Brauer diagram centralizers for full orthogonal and symplectic groups

Put $t=a+b$. The snake identities in [F2] bend all input factors into output factors, giving inverse isomorphisms on diagram morphism spaces and on equivariant linear-map spaces. Identify invariant tensors with invariant multilinear forms by the nondegenerate form on each factor. Each bent basis diagram is then its matching contraction times a sign in $\{1,-1\}$, because crossings act by $ϵ$ times the flip and caps by the form. The signs are units and the matching assignment is a bijection: bending changes boundary labels and introduces no closed component. Thus the linear map from the free matching space to the invariant forms is exactly the map of [F1], after invertible diagonal sign and boundary-order changes.

If $t$ is odd the diagram space is zero, and [F1] makes the invariant target zero as well. If $t$ is even, [F1] says that every invariant form is a sum of matching contractions. Bend their diagrams back using step 1.1 to obtain a preimage of every intertwiner. A linear combination maps to zero exactly when its bent contraction combination belongs to the determinant/Pfaffian relation span in [F1]. Undoing the same bend gives exactly the kernel asserted in the Statement. This holds for all $a,b$, including $a=0$ or $b=0$.

At $a=b=r$, the map is an algebra map because [F2] is a functor. The first isomorphism theorem gives $B_r(ϵm)\mathrm{/}\ker {\rho }_r\cong {\mathrm{End}}_G(V^{\otimes r})$. Injectivity is equivalent, by the invertible bends in step 1.1, to independence of the contractions on $2r$ labels. By [F1] this holds exactly for $r\le m$ in the orthogonal branch and $r\le n$ in the symplectic branch. The empty case $r=0$ is the scalar identity. The kernels in all boundary types form a tensor ideal since the functor sends a composite or tensor product containing a zero image to zero. Within each fixed algebra the described span is therefore automatically a two-sided ideal.