The rank-one tilting quotient matches the Temperley-Lieb ribbon convention

By [F1], the genuine full-algebra intertwiners $w,b$ satisfy both snakes, $b(w)=-\delta$ and $K{B}^t=-B$ for their bilinear form. Thus the self-duality sign is $-1$, independently of rescaling. Put $e=wb$ and $U=-e=\delta P_0$. The snakes give $U_i^2=\delta U_i$, $U_i{U}_{i+1}{U}_i=U_i$ and distant commutation, so they give every ${\rho }_n$. The two channels and their twists/crossings are exactly [F1]. In particular $T=qU-1=-s{c}^{-1}$, so $q^2$ belongs to the trivial channel, not to the highest-weight-$2$ channel.

The rigid tensor functor from $\mathrm{T}\mathrm{L}(-\delta )$ sends cup to $w$, cap to $b$, and $e_i$ to $w_i{b}_i$. The two snakes and negative loop verify its defining matching relations. The algebra identification $U_i\mapsto -e_i$ preserves the JW projectors: the recursion quantum integers for loop $-\delta$ are $(-1{)}^{j-1}[j]$, and these signs cancel against the generator sign in each recursive correction. Thus the evaluated projector $p_a$ is annihilated by the matrices $U_i$. For $a\le \mathrm{\ell }-1$, their common kernel in each weight space is one-dimensional: $U_i$ imposes that the coefficient of $-+$ is $q$ times that of $+-$, so adjacent swaps determine every coefficient from one word. The vectors $F^{(j)}{v}_{+}^{\otimes a}$, $0\le j\le a$, belong to this kernel and are nonzero: the divided coproduct has one Laurent-monomial coefficient for every specified subset of $j$ changed positions. The projector acts as identity on its common kernel, since its nonidentity generator words kill that kernel. Its image is therefore precisely this $(a+1)$-dimensional highest-weight string, equal to $\mathrm{\Delta }(a)$ by [F4]. In particular $p_{\mathrm{\ell }-1}$ has a nonzero image of zero pivotal dimension, and cannot lie in the kernel of the unreduced map.

In the quantum negligible quotient the boundary image is zero by [F4]. Transport [F2]'s path splits to the negative-loop category. Its down normalization is $-{\lambda }_a$ rather than ${\lambda }_a=[a+1]\mathrm{/}[a]$; replacing its down cap map by its negative yields exactly the positive-algebra normalization for $U=-e$. The transported path units therefore have the same multiplication and positive-$K$ trace weights $[a+1]$. They decompose $V^{\otimes n}$ into all retained $\mathrm{\Delta }(a)$ copies. Hom orthogonality in [F4] shows their rectangular blocks are the entire quantum quotient Hom spaces, not a proper subalgebra. Their positive trace weights show independence, while [F2] identifies the diagram kernel as exactly $I_n$. This proves the displayed centralizer isomorphism. On a chosen retained sum $R_n$, composites through discarded negligible summands are zero because its weighted matrix-trace pairing is nondegenerate. Thus projecting to $R_n$ realizes the same algebra as ${\mathrm{End}}_U(R_n)$; the quotient formulation is independent of that choice.

The same path decomposition on both ends gives fullness and faithfulness for Hom spaces between unequal tensor degrees as well. Every tilting is a sum of summands of powers of $V$ by [F4]. Consequently the functor extends from the additive Karoubi boundary quotient of $\mathrm{T}\mathrm{L}(-\delta )$ to a tensor equivalence with the whole tilting negligible quotient. In the source let $j_0$ be its planar pivotal structure. Its dimension at $p_a$ is $(-1{)}^a[a+1]$. The parity map ${ϵ}_n=(-1{)}^n{1}_n$ is a monoidal natural automorphism: Hom spaces between different parities are zero. It descends to summands with ${ϵ}_{p_a}=(-1{)}^a$ and commutes with duality. Thus $j=ϵj_0$ has the positive dimensions $[a+1]$. More precisely, the equality $K{B}^t=-B$ in step 1.1 says the functor's transported planar pivotal map on $V$ is $-K$; multiplying it by ${ϵ}_V=-1$ gives exactly $K$. Monoidality, direct sums and summands prove equality of pivotal structures on every object, not merely equality of dimensions.

The source crossing $c=s1+s^{-1}e$ is precisely the actual finite-$R$ crossing by [F1]. Its inverse and adjacent triple relations hold by the TL relations. Naturality with cups/caps can also be read directly: for $W_1=w\otimes 1$, $W_2=1\otimes w$, the snakes give $e_1{W}_1=-\delta W_1$, $e_2{W}_1=W_2$, $e_1{W}_2=W_1$, so $c_1{c}_2{W}_1=(s^2-\delta +s^{-2})W_1+W_2=W_2$; the reflected and reversed identities give the other three slides. Thus the functor preserves the full braiding. With $j_0$, the generator curl is $s(-s^2-s^{-2})+s^{-1}=-s^3$. Since $\theta =u^{-1}j$, changing $j_0$ to $ϵj_0$ changes this to $+s^3$. It gives ${\theta }_{p_a}=s^{a(a+2)}$ by balancing on the highest tensor component, as already proved in [F1]. Parity commutes with duality, so the twist remains dual-compatible. The pivotal tensor equivalence therefore preserves braiding and twist globally, hence is a ribbon equivalence. At $\mathrm{\ell }=3$, the removed $a=2$ channel is treated exactly as [F1] specifies.

The scalar composite of the fixed self-dual cup and cap is still $-\delta$; re-pivotalization changes the pivotal closing map, not that composite. Its corrected ribbon unknot closure is instead $\mathrm{tr}(K{\mathrm{\mid }}_V)=+\delta$, and one framing twist acts by $s^3$. Because the equivalence preserves evaluation, coevaluation, braiding and twist, it preserves every composite framed colored tangle and its closure, not just two channel scalars. The evaluation and ribbon-trace suppliers [F5,A1] apply with their stated countable-choice hypothesis only at this last universality step. The preceding steps establish all algebra, kernel, sign and ribbon assertions independently of that axiom and finish the proof.