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?


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.


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.)

  • $\begingroup$ 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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