The pointed Z/2 semion metric group

Normalization holds because every exponent vanishes if any argument is $0$. For bits, $b+c$ in the cocycle equations is reduced modulo $2$. The pentagon signs have exponents $abc+a(b+c)d+bcd=acd+bcd+abc+abd(\mathrm{m}\mathrm{o}\mathrm{d}2),$ so its two sides agree. For the first hexagon, the ratio of the $r$-terms is $i^{-2abc}=(-1{)}^{abc}$, which is exactly the ratio of the $\omega$-terms. For the second hexagon, the ratio of the $r$-terms is again $i^{-2abc}=(-1{)}^{abc}$, equal to $\omega (a,b,c)\omega (c,a,b)\omega (a,c,b{)}^{-1}=(-1{)}^{abc}$. The twist equation is trivial if either bit is zero; at $(1,1)$ its right side is $r(1,1{)}^{2}q(1{)}^2=i^2{i}^2=1=q(0)$. Also $q(-g)=q(g)$. Thus all specified cocycle and twist identities hold.

Supply the dual and pivotal data required by [F1]. The left dual of $g$ is $g$; its evaluation scalar is $1$, and its coevaluation scalar is $\omega (g,g,g{)}^{-1}=(-1{)}^g$. Both snake composites are one, since the associator contributes the same sign. Formula [F3], with reassociation from $g\otimes (g\otimes g)$ to $(g\otimes g)\otimes g$, gives $u_0=1$ and $u_1=(-1)(-1)i=i$. Put $j_g=u_{g}q(g)$, so $j_0=1$ and $j_1=-1$, and extend by direct sums. The tensor identity in [F3] and the balancing identity checked in step 1.1 prove monoidality of $j$, including all associators and double-dual comparisons. Its pivotal trace on a simple is the scalar ${\mathrm{ev}}_{g^{\vee }}(j_g\otimes 1){\mathrm{coev}}_g=(-1{)}^g(-1{)}^g=1$. The dual grade is the same, so the left and right simple traces agree; additivity gives sphericality on all finite direct sums. The induced twist $u^{-1}j$ is exactly $q$. Thus the full dimension-one pointed ribbon input, rather than just its scalar cocycle, has been constructed.

The double-braiding values are $b(0,g)=b(g,0)=1$ and $b(1,1)=i^2=-1$. Hence $s=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\mathrm{.}$ The only nonzero element $1$ is not in the radical because $b(1,1)=-1\mathrm{\ne }1$, so $\mathrm{rad}(b)=\{0\}$. By [F2] the category is modular; directly, $\det (s)=-2$ and $D^2=2$.