subalgebra $Z_0(Par_t)$ of $Z(Par_t)$ which acts by different scalars
on the irreducible modules $L(\lambda)$ and $L(\mu)$.
It follows in these cases that $Z_0(Par_t)$ is a dense subalgebra of
the pseudo-compact topological algebra  $Z(Par_t)$[
These algebras are certainly not equal since $Z(Par_t) \cong \prod_{\lambda \in \P} \kk_\lambda$ is of uncountable dimension.].
We do not expect that this is the case when $t \in \N$, but nevertheless
$Z_0(Par_t)$ is still sufficiently large to separate blocks.
This will be established in <ref> below, which shows
for any value of $t$
that  the relations $\sim_t$ and $\approx_t$ coincide, so that “blocks” are blocks, and <ref> is always the same decomposition as <ref>; see also

§.§ Special projective functors

From now on, we will primarily be
interested in parameter values $t \in \N$, so that
$Par_t$ is not semisimple.
Consider the atypical block
$\{\kappa^{(0)}, \kappa^{(1)}, \kappa^{(2)},\dots\}$ associated to
$\kappa \in \P_t$.
From <ref>, it follows that
$\kappa^{(n)}$ is obtained from $\kappa^{(n-1)}$ by adding
$\kappa_n - \kappa_{n+1}+1$ nodes to the $n$th row of its Young diagram, leaving all other rows unchanged.

The partition $\kappa^{(0)}$ is the smallest
of all of the $\kappa^{(n)}$, hence, it is maximal in the highest weight ordering from <ref>.
It follows that

\begin{align}\label{topinblock}
P(\kappa^{(0)}) &\cong \Delta(\kappa^{(0)}).
\end{align}

The indecomposable projectives
$\Delta(\kappa^{(0)})$ are exactly the ones of non-zero
categorical dimension mentioned already in <ref>, with
the irreducible $\kk S_t$-module associated to the image of $\Delta(\kappa^{(0)})$
under the equivalence $\overline{\psi}_t$ between the semisimplification of $\Kar(\Par_t)$ and
$\kk S_t\Modfd$
being the Specht module $S(\kappa)$.
It is also useful to note for $t \in \N$ and $\kappa \in \P_t$ that
the associated block $\{\kappa^{(0)}, \kappa^{(1)},\dots\}$ is
the set $S(\gamma)$ from <ref> for

\begin{equation}\label{kappatogamma}
\gamma := (\eps_{\kappa_1} - \eps_t) + (\eps_{\kappa_2-1}-\eps_{-1})
+\cdots + (\eps_{\kappa_t-t+1} - \eps_{-t}) \in P.
\end{equation}

This is follows easily using <ref>.

In order to understand the structure of the atypical blocks more fully,
we are going to use the endofunctor
$\mid\,\star:\Par_t\rightarrow \Par_t$.

\begin{align}
D &:= \res_{\mid\,\star}= 1_{\mid\,\star}Par_t  \otimes_{Par_t}:Par_t\Mod\rightarrow Par_t\Mod
%  D^\vee &:= \ind_{|\,\star}= Par_t 1_{|\,\star} \otimes_{Par_t}:Par_t\Mod\rightarrow Par_t\Mod
\end{align}

be the corresponding restriction functor from
<ref>. This obviously preserves locally finite-dimensional modules.
The object $\mid$ is self-dual so, by <ref>,
the restriction functor $D$ is isomorphic to the induction functor
By <ref>, $D$ is a
self-adjoint projective functor,
so it preserves finitely generated projectives (and finitely cogenerated injectives).
To make the canonical adjunction as explicit as possible, we note that
its unit and counit
are induced by the bimodule homomorphisms

