The q-wedge relations coincide with the type-A energy braiding kernel

The left fraction and loop swap. From [F1], since $(a-1)Y_2=Y_1-Y_2$, the numerator simplifies as an operator to $(a-1)S+Q-1=[(a-1)S_0+Q-1]s$. Thus $R_{\mathrm{f}\mathrm{i}\mathrm{x}}=R_0(a)s$ with $R_0(a)=((a-1)S_0+Q-1)\mathrm{/}(Q-a)$, proving $R_{\mathrm{f}\mathrm{i}\mathrm{x}}f=s(f)R_{\mathrm{f}\mathrm{i}\mathrm{x}}$ for every Laurent loop function $f$. In particular it swaps $Y_1,Y_2$. All denominators are LEFT coefficients; no commutation of $S$ with $a$ is used. The inverse is $-a^{-1}{\sum }_{r\ge 0}(Q\mathrm{/}a{)}^r$, defined in the separated $q$-adic loop-module completion. For an extremal diagonal residue tensor $w$, $S_{0}w=-w$ and the numerator is $(Q-a)w$, so $R_{\mathrm{f}\mathrm{i}\mathrm{x}}w=w$; common loop shifts give the same conclusion for each $u\otimes u$. By [F2] the original Hecke action and loop functions commute with $U_q^{\mathrm{\prime }}$; ratios have total degree zero, so the same operator intertwines the full affine action.

The exact original-space kernel. Subtracting the identity with LEFT denominator gives $R_{\mathrm{f}\mathrm{i}\mathrm{x}}-1=(Q-a{)}^{-1}(a-1)(S+1)$. Multiplication on the left by ${\psi }_{\mathrm{f}\mathrm{i}\mathrm{x}}=-(Q-a)\mathrm{/}a$ gives exactly $(a^{-1}-1)(S+1)$. The formal $q$-adic coefficient ${\psi }_{\mathrm{f}\mathrm{i}\mathrm{x}}$ is invertible, and multiplication by the nonzero Laurent polynomial $a^{-1}-1$ is injective on the original free Laurent-loop module by [F2]. Hence an ORIGINAL Laurent-loop vector is fixed by $R_{\mathrm{f}\mathrm{i}\mathrm{x}}$ exactly when it lies in $\ker (S+1)$. No localization of the original kernel and no finite-dimensional weight-space assumption is needed.

Yang--Baxter. Use the FINITE residue operators $S_{0,1},S_{0,2}$ with commuting external variables $x,y$, and put $B_i(x)=(x-1)S_{0,i}+Q-1$. Expand $B_1(x)B_2(xy)B_1(y)$ and $B_2(y)B_1(xy)B_2(x)$. Reduce each square with $S_{0,i}^2=(Q-1)S_{0,i}+Q$ and each three-letter braid with $S_{0,1}{S}_{0,2}{S}_{0,1}=S_{0,2}{S}_{0,1}{S}_{0,2}$. The coefficients of the remaining Hecke words agree, using $(x-1)(y-1)+(x-1)+(y-1)=xy-1$ for the one-letter terms. The denominator is the common commuting product $(Q-x)(Q-xy)(Q-y)$. In step 1.1 the loop swaps supply precisely $x,y,xy$ on the three factors, so this finite-residue identity proves $R_{12}{R}_{23}{R}_{12}=R_{23}{R}_{12}{R}_{23}$ in the stated completion. It is not an expansion of full affine $S_i$ against noncommuting loop coefficients.

The opposite channel and its leading term. Define $R_{\mathrm{o}\mathrm{p}}=P{R}_{\mathrm{f}\mathrm{i}\mathrm{x}}P$ with the full loop-and-factor flip. It intertwines the opposite coproduct because $P$ transports the checked fixed action, and its denominator is $1-Qa$. Evaluate $R_0(a)$ from step 1.1 on the three residue cases. At $q=0$ its pre-loop-swap factor is $a^{-1}$ for $j<k$, and $1$ for $j=k$ or $j>k$. Full flip exchanges both the residue order and $a$ with $a^{-1}$; hence the opposite pre-swap factors are $a,1,1$ for $j>k,j=k,j<k$. On $z^a{v}_j\otimes z^b{v}_k$ the loop swap adds the exponent difference $b-a$, giving $H_{\mathrm{o}\mathrm{p}}=\chi (j>k)+b-a$ exactly as in [F3]. The opposite normalization and Yang--Baxter identities follow by the same full flip.

The full kernel generators. Step 2.1 and flip give $\ker (R_{\mathrm{o}\mathrm{p}}-1)=P\ker (S+1)$. The actual two-factor Laurent-loop kernel is generated over symmetric loop functions by the diagonal tensors and, for $j<k$, $h_{jk}=v_j\otimes v_k+q^{-1}{v}_k\otimes v_j$ and $h_{jk}^{\mathrm{\prime }}=Y_1{v}_k\otimes v_j+q^{-1}{Y}_2{v}_j\otimes v_k$ (The q-antisymmetrizer splits the tensor power and its kernel is the sum of the Hecke kernels). Their flips, after multiplication by the units $q$ and $(Y_1{Y}_2{)}^{-1}$, are exactly the two order cases of the displayed $C_{j,k}$. Thus the displayed $C$ generate the entire original opposite kernel with symmetric loop shifts. To verify the full quantum orbit explicitly in the fixed channel, write $d_j=v_j\otimes v_j$. For $1\le j<e$, $\mathrm{\Delta }{F}_j{d}_j=q{h}_{j,j+1}$ and $(\mathrm{\Delta }{F}_j{)}^2{d}_j=(q+q^{-1})d_{j+1}$; hence all diagonals and all constant $h_{jk}$ are reached from $d_1$, extending the second index by the intervening finite lowering operators. Also $\mathrm{\Delta }{E}_0{d}_1=Y_2(v_1\otimes v_e)+q{Y}_1(v_e\otimes v_1)=q{h}_{1e}^{\mathrm{\prime }}$. For $j+1<k$, $\mathrm{\Delta }{F}_j{h}_{jk}^{\mathrm{\prime }}=h_{j+1,k}^{\mathrm{\prime }}$, and for $j<k-1$, $\mathrm{\Delta }{E}_{k-1}{h}_{jk}^{\mathrm{\prime }}=h_{j,k-1}^{\mathrm{\prime }}$; the partner factor has toral eigenvalue $1$ in both cases. These paths reach every $h_{jk}^{\mathrm{\prime }}$ with $j<k$, including $e=2$ where the first affine step already reaches the only such pair. Since symmetric Laurent multiplication commutes with the derived action, this proves that its symmetric-loop orbit contains the entire kernel basis. Conversely every such orbit element remains in the kernel by intertwining and symmetric-loop stability. Flipping proves the exact opposite quantum/symmetric-loop orbit asserted for the $C$ generators. Common loop shifts and symmetric powers produce every Laurent gap. This is the stated quantum AND symmetric-loop orbit, not arbitrary individual loop multiplication of a kernel vector. Flip back therefore gives precisely the full finite straightening quotient.

Conclusion. The checked completed left fraction swaps loops, fixes extremal diagonals, intertwines the fixed action and satisfies normalized Yang--Baxter. Its polynomial-cleared identity has injective Laurent multiplier on original vectors, giving the exact original-space kernel. The opposite flip has the stated energy, denominator and full symmetric-loop/quantum kernel generators, and transports back to the fixed finite-wedge quotient.