op}$ both define as $\mathscr{C}\times \mathscr{C}\to \mathscr{C}$.

In what sense braiding structure respects the tensor product structure? I highly doubt that $\sigma$ can be defined as a monoidal functor! The correct way, I think, should be treating $\sigma$ as a monoidal natural transformation. But how exactly?

I tried but cannot find anything reasonable. Can anyone shed some light on this?

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You're asking two basically independent questions here, so they really should be separated, but yes, taking duals is monoidal. The argument generalizes to the observation that taking adjoints of 1-morphisms in 2-categories respects composition.

As for braidings, one way to define a braided monoidal category is as a "monoidal monoidal category" (or "$E_2$ category" for short): that is, it's a category equipped with two monoidal structures $\otimes_1, \otimes_2$ both of which are monoidal with respect to each other. The braiding comes from applying the Eckmann-Hilton argument to this situation. You get that the two monoidal structures are equivalent, but along the way to writing down this equivalence you end up writing down a braiding.

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You're right about the first thing. The easiest way to see his is to. It's the monoidal category as a bicategorywith a single object. Then a dual is just an adjoint, and it's well known that adjoint said are functorial in this way.

On the other hand, to call the braiding a monoidal natural transformation would be the wrong approach. Try to write out what that means: it involves making $\otimes^{op}$ into a monoidal functor! Rather you want to know how it works when you braid a thread of several strands in various orders, e.g. how the paths $abc \to cba$ compare. This is succinctly summarized in the condition that the braiding should induce an action of the braid group on each list of $n$ objects, so you can find the necessary relations by looki up the braid group, or by visualizing which twists of adjacent strands in a braid affect each other. (This graphical calculus is really indispensable for studying structured monoidal categories.)

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  • $\begingroup$ Actually I asked this question when I observed how there is no way $\otimes^{op}$ can be turned into a monoidal functor. I'm quite familiar with braid group and I know what is going on in a braid category. My question is, in "pure abstraction", how do you define braiding as a compatible structure. It is not a monidal functor, it is not a monoidal natural transformation, so what in the world is it? $\endgroup$ – Hamed Jan 28 '16 at 3:51
  • $\begingroup$ It's a natural transformation such that (all the braid relations.) $\endgroup$ – Kevin Carlson Jan 28 '16 at 4:03
  • $\begingroup$ Oh I see. From the definition of braided monoidal that people usually give: all braid relations can be obtained from functoriality of braiding and braiding axiom (hexagon equations). So I guess one should try and construct it the other way around to see what I ask more clearly. $\endgroup$ – Hamed Jan 28 '16 at 4:08
  • $\begingroup$ I don't quite understand whether you have another question. $\endgroup$ – Kevin Carlson Jan 28 '16 at 4:24

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