The quasi-$R$ intertwiner determines its inverse and the coproduct factorizations

The barred equation. Bar the intertwining equation with $B=\psi \otimes \psi$: from $\mathrm{\Delta }(u)\mathrm{\Theta }=\mathrm{\Theta }\bar{\mathrm{\Delta }}(u)$ and [F2] one gets $\bar{\mathrm{\Delta }}(\psi (u))\bar{\mathrm{\Theta }}=\bar{\mathrm{\Theta }}\mathrm{\Delta }(\psi (u))$ for every $u$, i.e. after replacing $u$ by $\psi (u)$, $\bar{\mathrm{\Delta }}(u)\bar{\mathrm{\Theta }}=\bar{\mathrm{\Theta }}\mathrm{\Delta }(u)$. Inverting the original equation in the convolution algebra [F3] gives ${\mathrm{\Theta }}^{-1}\mathrm{\Delta }(u)=\bar{\mathrm{\Delta }}(u){\mathrm{\Theta }}^{-1}$; applying the algebra automorphism $B$ and using [F2] turns this into $\mathrm{\Delta }(\psi (u))B({\mathrm{\Theta }}^{-1})=B({\mathrm{\Theta }}^{-1})\bar{\mathrm{\Delta }}(\psi (u))$.

The coproduct factorizations by duality. Fix degrees $\gamma ,\eta$. In the corresponding component of $\mathrm{\Delta }x$, $x\in U^{+}[\gamma +\eta ]$, remove the middle torus to write the coefficient tensor as $x_1\otimes K_{-\eta }{x}_2$, with $\mathrm{\mid }{x}_1\mathrm{\mid }=\eta$, $\mathrm{\mid }{x}_2\mathrm{\mid }=\gamma$. The Hopf adjunction for $D$ gives $\sum D(x_1,z)D(x_2,y)=q^{(\eta ,\gamma )}D(x,yz)$ for $y\in U^{-}[-\gamma ]$, $z\in U^{-}[-\eta ]$: in ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}x$ the toral pairing is $D(K_{-\eta },K_{\gamma })=q^{-(\eta ,\gamma )}$. Contract the first two root slots of $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Theta }$ with these two functionals; the result is $q^{(\eta ,\gamma )}yz$. The proposed right side has the same contraction, since $x_2{K}_{-\eta }=q^{(\eta ,\gamma )}{K}_{-\eta }{x}_2$ and the two coevaluations contract to $y,z$. This proves the first identity by perfectness. For the second, the degree component of $\mathrm{\Delta }u$ for $u\in U^{-}[-\gamma -\eta ]$ is $u_1{K}_{\eta }\otimes u_2$. The other Hopf adjunction gives $\sum D(a,u_1)D(b,u_2)=D(ab,u)$ for $a\in U^{+}[\gamma ]$, $b\in U^{+}[\eta ]$; the first pairing with $u_1{K}_{\eta }$ has no additional scalar, by the counit term of ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}a$. Contracting the last two root slots on each proposed side therefore gives $ab$. This proves the second identity. All tests are in finite homogeneous components; the toral directions are retained, not discarded.

Identification of the inverse. Put $Y:=B({\mathrm{\Theta }}^{-1})=\overline{{\mathrm{\Theta }}^{-1}}$. Step 1.1 shows $\mathrm{\Delta }(v)Y=Y\bar{\mathrm{\Delta }}(v)$ for every $v$, and $Y_0=B({\mathrm{\Theta }}_0)=1$; by the uniqueness clause of [F1], $Y=\mathrm{\Theta }$, hence ${\mathrm{\Theta }}^{-1}=B(\mathrm{\Theta })=\bar{\mathrm{\Theta }}$. The inverse is taken in the same positive-first/negative-second completion, because the convolution algebra of [F3] is closed under inversion and $B$ is an automorphism of it; no opposite tensor completion is used.

