The quantum and Hecke tensor actions commute

For each $j$, $L_j\otimes L_j$ acts on $(a,b)$ by the scalar $q^{{\delta }_{j,a}+{\delta }_{j,b}}$. Every term in $T(a,b)$ has the same multiset of labels as $(a,b)$, so it has the same scalar. Hence $T(L_j\otimes L_j)=(L_j\otimes L_j)T$; the inverse $L_j^{-1}\otimes L_j^{-1}$ commutes as well.

Fix $i$ and write $(a,b)$ for $e_a\otimes e_b$. If the pair contains no $i+1$, both $T{\mathcal{E}}_i$ and ${\mathcal{E}}_{i}T$ vanish by [F2]–[F3]. For the remaining labels, the complete cases are: on $(i+1,i+1)$ both composites give $(i,i+1)+q(i+1,i)$; on $(i+1,i)$ both give $q^2(i,i)$; on $(i,i+1)$ both give $q(i,i)$; for $r<i$, the inputs $(i+1,r)$ and $(r,i+1)$ give respectively $(r,i)+c(i,r)$ and $(i,r)$ under either composite; for $r>i+1$, those same two orientations give respectively $(r,i)$ and $(i,r)+c(r,i)$. These cases exhaust every pair containing $i+1$, including both orientations of each outside label, so $T{\mathcal{E}}_i={\mathcal{E}}_{i}T$. The only coefficient simplification used in the repeated-label case is $q^{-1}+c=q$.

If a pair contains no $i$, both composites with ${\mathcal{F}}_i$ vanish. The remaining complete cases are: on $(i,i)$ both give $(i,i+1)+q(i+1,i)$; on $(i,i+1)$ both give $q(i+1,i+1)$; on $(i+1,i)$ both give $q^2(i+1,i+1)$; for $r<i$, the inputs $(i,r)$ and $(r,i)$ give respectively $(r,i+1)+c(i+1,r)$ and $(i+1,r)$ under either composite; for $r>i+1$, they give respectively $(r,i+1)$ and $(i+1,r)+c(r,i+1)$. Thus every pair containing $i$ is covered and $T{\mathcal{F}}_i={\mathcal{F}}_{i}T$. The repeated-label coefficients use $q^{-1}+c=q$ and $1+cq=q^2$.

Coassociativity and the coproduct formulas in [F2] give on $n$ positions ${\mathrm{\Delta }}_n(E_i)={\sum }_{r=1}^n{1}^{\otimes (r-1)}\otimes E_i\otimes K_i^{\otimes (n-r)}$, ${\mathrm{\Delta }}_n(F_i)={\sum }_{r=1}^n(K_i^{-1}{)}^{\otimes (r-1)}\otimes F_i\otimes 1^{\otimes (n-r)}$, and ${\mathrm{\Delta }}_n(L_j)=L_j^{\otimes n}$. For positions $a,a+1$, the two $E_i$ summands at those positions restrict there to $E_i\otimes K_i+1\otimes E_i$, and the two $F_i$ summands restrict to $F_i\otimes 1+K_i^{-1}\otimes F_i$; every earlier summand restricts to $K_i\otimes K_i$ or $K_i^{-1}\otimes K_i^{-1}$, and every later summand to the identity. Step 1.1 handles these Cartan factors and steps 1.2–1.3 handle the two-site sums; therefore $T_a$ commutes with each iterated quantum generator. Commuting with the generators implies commuting with their products and linear combinations, which proves $B_F\subseteq {\mathrm{End}}_{A_F}(V_N^{\otimes n})$.