\begin{align}\label{etaadj}
\eta&:Par_t\to 1_{\mid\,\star} Par_t\otimes_{Par_t} 1_{\mid\,\star} Par_t,
\draw[-,thick] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (.4,-.6) {$\stringlabel{1}$};
\node at (-.4,-.6) {$\stringlabel{n}$};
\node at (.4,.6) {$\stringlabel{1}$};
\node at (-.4,.6) {$\stringlabel{m}$};
\end{tikzpicture}\mapsto\begin{tikzpicture}[scale=1.2,anchorbase]
\draw[-,thick] (-0.7,-0.5)to(-0.7,.5);
\draw[-,thick] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (.4,-.6) {$\stringlabel{1}$};
\node at (-.4,-.6) {$\stringlabel{n}$};
\node at (-.7,-.6) {$\stringlabel{n\!+\!1}$};
\node at (.4,.6) {$\stringlabel{1}$};
\node at (-.4,.6) {$\stringlabel{m}$};
\node at (-.7,.6) {$\stringlabel{m\!+\!1}$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[anchorbase,scale=1.2]
\draw[-,thick] (-1.1,0.5)to(-1.1,0)to[out=down,in=left](-0.9,-0.2)to[out=right,in=down](-0.7,0)to(-0.7,0.5);
\draw[-,thick] (-0.4,0.5)--(-0.4,-0.5);
\draw[-,thick] (0.4,0.5)--(0.4,-0.5);
\node at (-0.2,0.05) {$\cdot$};
\node at (-0,0.05) {$\cdot$};
\node at (0.2,0.05) {$\cdot$};
\node at (.4,.6) {$\stringlabel{1}$};
\node at (-.4,.6) {$\stringlabel{n}$};
\node at (.4,-.6) {$\stringlabel{1}$};
\node at (-.4,-.6) {$\stringlabel{n}$};
\node at (-.72,.6) {$\stringlabel{n\!+\!1}$};
\node at (-1.12,.6) {$\stringlabel{n\!+\!2}$};
\end{tikzpicture}\ ,\\\label{epsadj}
\varepsilon&:1_{\mid\,\star}Par_t \otimes_{Par_t} 1_{\mid\,\star} Par_t\rightarrow Par_t, &
\begin{tikzpicture}[anchorbase,scale=1.2]
\draw[-,thick] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.45,0.2)--(-0.45,0.5);
\draw[-,thick] (-0.25,0.2)--(-0.25,0.5);
\draw[-,thick] (0.45,0.2)--(0.45,0.5);
\node at (-0.05,0.4) {$\cdot$};
\node at (.1,0.4) {$\cdot$};
\node at (0.25,0.4) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (.4,-.6) {$\stringlabel{1}$};
\node at (-.4,-.6) {$\stringlabel{n}$};
\node at (.45,.6) {$\stringlabel{1}$};
\node at (-.25,.6) {$\stringlabel{m}$};
\node at (-.5,.6) {$\stringlabel{m\!+\!1}$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[anchorbase,scale=1.2]
\draw[-,thick] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (-0.2,0.2)--(-0.2,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (0.1,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (.4,.6) {$\stringlabel{1}$};
\node at (-.2,.6) {$\stringlabel{n}$};
\node at (.4,-.6) {$\stringlabel{1}$};
\node at (-.4,-.6) {$\stringlabel{n\!+\!1}$};
\node at (-.45,.6) {$\stringlabel{n\!+\!1}$};
\end{tikzpicture}
\mapsto\begin{tikzpicture}[anchorbase,scale=1.2]
\draw[-,thick] (-0.5,-0.5)--(-0.5,-0.1)--(0.5,-0.1)--(0.5,-0.5)--(-0.5,-0.5);
\draw[-,thick] (-0.5,0.5)--(-0.5,0.1)--(0.5,0.1)--(0.5,0.5)--(-0.5,0.5);
\draw[-,thick] (-0.4,-0.5)--(-0.4,-0.7);
\draw[-,thick] (0.4,-0.7)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.1)--(-0.2,-0.1);
\draw[-,thick] (-0.4,-.1) to[out=120,in=down] (-0.7,.3) to (-.7,.5) to[out=up,in=up,looseness=2] (-.45,.5);
\draw[-,thick] (0.4,0.1)--(0.4,-0.1);
\draw[-,thick] (0.45,0.7)--(0.45,0.5);
\draw[-,thick] (-0.25,0.7)--(-0.25,0.5);
\node at (0,0.3) {$\scriptstyle{f}$};
\node at (0,-0.3) {$\scriptstyle{g}$};
\node at (-0.1,-0.03) {$\cdot$};
\node at (0.05,-0.03) {$\cdot$};
\node at (0.2,-0.03) {$\cdot$};
\node at (-0.2,-0.62) {$\cdot$};
\node at (0,-0.62) {$\cdot$};
\node at (0.2,-0.62) {$\cdot$};
\node at (-0.05,0.62) {$\cdot$};
\node at (.1,0.62) {$\cdot$};
\node at (0.25,0.62) {$\cdot$};
\node at (.4,-.8) {$\stringlabel{1}$};
\node at (-.4,-.8) {$\stringlabel{n\!+\!1}$};
\node at (.4,.8) {$\stringlabel{1}$};
\node at (-.25,.8) {$\stringlabel{m}$};
\end{tikzpicture}\ .
\end{align}

Using <ref>, it follows that
$D$ commutes with the duality
$?^\sigmadual$ on $Par_t\Modlfd$.

