The specialized Hecke Markov trace descends to the diagram quotient

Since $v+1=q^2+1=q\delta \mathrm{\ne }0$, the assignment $T_i\mapsto q{U}_i-1$ is defined. Its quadratic relation follows from $(q{U}_i-1-v)(q{U}_i)=q^2(U_i-\delta )U_i=0$, using $v=q^2$ and $U_i^2=\delta U_i$. For adjacent generators, expanding both braid words gives difference $q^3(U_i{U}_{i+1}{U}_i-U_{i+1}{U}_i{U}_{i+1})-q^2\delta (U_i-U_{i+1})+q(U_i-U_{i+1})=0$, since $q^2\delta =q^3+q$. Distant commutation is inherited, so the assignment defines a tower homomorphism; it is surjective by [F4].

Put $E_i={\sum }_{w\in ⟨s_i,s_{i+1}⟩}{T}_w$, the six-term expression in the Statement. Expanding with the Hecke quadratic and braid relations gives $(T_i+1)(T_{i+1}+1)(T_i+1)=E_i+v(T_i+1)$. Under $T_j+1=q{U}_j$, the left side maps to $q^3{U}_i{U}_{i+1}{U}_i=q^3{U}_i$ and the second term maps to $q^3{U}_i$, so ${\varphi }_n(E_i)=0$. Since $J_n$ is the two-sided ideal generated by the $E_i$, ${\varphi }_n(J_n)=0$ and ${\varphi }_n$ factors through $H(n)\mathrm{/}{J}_n$.

The pullback ${\tilde{\tau }}_n:={\tau }_n\circ {\varphi }_n$ is cyclic and has ${\tilde{\tau }}_n(1)=1$ by [F3]. For $n\ge 1$ and $x\in H(n)$, its image in ${\mathrm{T}\mathrm{L}}_{n+1}$ is ${\varphi }_n(x)\otimes 1$, and $T_n$ maps to $q{U}_n-1$, so ${\tilde{\tau }}_{n+1}(x{T}_n)=(q\mathrm{/}\delta -1){\tau }_n({\varphi }_n(x))=-{\tilde{\tau }}_n(x)\mathrm{/}(q\delta )=-{\tilde{\tau }}_n(x)\mathrm{/}(q^2+1)=z_0{\tilde{\tau }}_n(x)$. The tower restriction rule follows from [F3]; hence $\tilde{\tau }$ satisfies the specialized Markov rules.

The presentation [F4] gives ${\psi }_n:{\mathrm{T}\mathrm{L}}_n\to H(n)\mathrm{/}{J}_n$ by $U_i\mapsto q^{-1}(T_i+1)$. The Hecke quadratic relation gives $U_i^2=\delta U_i$ for these images, while step 2.1 in the quotient gives $(T_i+1)(T_{i+1}+1)(T_i+1)=q^2(T_i+1)$ and hence $U_i{U}_{i+1}{U}_i=U_i$; the other adjacent relation follows by interchanging indices and distant commutation is inherited. Thus ${\psi }_n$ is well defined. Its composites with the map induced by ${\varphi }_n$ fix each generator: $U_i\mapsto q^{-1}(q{U}_i-1+1)=U_i$ and $T_i\mapsto q(q^{-1}(T_i+1))-1=T_i$. Therefore $H(n)\mathrm{/}{J}_n\cong {\mathrm{T}\mathrm{L}}_n$, proving the quotient isomorphism.

The decomposition in [F5] writes every element of $H(n+1)$ as a sum of an element of $H(n)$ and terms $x{T}_{n}y$ with $x,y\in H(n)$. For any Markov trace, restriction determines the first term; cyclicity and (M4) force ${\mathrm{tr}}_{n+1}(x{T}_{n}y)=z{\mathrm{tr}}_n(xy)$ on each remaining term. Induction from $H(1)=\mathbb{C}$ therefore shows at most one specialized trace can satisfy (M1)--(M4), even when $v,z$ have been specialized: this uses only the free decomposition [F5], not division by a parameter. By step 2.2 the pullback $\tilde{\tau }$ is such a trace, so it is the specialized Ocneanu trace. Since ${\varphi }_n(J_n)=0$, the trace vanishes on all of the two-sided ideal $J_n$. The quotient isomorphism of step 3.1 identifies the descended functional with the normalized closure trace ${\tau }_n$.