Rank-one check in every degree. Put $t_n=[n{]}_{q_i^2}!=q_i^{n(n-1)\mathrm{/}2}[n{]}_i!$. The Gaussian coproduct expansions are $\mathrm{\Delta }(E_i^{(n)})={\sum }_{\gamma +\eta =n}{q}_i^{\gamma \eta }{E}_i^{(\eta )}\otimes K_i^{-\eta }{E}_i^{(\gamma )}$ and $\mathrm{\Delta }(F_i^{(n)})={\sum }_{\gamma +\eta =n}{q}_i^{-\gamma \eta }{F}_i^{(\gamma )}{K}_i^{\eta }\otimes F_i^{(\eta )}$. Also ${\left(\genfrac{}{}{0px}{}{n}{\gamma }\right)}_i{t}_{\gamma }{t}_{\eta }=q_i^{-\gamma \eta }{t}_n$. In (ii)'s first right side, toral crossing $E_i^{(\gamma )}{K}_i^{-\eta }=q_i^{2\gamma \eta }{K}_i^{-\eta }{E}_i^{(\gamma )}$ and divided-power multiplication produce the coefficient $(q_i-q_i^{-1}{)}^n{t}_n{q}_i^{\gamma \eta }$. In the second right side, no toral crossing is needed and the coefficient is $(q_i-q_i^{-1}{)}^n{t}_n{q}_i^{-\gamma \eta }$. These are precisely the two left-side coproduct coefficients.

The barred coproduct identity. Let $r(u)$ be the negative-half twisted coproduct obtained from $\mathrm{\Delta }u$ by removing $K$-factors, with tensor multiplication twisted by $q^{-(\beta ,\gamma )}$. Its bar counterpart satisfies $\bar{r}(u{)}_{\eta ,\gamma }=q^{(\eta ,\gamma )}\mathrm{flip}(r(u{)}_{\gamma ,\eta })$: both sides send $F_i$ to $F_i\otimes 1+1\otimes F_i$, and the scalar twist verifies multiplicativity on homogeneous products. In $\bar{\mathrm{\Delta }}u$, moving the torus to the left contributes $q^{-(\eta ,\gamma )}$ and cancels this factor. Thus, after removing the middle torus $K_{-\gamma }$, the $(\eta ,\gamma )$ coefficient of $\bar{\mathrm{\Delta }}u$ is the flip of $r(u{)}_{\gamma ,\eta }$. The Hopf multiplication adjunction used in step 1.2 now shows that contraction of the last two root slots of $(\mathrm{i}\mathrm{d}\otimes \bar{\mathrm{\Delta }})\mathrm{\Theta }$ gives $ab$, with $a$ testing degree $\gamma$ and $b$ degree $\eta$. The same contraction of $({\mathrm{\Theta }}_{\gamma }{)}_{13}{K}_{-\gamma }^{(2)}({\mathrm{\Theta }}_{\eta }{)}_{12}$ gives $ab$. Perfectness proves (iv).

The cocycle identity. Summing (ii)'s first identity, and noting that the middle torus commutes with every outer $({\mathrm{\Theta }}_{\eta }{)}_{13}$, gives $((\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Theta }){\mathrm{\Theta }}_{12}={\mathrm{\Theta }}_{23}({\sum }_{\eta }({\mathrm{\Theta }}_{\eta }{)}_{13}{K}_{-\eta }^{(2)}){\mathrm{\Theta }}_{12}$. By (iv) the last two factors are $(\mathrm{i}\mathrm{d}\otimes \bar{\mathrm{\Delta }})\mathrm{\Theta }$. Applying $\mathrm{\Delta }(y)\mathrm{\Theta }=\mathrm{\Theta }\bar{\mathrm{\Delta }}(y)$ in the last two positions to each second-factor coefficient $y$ of $\mathrm{\Theta }$ gives ${\mathrm{\Theta }}_{23}(\mathrm{i}\mathrm{d}\otimes \bar{\mathrm{\Delta }})\mathrm{\Theta }=((\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Theta }){\mathrm{\Theta }}_{23}$. This proves (iii). All operations are coefficientwise in the threefold completion, or finite on the stated admissible vectors. No middle torus is commuted through a $({\mathrm{\Theta }}_{\gamma }{)}_{23}$ term.

Conclusion and normalization. Steps 1.1 and 2.1 prove (i); step 1.2 proves (ii) with the degree-${\alpha }_i$ instances and step 2.2 verifies it independently in rank one to all degrees; step 2.3 proves (iv); step 3.1 proves the cocycle identity (iii). The equation $\mathrm{\Delta }(u)X=X\bar{\mathrm{\Delta }}(u)$ is homogeneous and linear in $X$, so every nonzero scalar multiple of $\mathrm{\Theta }$ satisfies it with ${\mathrm{\Theta }}_0$ scaled equally, and only the normalization ${\mathrm{\Theta }}_0=1$ selects $\mathrm{\Theta }$; all computations are at the field level and involve no integrality, while the finite-support hypotheses of [F3] are checked before any completed series is evaluated.