The cartesian swap braiding is a symmetry

For each pair $X,Y$, the two coordinate projections from $X\times Y$ define a unique morphism ${\tau }_{X,Y}:X\times Y\to Y\times X$ with first projection ${\pi }_2$ and second projection ${\pi }_1$. The same universal property defines its inverse ${\tau }_{Y,X}$, so the family is a natural isomorphism.

To check the first hexagon, compare both composites from $X\times (Y\times Z)$ to $(Y\times Z)\times X$ after composing with the three product projections. Each route sends $(x,y,z)$ to $(y,z,x)$, so the two maps are equal. The second hexagon is the same coordinate permutation written on $((X\times Y)\times Z)$ and is checked in the same way.

Swapping twice returns every pair to itself, so ${\tau }_{Y,X}\circ {\tau }_{X,Y}=1_{X\times Y}$. By [L2], the cartesian swap braiding is therefore a symmetry.