Vacuum-channel anyonic projectors satisfy Temperley–Lieb relations

Restrict three adjacent $a$ sites to total charge $a$. The subspace where the first pair fuses to $\mathbf{1}$ is one-dimensional, since $N_{aa}^{\mathbf{1}}=1$ and $\mathbf{1}\otimes a\cong a$. The same holds for the subspace where the last pair fuses to $\mathbf{1}$. In the two orthonormal tree bases their unit vectors have overlap $(F_a^{aaa}{)}_{\mathbf{1},\mathbf{1}}$. The product of rank-one projections onto unit vectors $e,f$ is $P_e{P}_f{P}_e=\mathrm{\mid }⟨e,f⟩{\mathrm{\mid }}^2{P}_e$, so [F3] gives $P_i{P}_{i+1}{P}_i=d_a^{-2}{P}_i$ on this total-$a$ block. Outside total $a$, both vacuum-pair projections vanish because the remaining site is $a$; hence the identity holds on the full three-site space and, after tensoring with spectator identities, on the path space.

By [F2], $P_i^2=P_i$, so $U_i^2={\delta }_a^2{P}_i={\delta }_a{U}_i$. The adjacent relation follows from step 1.1: $U_i{U}_{i+1}{U}_i={\delta }_a^3{d}_a^{-2}{P}_i={\delta }_a{P}_i=U_i$, since ${\delta }_a^2=d_a^2$.

If $\mathrm{\mid }i-j\mathrm{\mid }>1$, choose a parenthesization grouping the two disjoint adjacent pairs as separate tensor factors, retaining the intervening and outer factors. In that parenthesization their maps have the form $A\otimes 1$ and $1\otimes B$ on those factors; bifunctoriality gives $(A\otimes 1)(1\otimes B)=A\otimes B=(1\otimes B)(A\otimes 1)$. Tensoring with the other identities and transporting back to the fixed parenthesization preserves this equality by [F4]. Thus all displayed Temperley–Lieb relations hold.