Braided and Symmetric Monoidal Categories - Examples

The category $\mathbf{S}\mathbf{e}\mathbf{t}$ has finite products, namely cartesian products of sets.

Therefore [L1] applies with $\mathcal{C}=\mathbf{S}\mathbf{e}\mathbf{t}$, and the resulting braiding is exactly the swap map $(x,y)\mapsto (y,x)$.

Specializing [L1] to $n=3$ leaves two generators and only one braid relation, because there is no pair $i,j$ with $\mathrm{\mid }i-j\mathrm{\mid }>1$ among $\{1,2\}$. This gives the displayed presentation of $B_3$.

Specializing [L2] to $n=3$ gives the quotient map $B_3\to S_3$ with ${\sigma }_1\mapsto (12)$ and ${\sigma }_2\mapsto (23)$.

Write the source of the first hexagon as $X\times (Y\times Z)$ and evaluate the left-hand route on $(x,y,z)$. The inner swap sends $(x,(y,z))$ to $((y,z),x)$ after the evident rebracketing.

Evaluate the right-hand route: first swap $x$ past $y$, then swap $x$ past $z$. The result is again $(y,z,x)$ after the same rebracketing identifications.

Since the two routes agree on every triple $(x,y,z)$, the first hexagon commutes in this cartesian case. The second hexagon is checked by the same coordinate calculation with the factors regrouped as $((x,y),z)$.

The displayed map is natural and invertible: applying it twice multiplies a pure tensor by $(-1{)}^{\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }}(-1{)}^{\mathrm{\mid }w\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }}=(-1{)}^{2\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }}=1$, so $c_{W,V}\circ c_{V,W}=1_{V\otimes W}$.

On a triple tensor $u\otimes v\otimes w$, each route around a braiding hexagon contributes the sign $(-1{)}^{\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }+\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }}=(-1{)}^{\mathrm{\mid }u\mathrm{\mid }(\mathrm{\mid }v\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid })}$, because moving $u$ past $v\otimes w$ is the same as moving it past $v$ and then past $w$. Thus both hexagons commute.

By [L1], these two checks show that $\mathbf{S}\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}$ carries a symmetric braiding, namely the sign braiding.

By [L1], the braiding on $2$ is the element ${\sigma }_1$.

By [L2], all nonzero powers of ${\sigma }_1$ are distinct from the identity. Hence the braiding on $2$ has infinite order.

The two displayed composites have underlying braids ${\sigma }_1{\sigma }_2$ and ${\sigma }_2{\sigma }_1$ in $B_3$.

These braids are not equal: their images in $S_3$ are $(123)$ and $(132)$, which are distinct permutations.

Therefore [L1] implies that the two canonical composites do not agree. In particular, they are distinct endomorphisms of the object $3$ in the braid category from [L2].

By [L1], a monoid object in $(\mathbf{S}\mathbf{e}\mathbf{t},\times )$ is a set $M$ with a multiplication map $m:M\times M\to M$ and a unit element satisfying the usual associative and unital equations.

The symmetry on $\mathbf{S}\mathbf{e}\mathbf{t}$ is the swap map $(a,b)\mapsto (b,a)$, so the categorical commutativity condition says $m(a,b)=m(b,a)$ for all $a,b\in M$. By [L2], that is exactly the ordinary commutativity law.