For $\lambda \in \P$, there is
a filtration
$0=V_0 \subseteq V_1 \subseteq V_2 \subseteq V_3 = D \Delta(\lambda)$
such that

\begin{align*}
V_3/V_2 &\cong \bigoplus_{a \in \add(\lambda)} \Delta\left(\lambda+\boxed{a}\right),\\
V_2 / V_1 &\cong \Delta(\lambda)\oplus
\bigoplus_{b \in \rem(\lambda)}
\bigoplus_{a \in \add(\lambda - \boxed{b})}
\Delta\left((\lambda - \boxed{b})+\boxed{a}\right),\\
V_1 / V_0 &\cong \bigoplus_{b \in \rem(\lambda)} \Delta\left(\lambda-\boxed{b}\right).
\end{align*}

By <ref>, $D$ is isomorphic to the functor
$Q(\boxempty) \ostar?$ defined by taking the induction product with
the projective module $Q(\boxempty)$.
By <ref>(iii) and <ref>,
$Q(\boxempty)$ has a $\Delta$-flag of length two with sections
$\Delta(\boxempty)$ at the top and $\Delta(\varnothing)$ at the bottom.
Applying <ref>, we deduce that
$D \Delta(\lambda) \cong Q(\boxempty) \ostar \Delta(\lambda)$
has a $\Delta$-flag with one section
$\Delta(\varnothing) \ostar \Delta(\lambda) \cong \Delta(\lambda)$
and other sections arising from the $\Delta$-flag
of $\Delta(\boxempty)\ostar \Delta(\lambda)$ described in that theorem.
For this, one just needs to know the values of the
reduced Kronecker coefficients $\overline{G}_{\boxvoid,\lambda}^{\mu}$ which were worked out in <ref>.
The $\Delta$-flag can be ordered in the way described
since $\Ext^1(\Delta(\mu),\Delta(\nu)) = 0$ if $|\mu| \leq |\nu|$.

<ref> also follows from <ref> below, which
constructs the filtration explicitly. The proof of <ref> is also valid over fields of positive characteristic.

Now we are going to use the affine partition category $\APar$
to decompose the endofunctor $D$ as a direct sum of special projective functors
$D_{b|a}$. The approach here is analogous to the way the affine symmetric category
$\ASym$ was used to decompose
$E$ and $F$ as direct sums of
$E_a$ and $F_b$ in <ref>.
As noted at the end of <ref>,
$\Par_t$ is isomorphic to the quotient of $\APar$ by a left tensor ideal.
Hence, $\Par_t$ is a strict $\APar$-module category.
The self-adjoint functor $D$ is also the restriction functor $\res_{\mid\,\star}$
arising from this categorical action of $\APar$ on $\Par_t$.
Now the left and right dots give us natural transformations

\begin{align*}
\alpha &:= \begin{tikzpicture}[anchorbase,scale=1.2]
\draw[-,thick](0,0)--(0,0.4);
\draw[-,thick](0,0.2)--(0.15,0.2);
\node  at (0.15,0.2) {$\bullet$};
\node at (.4,.2) {$\star$};
\end{tikzpicture}:\mid\,\star \Rightarrow \mid\,\star,&
\beta &:= \begin{tikzpicture}[anchorbase,scale=1.2]
\draw[-,thick](0,0)--(0,0.4);
\draw[-,thick](0,0.2)--(-0.15,0.2);
\node  at (-0.15,0.2) {$\bullet$};
\node at (.2,.2) {$\star$};
\end{tikzpicture}:\mid\,\star\Rightarrow\mid\,\star.
\end{align*}

Applying the general construction from <ref> to these,
we obtain commuting endomorphisms

\begin{align}
x &:= \res_\alpha:D \Rightarrow D,
y &:= \res_\beta:D \Rightarrow D.
\end{align}

