The fundamental quantum sl2 ribbon trace matches the Hecke and bracket evaluations

In the ordered basis of [F1], direct subtraction gives $e:=A(c_{V,V}-AI)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& -A^2& 1& 0\\ 0& 1& -A^{-2}& 0\\ 0& 0& 0& 0\end{array}\right)$. Writing $U(1)=-A^2{v}_{+}\otimes v_{-}+v_{-}\otimes v_{+}$ and $B(v_{+}\otimes v_{-})=1$, $B(v_{-}\otimes v_{+})=-A^{-2}$ shows entrywise that $e=UB$. Hence $B(U(1))=-A^2-A^{-2}=\delta$ and $e^2=\delta e$. The explicit snake equalities in [F2] give the adjacent tensor relations: $e_{12}{e}_{23}{e}_{12}=UB\otimes 1$ because the middle composites $(1\otimes B)(U\otimes 1)$ and then $(B\otimes 1)(1\otimes U)$ are identities; the reversed calculation gives $e_{23}{e}_{12}{e}_{23}=e_{23}$. Distant placements commute because they act on disjoint tensor factors.

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. Thus the A/B crossing expansion is $C_i=AI+A^{-1}{e}_i$. Its inverse is $C_i^{-1}=A^{-1}I+A{e}_i$, since their product is $I+(A^2+A^{-2}+\delta )e_i=I$. By [A1] and [F8]–[F9], the categorical ribbon trace is the framed tangle evaluation of this braid closure, and that evaluator is monoidal with the stated crossing and duality maps. Expanding a braid word into the two smoothing choices therefore gives the bracket weights in [F4]. Because the closure has $n\ge 1$ strands, each complete smoothing is a nonempty closed 1-manifold, hence a union of at least one circle. Its categorical value is ${\delta }^{\mathrm{\mid }S\mathrm{\mid }}$ by monoidality in [F8] and the loop value [F2], whereas the bracket state sum uses ${\delta }^{\mathrm{\mid }S\mathrm{\mid }-1}$; the extra single factor is independent of the state. Thus $t_n(\beta )=\delta ⟨\widehat{\beta }⟩$.

The matrix trace $t_n$ in [F3] closes the tensor-word braid with the duality maps $B,U$ of [F2]. Substituting each crossing into the trace expansion from step 2.1 therefore gives the first identity in the Statement. The Hecke formula [F6] then yields $t_n(\beta )={\delta }^n(-A{)}^{e(\beta )}{\mathrm{tr}}_n({\pi }_n(\beta )){\mid }_{v=t,z=z_0}$. By [F7], $e(\beta )=w(\beta )$. The twist in [F1] is $\lambda =-A^3$, so multiplying $t_n=\delta ⟨\widehat{\beta }⟩$ by ${\lambda }^{-w(\beta )}$ gives the second displayed identity. The only choice assumption is [A1], used in step 2.1 to invoke the published trace-to-closure identification; no stronger choice principle or division by $\delta$ is used. Since $A$ is indeterminate, the specialization is taken in the rational-function field where $t+1\mathrm{\ne }0$.