F-symbol unitarity

For either tree domain $X$, let $(u_j)$ be its complete tree basis from [F4]. Composing the orthonormal vertex identities of [F3] along the tree, as in [F4], gives $u_j\circ u_k^{†}={\delta }_{jk}{\mathrm{id}}_d$. Consequently, for $f={\sum }_j{f}_j{u}_j$ and $g={\sum }_j{g}_j{u}_j$, conjugate-linearity of the dagger gives $f\circ g^{†}=({\sum }_j{f}_j\overline{g_j}){\mathrm{id}}_d$. Thus the coordinate inner product satisfies $⟨f,g⟩{\mathrm{id}}_d=f\circ g^{†}$ on both tree Hom spaces, although $X$ need not be simple.

Let $T(f)=f\circ {\alpha }_{a,b,c}^{-1}$. For maps $f,g$ in the left-tree Hom space, step 1.1 and unitarity of $\alpha$ give $⟨T(f),T(g)⟩{\mathrm{id}}_d=(f\circ {\alpha }^{-1})\circ (g\circ {\alpha }^{-1}{)}^{†}=f\circ {\alpha }^{-1}\circ \alpha \circ g^{†}=⟨f,g⟩{\mathrm{id}}_d$. Since $d$ is a nonzero simple object, its identity is nonzero, so $T$ preserves inner products.

The associator is an isomorphism, so $T$ is bijective. By [F1], its coordinate matrix in the orthonormal bases of [F4] is $F_d^{abc}$. A bijective inner-product-preserving linear map has a matrix $F$ satisfying $F^{†}F=F{F}^{†}=I$ in orthonormal bases; if the space is zero, $F$ is the empty identity matrix. Hence this F-symbol is unitary.