Let $D_{b|a}$ be the summand of $D$ that is the
simultaneous generalized eigenspace of $x$ and $y$ of eigenvalues $a$ and $b$, respectively.
Explicitly, $D = \res_{\mid\,\star}$ is defined by tensoring
with the bimodule
$1_{\mid\,\star} Par_t$, and
the endomorphisms $x$ and $y$ of $D$
are induced by the bimodule endomorphisms $\rho$ and $\lambda$ of $1_{\mid\,\star} Par_t$
given by left multiplication by $x^R_{m+1}$ and $x^L_{m+1}$, respectively,
on the summand $1_{m+1}  Par_t$ of $1_{\mid\,\star} Par_t$ for each $m \geq 0$.
Then, $D_{b|a}$ is the functor defined by tensoring with the summand of
$1_{\mid\,\star} Par_t$ that is the
simultaneous generalized eigenspaces
of $\rho$ and $\lambda$ for
the eigenvalues $a$ and $b$, respectively.
As $1_{m+1} Par_t = \bigoplus_{n \geq 0} 1_{m+1} Par_t 1_n$
with each $1_{m+1} Par_t 1_n$ being finite-dimensional,
these endomorphisms are locally finite, so we have that

\begin{equation}\label{summertime}
D = \bigoplus_{a,b \in \kk} D_{b|a}.
\end{equation}

For $a,b \in \kk$, the endofunctor $D_{b|a}$ commutes with the duality $?^\sigmadual$, i.e., $D_{b|a} \circ ?^\sigmadual \cong ?^\sigmadual \circ D_{b|a}$.

This follows
from the fact that $D$ commutes with the duality $?^\sigmadual$, and $\sigma$ fixes both the left dot and the right dot.

For $a,b \in \kk$, the endofunctors $D_{b|a}$ and $D_{a|b}$ are biadjoint.

The adjunction $(D_{a|b}, D_{b|a})$ is induced by the self-adjunction
of $D$. The unit $\bar \eta$ of adjunction comes from the bimodule homomorphism
that is the composition of the unit $\eta$ from <ref> with the projection onto the generalized $a$ and $b$ eigenspaces of $\rho$ and $\lambda$ on the left tensor factor and the generalized $b$ and $a$ eigenspaces of $\rho$ and $\lambda$ on the right tensor factor.
The counit $\bar\eps$ of adjunction comes from the composition of the counit $\eps$ from <ref> with the inclusion
of the generalized $b$ and $a$ eigenspaces of $\rho$ and $\lambda$ on the left tensor factor and the generalized $a$ and $b$ eigenspaces of $\rho$ and $\lambda$ on the right tensor factor. To check the zig-zag identities, one just needs to use the relations

\begin{align*}
\begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0.5) to (0,0.35) to[out=down,in=left] (0.35,0) to[out=right,in=down] (0.7,0.35) to (0.7,0.5);
\draw[-,thick] (0.35,0.35)--(0.7,0.35);
\node at (0.35,0.35) {$\bullet$};
\end{tikzpicture}\ &=\ \begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0.5) to (0,0.35) to[out=down,in=left] (0.35,0) to[out=right,in=down] (0.7,0.35) to (0.7,0.5);
\draw[-,thick] (0.35,0.35)--(0,0.35);
\node at (0.35,0.35) {$\bullet$};
\end{tikzpicture}\ ,&\begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0) to (0,0.15) to[out=up,in=left] (0.35,0.5) to[out=right,in=up] (0.7,0.15) to (0.7,0);
\draw[-,thick] (0.35,0.15)--(0.7,0.15);
\node at (0.35,0.15) {$\bullet$};
\end{tikzpicture}\ &=\ \begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0) to (0,0.15) to[out=up,in=left] (0.35,0.5) to[out=right,in=up] (0.7,0.15) to (0.7,0);
\draw[-,thick] (0.35,0.15)--(0,0.15);
\node at (0.35,0.15) {$\bullet$};
\end{tikzpicture}\ ,&\begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0.5) to (0,0.35) to[out=down,in=left] (0.35,0) to[out=right,in=down] (0.7,0.35) to (0.7,0.5);
\draw[-,thick] (0.95,0.35)--(0.7,0.35);
\node at (0.95,0.35) {$\bullet$};
\end{tikzpicture}\ &=\ \begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0.5) to (0,0.35) to[out=down,in=left] (0.35,0) to[out=right,in=down] (0.7,0.35) to (0.7,0.5);
\draw[-,thick] (-0.25,0.35)--(0,0.35);
\node at (-0.25,0.35) {$\bullet$};
\end{tikzpicture}\ ,&\begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0) to (0,0.15) to[out=up,in=left] (0.35,0.5) to[out=right,in=up] (0.7,0.15) to (0.7,0);
\draw[-,thick] (0.95,0.15)--(0.7,0.15);
\node at (0.95,0.15) {$\bullet$};
\end{tikzpicture}\ &=\ \begin{tikzpicture}[anchorbase]
\draw[-,thick] (0,0) to (0,0.15) to[out=up,in=left] (0.35,0.5) to[out=right,in=up] (0.7,0.15) to (0.7,0);
\draw[-,thick] (-0.25,0.15)--(0,0.15);
\node at (-0.25,0.15) {$\bullet$};
\end{tikzpicture}\ ,
\end{align*}

i.e., the fact that the left and right dots are duals.

