Proving that $(\sigma_{V,W})^* = \sigma_{V^*,W^*}$ in a braided tensor category.

Question

This question is about a proposition that appears in Lectures on Tensor Categories and Modular Functors by Bakalov and Kirillov. They present a graphical calculus for morphisms and a using that prove that in a ribbon category $(\sigma_{V,W})^* = \sigma_{V^*,W^*}$. They state that the proof from axioms is short and not hard but I cannot see it. Could anyone help? Here's the graphical proof, plus we apply rigidity:

Answer 1

I'll use the notation from the Bakalov & Kirillov book. Let's first see what do we exactly mean by a map $W^* \xrightarrow{f^*} V^*$ dual to $V \xrightarrow{f} W$. We define $f^*$ by $$W^* \xrightarrow{\mathrm{id}_W \otimes i_V} W^* \otimes V \otimes V^* \xrightarrow{\mathrm{id}_{W^*} \otimes f \otimes \mathrm{id}_{V^*}} W^* \otimes W \otimes V^* \xrightarrow{e_{W^*} \otimes \mathrm{id}_{V^*}} V^*.$$

Now let's write the morphism that is defined by the left picture. For simplicity, I'll suppress tensor products with identity morphisms and just write the main ones. I hope it's possible to follow what's going on. OK, left picture gives:

\begin{align}V^* \otimes W^* &\xrightarrow{i_V} V^* \otimes W^* \otimes V \otimes V^* \xrightarrow{i_W} V^* \otimes W^* \otimes V \otimes W \otimes W^*\otimes V^* \\\\ &\xrightarrow{\sigma_{VW}} V^* \otimes W^* \otimes W \otimes V \otimes W^*\otimes V^* \xrightarrow{e_{W}} V^* \otimes V \otimes W^* \otimes V^* \\\\ &\xrightarrow{e_V} W^* \otimes V^*.\end{align}

Middle picture gives: \begin{align}V^* \otimes W^* &\xrightarrow{i_V} V^* \otimes W^* \otimes V \otimes V^* \xrightarrow{i_W} V^* \otimes W^* \otimes V \otimes W \otimes W^*\otimes V^* \\\\ &\xrightarrow{\sigma_{VW}} V^* \otimes W^* \otimes W \otimes V \otimes W^*\otimes V^* \xrightarrow{\sigma_{V^*, W^* \otimes W}} W^* \otimes W \otimes V^* \otimes V \otimes W^* \otimes V^* \\\\ &\xrightarrow{e_V} W^* \otimes W \otimes W^* \otimes V^* \xrightarrow{e_W} W^* \otimes V^*. \end{align}

First, notice that the difference is entirely in the middle row.

In the left picture, after $\sigma_{VW}$, we do $e_W$, then $e_V$. In the middle picture, after $\sigma_{VW}$ we jump with leftmost $V^*$ over $W^* \otimes W$, then do $e_V$, then $e_W$.

So, why are these two morphisms same? In my edition of the book, authors call upon relation (2.3.7.) that says that morphisms $$A \otimes B^* \otimes B \xrightarrow{\mathrm{id}_A \otimes e_B} A $$ and $$A \otimes B^* \otimes B \xrightarrow{\sigma_{A, B^* \otimes B}} B^* \otimes B \otimes A \xrightarrow{e_B \otimes \mathrm{id}_A} A $$ are equal.

How do we know that this relation holds? By functoriality of $\sigma_{A, -}$ (that's one of the axioms)!

I suppose you now see why people came up with the graphical calculus to write down these kind of things. The equality between the middle and the right picture also comes from (2.3.7). Try writing out how the morphism defined by the right picture looks, and hopefully it will be easy to see where relation (2.3.7) is used.