TITLE: Constructing the inverse of a braiding in a braided pivotal category
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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).