When $a \neq b$, <ref> can also be proved
a bit more easily using the description of $D_{b|a}$ given in the following lemma,
since the projection functor $\PR_\gamma$ commutes with $?^\sigmadual$ thanks to <ref>.

Let $\PR_\gamma$ be the projection functor defined by <ref>.
If $a \neq b$ then

D_{b|a} \cong \bigoplus_{\gamma\in P} \PR_{\gamma + \alpha_a-\alpha_b}
\circ D \circ \PR_\gamma.

Also $\bigoplus_{\gamma \in P} \PR_\gamma \circ D \circ \PR_\gamma\cong\bigoplus_{a \in \kk} D_{a|a}$.

Take a module $V$ in the “block” parametrized by $\gamma \in P$,
so that $\wt_t(\lambda) = \gamma$ for all irreducible subquotients of $V$.
We need to show that $D_{b|a} V$ is in the “block” parametrized
by $\gamma + \alpha_a - \alpha_b$.
Since $D_{b|a}$ is exact, we may assume that $V$ is irreducible, so $V = L(\lambda)$ for $\lambda \in \P$ with $\wt_t(\lambda) = \gamma$.
The module $D V = 1_{\mid\,\star} Par_t \otimes_{Par_t} V \cong
1_{\mid\,\star} V$ is generated by the finite-dimensional vector spaces
$1_{m+1} V$ for all $m \geq 0$.
Hence, $D_{b|a} V$ is generated by the simultaneous generalized eigenspaces of
$x_{m+1}^R$ and $x_{m+1}^L$ on $1_{m+1} V$
of eigenvalues $a$ and $b$, respectively. Consequently, if $L(\mu)$ is an irreducible subquotient of
$D_{b|a} V$, then $c(u)$ must act on $L(\mu)$ in the same way as
$\mid\star c_{m}(u)$ acts on a simultaneous eigenvector $v \in 1_{m+1} V$ for
$x_{m+1}^R$ and $x_{m+1}^L$ of eigenvalues $a$ and $b$.
Also $c_{m+1}(u)$ acts on $v \in V$ as multiplication by $\chi_\lambda(c(u))$,
the rational function displayed on the right hand side of <ref>.
Using <ref>, we deduce that

\chi_\mu(c(u))= \frac{\alpha_a(u)}{\alpha_b(u)} \times \chi_\lambda(c(u)).

Hence, $\wt_t(\mu) = \wt_t(\lambda)+\alpha_a - \alpha_b$.

Our main combinatorial result about the functors $D_{b|a}$ is as follows.

For $\lambda \in \P$ and $a,b \in \kk$, there is
a filtration
$0=V_0 \subseteq V_1 \subseteq V_2 \subseteq V_3 = D_{b|a} \Delta(\lambda)$
such that

\begin{align*}
V_3/V_2 &\cong \left\{\begin{array}{l}\Delta(\lambda+\boxed{a})\\
\text{if $a \in \add(\lambda)$ and $b=t-|\lambda|$}\\
\text{otherwise,}\end{array}\\
V_2 / V_1 &\cong
\left\{\begin{array}{ll}
\Delta(\lambda)\oplus \Delta(\lambda)\\
\Delta(\lambda)\\
\Delta\left((\lambda - \boxed{b})+\boxed{a}\right)\\
\begin{array}{l}
\text{if $t-|\lambda| = a=b \in \rem(\lambda)$}\\
\text{if $t-|\lambda| \neq a=b \in \rem(\lambda)$ or $t-|\lambda|=a=b\notin\rem(\lambda)$}\\
\text{if $a \neq b \in \rem(\lambda)$ and $a \in \add(\lambda-\boxed{b})$}\\
\text{otherwise},
\end{array}\\
V_1/V_0 &\cong \left\{
\begin{array}{l}\Delta(\lambda-\boxed{b})\\
&&\begin{array}{l}\text{if $a=t-|\lambda|+1$ and $b \in \rem(\lambda)$}\\
\text{otherwise.}\end{array}
\end{align*}

In particular, when $t \in \Z$, the functor $D_{b|a}$ is zero unless both
$a$ and $b$ are integers.

