Right closure is a trace-preserving diagram-tower expectation

On a basis diagram, appending a rightmost vertical strand leaves all existing matching pairs unchanged. Distinct diagrams remain distinct after appending it, so freeness in [F1] makes ${\iota }_n$ injective. Stacking two such appended diagrams creates no new closed component and is the appended version of their original stack, so ${\iota }_n$ preserves products and the unit. When $\delta$ is real, horizontal reflection also commutes with appending, so it preserves the involution. Right-closing the newly appended vertical strand creates exactly one circle, so $c_n({\iota }_n(D))=\delta D$ and $E_n{\iota }_n(D)=D$. Linearity gives $E_n{\iota }_n=\mathrm{i}\mathrm{d}$.

For basis diagrams $a,b$ on $n$ strands and $x$ on $n+1$ strands, stack the three diagrams and attach the right closure arc at the right boundary. Every path or cycle in this finite graph is unchanged if the left $n$-strand stacks are performed before the right closure: the closure arc meets only the two rightmost boundary endpoints, and the factors $a,b$ attach only to the left $n$ interfaces. The exterior matching is therefore the product $a{c}_n(x)b$, with the same number of removed cycles and the same powers of $\delta$. Thus $c_n({\iota }_n(a)x{\iota }_n(b))=a{c}_n(x)b$. Multiplication by ${\delta }^{-1}$ proves bimodularity on basis diagrams; bilinearity extends it.

Closing all remaining strands of $c_n(x)$ produces exactly the same closed matching graph as closing all $n+1$ strands of $x$. The normalized factor for $E_n$ is ${\delta }^{-1}$, so the total normalization changes from ${\delta }^{-n}$ to ${\delta }^{-(n+1)}$, exactly the normalization in [F3] for rank $n+1$. Hence ${\tau }_n(E_n(x))={\tau }_{n+1}(x)$ on basis diagrams and then on all $x$. The retraction identity also gives $({\iota }_n{E}_n{)}^2={\iota }_n{E}_n$ as endomorphisms of ${\mathrm{T}\mathrm{L}}_{n+1}$.

Right-closing ${\iota }_n(x)d_n$ connects the last cap and cup to the exterior closure arc; the result is the diagram $x$ on the first $n$ strands, with no new circle. Thus $c_n({\iota }_n(x)d_n)=x$, so $E_n({\iota }_n(x)d_n)={\delta }^{-1}x$. Trace preservation and [F4] give ${\tau }_{n+1}({\iota }_n(x)d_n)={\tau }_n(E_n({\iota }_n(x)d_n))={\delta }^{-1}{\tau }_n(x)$.

Suppose $x\ge 0$ in the finite-dimensional $C^{\ast }$-algebra ${\mathrm{T}\mathrm{L}}_{n+1}$ and the traces are faithful and positive. On each basis matching, reflecting before closing produces the same retained $(n,n)$ matching and the same closed components as closing before reflecting; the scalar ${\delta }^{-1}$ is real. Hence $E_n(x^{\ast })=E_n(x{)}^{\ast }$, so $E_n(x)$ is self-adjoint. For every $b\in {\mathrm{T}\mathrm{L}}_n$, trace preservation and bimodularity give ${\tau }_n(b^{\ast }{E}_n(x)b)={\tau }_{n+1}({\iota }_n(b{)}^{\ast }x{\iota }_n(b))\ge 0$. If $E_n(x)$ had a negative spectral subspace, its nonzero spectral projection $p$ would satisfy ${\tau }_n(p{E}_n(x)p)<0$, since $p{E}_n(x)p$ is strictly negative on that subspace and ${\tau }_n$ is faithful; this contradicts the same inequality with $b=p$. Thus $E_n(x)\ge 0$. It is unital, idempotent, bimodular and trace-preserving by the preceding steps, so it is a positive conditional expectation in the stated $C^{\ast }$-positive regime.