Coherence in braided monoidal categories

Question

Let ($\mathcal{C}$,c) be a braided monoidal (tensor) category. Then c is compatible with the morphisms l,r associated with the unit object 1 of $\mathcal{C}$, in the sense that:

$l_X \circ c_{X,1}=r_X$, $\forall X \in \mathcal{C}$

I need a proof, or references of.

Answer 1

One can prove this axiom from one hexagon identity and the axioms of a monoidal category. It's a great exercise and goes like this:

• Take the second hexagon identity (e.g. from http://ncatlab.org/nlab/show/braided+monoidal+category#the_coherence_laws) and set $(X, Y, Z) = (I,I,X)$. Draw it very large, filling a whole sheet of paper.
• Draw the morphisms $l_{I \otimes X}: I \otimes (I \otimes X) \to X \otimes I$ and $l_{I\otimes X}: (I \otimes I) \otimes X \to I \otimes X$, inwards in the hexagon. Use the coherence theorem of monoidal categories to show that the resulting triangle commutes. Do the same on the opposite side with $r$.
• Use naturality in $c_{-,X}$ to draw the morphism $c_{I,X}$.
• Draw $l_{X \otimes I}: I \otimes (X \otimes I) \to X \otimes I$ (draw a new copy of $X \otimes I$ on the opposite side of the diagram). Complete the naturality square in $l$. Do the same for $r$ on the other side. You get two more $c_{I,X}$, one of which cancels the one from the previous step.
• Use coherence of monoidal categories again to get rid of the remaining associator and arrive at your result.

Answer 2

The full proof of this can be found as the 6-cell 'Lemma 4.4 Commutator' in the Globular workspace http://globular.science/1705.001, created to accompany the paper 'Coherence for braided and symmetric pseudomonoids'. (Use Chrome to view). This is an external proof, where the category is viewed as an braided pseudomonoid object in the symmetric monoidal 2-category Cat. However, this proof holds for general braided pseudomonoids in semistrict braided monoidal 2-categories.

To interpret the worksheet, note that a semistrict braided monoidal 2-category can be considered as a semistrict 4-category with one object and one 1-cell. As you will see in the Globular worksheet, this is the case here. There is one nontrivial 2-cell A; this is our category object. This has product and unit 3-cells, and associator, unitor and commutator 4-cells. Coherence equalities are given as invertible 5-cells. The braided monoidal structure of the ambient bicategory is given by the definition of a semistrict 4-category built into Globular. The syllepsis and its symmetry coherence isomorphism, which promote the ambient braided monoidal 2-category to a symmetric monoidal 2-category, are added in by hand.

Here are the instructions for viewing the proof:

Click on the 6-cell 'Lemma 4.4 Commutator' at the left side of the page. Go to the 'Project' box and go down to level 1. You will now see three 'slice' boxes, set to 0 0 0. The left slice box, which can take values 0 or 1, toggles between the proof statement (0) and the proof itself (1). The middle slice box runs through the stages of the proof. The final slice box runs through the 2-morphism at each stage of the proof.

We will run through the statement of the proof now. The toggle boxes should start at '0 0 0'. The diagram displayed is (1 (x) A), the red dot being a unit. (Remember we are in Cat).

We will now run through the 2-morphism '0 0 -'. Click on the right slice box to change it from 0 to 1. You will see that a left unitor has been performed. Now take the right slice box down to 0 again and set the middle slice box to 1. Moving the right slice box forward , you will see that a commmutator and then a right unitor are performed. The theorem therefore states that these are equal.

Now set middle and right slice boxes back to 0 and set the left slice box to 1. This allows us to view the proof. We detail the first few stages. With the middle slice box at 0, a left unitor is performed. With the middle slice box at 1, a pullthrough 2-cell and its inverse in Cat are performed (invisible internal to the monoidal category), and then the left unitor is performed. Clearly this is equal. With the middle slice box at 2, the pullthrough is performed, then a commutator and its inverse, then the inverse pullthrough, then the left unitor. Again, this is equal, since we have just inserted a 2-cell and its inverse. The proof continues like this until we eventually get to the target.