See <ref> below.

The following corollary
is an immediate consequence of
the theorem, but actually it has a much easier proof which we include below.

For $\lambda \in \P$ and $a,b \in \kk$ with $a \neq b$, there is
a filtration
$0=V_0 \subseteq V_1 \subseteq V_2 \subseteq V_3 = D_{b|a} \Delta(\lambda)$
such that

\begin{align*}
V_3/V_2 &\cong
\left\{\begin{array}{l}\Delta(\lambda+\boxed{a})\\
\begin{array}{l}\text{if $a \in \add(\lambda)$ and $b=t-|\lambda|$}\\
\text{otherwise,}\end{array}\\
V_2 / V_1 &\cong
\left\{\begin{array}{l}
\Delta\left((\lambda - \boxed{b})+\boxed{a}\right)\\
\text{if $b \in \rem(\lambda)$ and $a \in \add(\lambda-\boxed{b})$}\\
\text{otherwise},
\end{array}\\
V_1/V_0&\cong \left\{\begin{array}{l}    \Delta(\lambda-\boxed{b}) \\
&&    \begin{array}{l}\text{if $a=t-|\lambda|+1$ and $b \in \rem(\lambda)$}\\
\text{otherwise.}\end{array}
\end{align*}

Let $\gamma := \wt_t(\lambda)$.
By <ref>, we can compute $D_{b|a} \Delta(\lambda)$
by applying $\PR_{\gamma+\alpha_a-\alpha_b}$ to the $\Delta$-flag for $D \Delta(\lambda)$ from <ref>. This produces a module with a $\Delta$-flag consisting of all $\Delta(\mu)$ in the original $\Delta$-flag such that
$\wt_t(\mu) -\wt_t(\lambda)=\alpha_a - \alpha_b$.
It just remains to compute $\wt_t(\mu)-\wt_t(\lambda)$ for the various possible
If $\mu = \lambda + \boxed{c}$ for $c \in \add(\lambda)$
then, by a computation using the first equality from <ref>,
we have that $\wt_t(\mu)-\wt_t(\lambda) =
\alpha_c-\alpha_{t-|\lambda|}$; for this to equal $\alpha_a-\alpha_b$
we must have $b=t-|\lambda|$ and $c=a$.
If $\mu = \lambda - \boxed{d}$ for $d \in \rem(\lambda)$
then, by a similar computation, $\wt_t(\mu)-\wt_t(\lambda)
= \alpha_{t-|\lambda|+1}-\alpha_d$; for this to equal $\alpha_a-\alpha_b$
we must have $d=b$ and $a=t-|\lambda|+1$.
Finally if $\mu=(\lambda - \boxed{d})+\boxed{c}$ for $d \in \rem(\lambda)$
and $c \in \add(\lambda-\boxed{d})$
then $\wt_t(\mu)-\wt_t(\lambda) = \alpha_c-\alpha_d$;
for this to equal $\alpha_a-\alpha_b$ we must have $c=a$ and $d=b$.

§.§ Blocks

We assume throughout the subsection that $t \in \N$. We are going to
describe the structure of the atypical “blocks,” revealing in particular that
they are indecomposable, hence, they are actually blocks.
Recall from <ref> that the atypical “blocks” are parametrized by partitions $\kappa \in \P_t$, with the irreducible modules in the “block” being
the ones labelled by the partitions
$\{\kappa^{(0)}, \kappa^{(1)}, \dots\}$. This is the set
$S(\gamma)$ from <ref> where $\gamma \in P$
is obtained from $\kappa$ according to <ref>.

The first step is to show that all of the atypical “blocks” are equivalent to each other. The proof of this uses the special projective functors
$D_{b|a}$ with $a \neq b$. These are the ones which can be defined just using
information about central characters rather than requiring
the Jucys-Murphy elements;
cf. <ref>.
In view of <ref>,
this sort of information was already available to Comes and Ostrik
in an equivalent form, and indeed they were also able to prove
a similar result by an analogous argument; see
<cit.> and <cit.>.

Let $\kappa$ and $\tilde\kappa$ be partitions of $t$ such that $\tilde\kappa$ is obtained from $\kappa$ by moving a node from the first row of its Young diagram
to its $(r+1)$th row for some $r \geq 1$.
Let $a := \kappa_{r+1}-r+1$ and $b := \kappa_1$.
Then for all $n \geq 0$ we have that $D_{b|a} \Delta(\kappa^{(n)}) \cong \Delta(\tilde\kappa^{(n)})$ and $D_{a|b} \Delta(\tilde\kappa^{(n)}) \cong \Delta(\kappa^{(n)})$.

