Signed Brauer contraction functor and bending isomorphism

Choose a basis $b_1,\dots ,b_m$ and its $B$-dual basis ${\bar{b}}_1,\dots ,{\bar{b}}_m$, so $B({\bar{b}}_i,b_j)={\delta }_{ij}$, and set $c={\sum }_i{b}_i\otimes {\bar{b}}_i$. Nondegeneracy gives $v={\sum }_{i}B(v,b_i){\bar{b}}_i={\sum }_{i}B({\bar{b}}_i,v)b_i$, hence $(B\otimes \mathrm{id})(\mathrm{id}\otimes c)={\mathrm{id}}_V$ and $(\mathrm{id}\otimes B)(c\otimes \mathrm{id})={\mathrm{id}}_V$. Invariance of $B$ implies invariance of its inverse tensor $c$. Symmetry or alternation gives $BP=ϵB$ and $Pc=ϵc$, while $B(c)={\sum }_{i}B(b_i,{\bar{b}}_i)=ϵm$.

The maps $P$ satisfy $P^2=1$ and the braid relation because they permute tensor factors. Thus $ϵP$ satisfies the crossing-square and braid relations; $B(ϵP)=B$ and $(ϵP)c=c$ give the cap-crossing and cup-crossing relations. On $v\otimes w\otimes u$, both sides of the cap-slide relation $(A\otimes I)(I\otimes X)=(I\otimes A)(X\otimes I)$ equal $ϵB(v,u)w$. The dual slide relation $(I\otimes X)(U\otimes I)=(X\otimes I)(I\otimes U)$ sends $v$ on both sides to $ϵ{\sum }_i{b}_i\otimes v\otimes {\bar{b}}_i$. Reflections give the other slide conventions. Both snake identities from step 1.1 and $B(c)=ϵm$ give the straightening and loop relations. Therefore the assignments satisfy every relation in [F2] and extend to a monoidal functor with $\delta =ϵm$.

For a basis multi-index $I=(i_1,\dots ,i_a)$ write $b_I=b_{i_1}\otimes \cdots \otimes b_{i_a}$ and ${\bar{b}}_I={\bar{b}}_{i_1}\otimes \cdots \otimes {\bar{b}}_{i_a}$. Map a $G$-equivariant linear map $T:V^{\otimes a}\to V^{\otimes b}$ to ${\sum }_I{b}_I\otimes T({\bar{b}}_I)$. The inverse sends a tensor ${\sum }_I{b}_I\otimes w_I$ to $x\mapsto {\sum }_I{B}^{\otimes a}(x,b_I)w_I$; the dual-basis identities in step 1.1 show both composites are identities. The maps commute with $G$ because $B$ and $c$ are invariant, so this restricts to the stated invariant and intertwiner spaces.