For more information about Globular, see the nLab page https://ncatlab.org/nlab/show/Globular.

Answer 3

I'm going to present a full proof by diagram chasing.

One can get a similar proof from [Kock, Frobenius Algebras and 2D Topological Quantum Field Theories, 3.2.40]. There one needs to insert the associators appropriately and does a diagram chasing in a tetrahedron. One also needs $\lambda_{I}\otimes id_{X}\circ \alpha_{I,I,X}^{-1}=\lambda_{I\otimes X}$ and $\lambda_{X}\otimes id_{I}\circ \alpha_{I,X,I}^{-1}=\lambda_{X\otimes I}$ (a proof for them lies in the nLab entry "monoidal category").

Here goes my proof. One can find all the relevant definitions in the nLab entries "monoidal category" and "braided monoidal category".

I will use $\alpha$ for the associator, $\lambda$ for the left unitor, and $\rho$ for the right unitor. And I will use $\tau$ for the braiding, which is a natural isomorphism.

Note that $\tau$ being an isomorphism is necessary for the following proof. On the other hand, in the triangle identity that we want to prove, as $\lambda$ and $\rho$ are also natural isomorphisms, $\tau$ is forced to be a natural isomorphism. So if the component of $\tau$ is not an isomorphism, the triangle identity would not hold.

Here goes the diagram (with two redundant arrows anyway). (I'm not sure how to draw the diagram in MSE. Thanks to quiver for helping me.)

Then let us do the diagram chasing. Our goal is to show that the south west triangle of the above diagram is commutative, i.e., $$id_{I}\otimes\lambda_{X}=id_{I}\otimes\rho_{X}\circ id_{I}\otimes\tau_{I,X},$$ which would imply $$\lambda_{X}=\rho_{X}\circ \tau_{I,X}$$ via the natural isomorphism $\lambda$.

Since $\tau_{I,X}$ is an isomorphism, it is enough for us to show $$\tau_{I,X}\circ id_{I}\otimes\lambda_{X}=\tau_{I,X}\circ id_{I}\otimes\rho_{X}\circ id_{I}\otimes\tau_{I,X}.$$

Let's go from the left-hand side:

\begin{align} \tau_{I,X}\circ id_{I}\otimes\lambda_{X} & = \tau_{I,X}\circ \rho_{I}\otimes id_{X}\circ \alpha^{-1}_{I,I,X}=id_{X}\otimes\rho_{I}\circ \tau_{I\otimes I,X}\circ \alpha^{-1}_{I,I,X}\\ & = id_{X}\otimes\lambda_{I}\circ \tau_{I\otimes I,X}\circ \alpha^{-1}_{I,I,X}=\rho_{X\otimes I}\circ \alpha^{-1}_{X,I,I}\circ \tau_{I\otimes I,X}\circ \alpha^{-1}_{I,I,X}\\ & = \rho_{X\otimes I}\circ\tau_{I,X}\otimes id_{I}\circ \alpha^{-1}_{I,X,I}\circ id_{I}\otimes\tau_{I,X}=\tau_{I,X}\circ \rho_{I\otimes X}\circ \alpha^{-1}_{I,X,I}\circ id_{I}\otimes\tau_{I,X}\\ & =\tau_{I,X}\circ id_{I}\otimes\rho_{X} \circ id_{I}\otimes\tau_{I,X}, \end{align} which is just as our wish.

Here in the first line, we use one triangle equality from coherence and the naturality of $\tau$.

In the second line, we use the fact $\lambda_{I}=\rho_{I}$ (see the nLab entry "monoidal category" for a proof) and one more triangle equality from coherence.

In the third line, we use one hexagon identity from our braiding and the naturality of $\rho$.

Finally, we use $$\rho_{I\otimes X}\circ \alpha^{-1}_{I,X,I}= id_{I}\otimes\rho_{X},$$ which is as the results for $\lambda$ we mentioned at the beginning (also see the nLab entry "monoidal category" for a proof).