Let $\gamma, \tilde\gamma \in P$ be defined from $\kappa$ and $\tilde\kappa$
according to <ref>. From this formula it follows that
$\tilde\gamma = \gamma + \alpha_a - \alpha_b$ where $a = \kappa_{r+1}-r+1$ and $b = \kappa_1$ as in the statement of the lemma.
Note that $a \neq b$.
So we can apply <ref> to see that
$D_{b|a} \Delta(\kappa^{(n)}) = \PR_{\gamma + \alpha_a - \alpha_b} (D \Delta(\kappa^{(n)}))$
and $D_{a|b} \Delta(\tilde\kappa^{(n)}) = \PR_{\gamma - \alpha_a+\alpha_b}
(D \Delta(\tilde\kappa^{(n)}))$.
Now we use this description to show that
$D_{b|a} \Delta(\kappa^{(n)}) \cong \Delta(\tilde\kappa^{(n)})$. The
proof that $D_{a|b} \Delta(\tilde\kappa^{(n)}) \cong \Delta(\kappa^{(n)})$ is similar and we leave this to the reader.

Fix $n \geq 0$ and
let $B_n$ be the set of $\mu \in \P$ which are obtained from $\kappa^{(n)}$
by removing a node, removing a node then adding a different
node, or adding a node.
Bearing in mind that $a \neq b$,
the standard modules $\Delta(\mu)$ for $\mu \in B_n$ include all
of the ones which are sections of the $\Delta$-flag from <ref>
which could possibly be in the same block as $\Delta(\tilde\kappa^{(n)})$.
Now it suffices to show for $m \geq 0$
that $\tilde\kappa^{(m)} \in B_n$ if and only if $m=n$.
There are four cases to consider.

Case one: $n=0$. We have that $\kappa^{(0)} = (\kappa_2,\kappa_3,\dots,\kappa_{r+1},\dots)$
and $\tilde\kappa^{(0)} = (\kappa_2,\kappa_3,\dots,\kappa_{r+1}+1,\dots)$, which
is $\kappa^{(0)}$ with one node added to the $r$th row of
its Young diagram.
We definitely have that $\tilde\kappa^{(0)} \in B_0$.
All other $\mu \in B_0$ satisfy
$|\mu| \leq |\tilde\kappa^{(0)}|$.
Since all $\tilde\kappa^{(m)}$ with $m > 0$
have $|\tilde\kappa^{(m)}| > |\tilde\kappa^{(0)}|$,
none of these belong to $B_0$.

Case two: $1 \leq n < r$.
We have that $\kappa^{(n)} = (\kappa_1+1,\kappa_2+1,\dots,\kappa_n+1,\dots,\kappa_{r+1},\dots)$ and $\tilde\kappa^{(n)} = (\kappa_1,\kappa_2+1,\dots,\kappa_n+1,\dots,\kappa_{r+1}+1,\dots)$, which is $\kappa^{(n)}$ with a node removed from the first row and a node added to the $r$th row of its Young diagram.
We definitely have that $\tilde\kappa^{(n)} \in B_n$.
For $m < n$,  $\tilde\kappa^{(m)}$ is of smaller size than
$\kappa^{(n)}$ and its $r$th row is of length $\kappa_{r+1}+1$. This cannot be obtained
from $\kappa^{(n)}$ by removing a node since $\kappa^{(n)}$ has $r$th row of length $\kappa_{r+1}$. So it does not belong to $B_n$.
For $m > n$, $\tilde\kappa^{(m)}$ is of greater size than
$\kappa^{(n)}$ and its first row is of length $\kappa_1$. This cannot be obtained from $\kappa^{(n)}$ by adding a node since $\kappa^{(n)}$ has first row of length $\kappa_1+1$. So again it does not belong to $B_n$.

