The inverse braiding is again a braiding

Because [L1] says each $c_{X,Y}$ is an isomorphism, the family $c_{X,Y}^{\mathrm{\prime }}=c_{Y,X}^{-1}$ is well defined. Inverting the naturality square for $c$ and swapping the variable names shows that $c^{\mathrm{\prime }}$ is natural in both variables.

Substitute $c_{X,Y}^{\mathrm{\prime }}=c_{Y,X}^{-1}$ into the first hexagon for $c^{\mathrm{\prime }}$. After reversing the arrows, this equation is exactly the second hexagon for $c$ from [L1] with the object names permuted. Hence the first hexagon holds for $c^{\mathrm{\prime }}$.

The same calculation with the roles of the two hexagons reversed shows that the second hexagon for $c^{\mathrm{\prime }}$ is the first hexagon for $c$ written backwards. Therefore $c^{\mathrm{\prime }}$ satisfies both hexagon identities and is a braiding.