Case three: $n=r$.
We have that $\kappa^{(n)} = (\kappa_1+1,\kappa_2+1,\dots,\kappa_{r}+1,\kappa_{r+2},\dots)$ and $\tilde\kappa^{(n)} = (\kappa_1,\kappa_2+1\dots,\kappa_{r}+1,\kappa_{r+2},\dots)$, which is $\kappa^{(n)}$ with a node removed from the first row of its Young diagram.
We definitely have that $\tilde\kappa^{(n)} \in B_n$.
The $\tilde\kappa^{(m)}$ with $m < n$ have $|\tilde\kappa^{(m)}|\leq|\tilde\kappa^{(n)}| -1= |\kappa^{(n)}|-2$ so are not elements of $B_n$.
The $\tilde\kappa^{(m)}$ with $m > n$ have
$(r+1)$th row of length $\kappa_{r+1}+2$,
so these are not elements of $B_n$ either since this is at least two more than the length of the $(r+1)$th row of $\kappa^{(n)}$.

Case four: $n > r$.
We have that $\kappa^{(n)} = (\kappa_1+1,\kappa_2+1,\dots,\kappa_{r+1}+1,\dots)$ and $\tilde\kappa^{(n)} = (\kappa_1,\kappa_2+1,\dots,\kappa_{r+1}+2,\dots)$, which is $\kappa^{(n)}$ with
a node removed from its first row and a node added to its $(r+1)$th row.
We definitely have that $\tilde\kappa^{(n)} \in B_n$.
The $\tilde\kappa^{(m)}$ with $m > n$ are of greater size than $\kappa^{(n)}$
and have first row of length $\kappa_1$; these cannot be obtained by adding a node to $\kappa^{(n)}$.
The $\tilde\kappa^{(m)}$ with $r+1\leq m < n$ are of smaller size than $\kappa^{(n)}$
and have $(r+1)$th row of length $\kappa_{r+1}+2$; these cannot be obtained by removing a node from $\kappa^{(n)}$.
The $\tilde\kappa^{(m)}$ with $ m \leq r$
have first row of length $\leq \kappa_1$ and $(r+1)$th row of length
$\kappa_{r+2}$, whereas these two rows of
$\kappa^{(n)}$ are of lengths $\kappa_1+1$ and $\kappa_{r+1}+1 > \kappa_{r+2}$,
so these are not elements of $B_n$.

Let $\kappa$ and $\tilde\kappa$ be partitions of $t$,
denoting the associated $\sim_t$-equivalence classes by $S := \{\kappa^{(0)}, \kappa^{(1)}, \dots\}$ and
$\widetilde{S} := \{\tilde\kappa^{(0)}, \tilde\kappa^{(1)}, \dots\}$. There is an equivalence of categories

\Sigma:1_S Par_t\Mod \rightarrow 1_{\widetilde{S}} Par_t\Mod

between the corresponding “blocks”
such that $\Sigma L(\kappa^{(n)}) \cong L(\tilde\kappa^{(n)})$
for all $n \geq 0$.
The functor $\Sigma$ is
a composition of the special projective functors $D_{b|a}\:(a \neq b)$, hence, it is
a projective functor.

We may assume that $\tilde\kappa$ is obtained from $\kappa$ by moving a node
from the first row of its Young diagram to its $(r+1)$th row for some $r \geq 1$. Thus, we are in the situation of <ref>. The lemma gives us functors
$D_{b|a}:1_S Par_t\Mod \rightarrow 1_{\widetilde{S}}Par_t\Mod$
and $D_{a|b}:1_{\widetilde{S}}Par_t\Mod\rightarrow 1_S Par_t\Mod$
such that $D_{b|a} \Delta(\kappa^{(n)}) \cong \Delta(\tilde\kappa^{(n)})$
$D_{a|b} \Delta(\tilde\kappa^{(n)}) \cong \Delta(\kappa^{(n)})$.
These functors are also biadjoint thanks to <ref>.
It follows easily that they are quasi-inverse equivalences of categories as claimed in the theorem.
In more detail, the unit and counit of one of the adjunctions
gives natural transformations $D_{a|b} \circ D_{b|a} \Rightarrow \Id$
and $\Id \Rightarrow D_{b|a}\circ D_{a|b}$. We claim that these natural transformations are isomorphisms. They are non-zero, hence, they are isomorphisms
on all standard modules. The functors are exact and indecomposable projectives have finite $\Delta$-flags,
so it follows that
the natural transformations
are isomorphisms on all indecomposable projectives.
Then we get that they are isomorphisms on an arbitrary module
by considering a two step projective resolution and applying the Five Lemma.

The next lemma does use the functors $D_{b|a}$ in the case $a=b$, i.e., it definitely
requires the full strength of <ref> rather than merely <ref>.

Let $\kappa \in \P_t$ and $S := \{\kappa^{(0)},\kappa^{(1)},\dots\}$
be the corresponding $\sim_t$-equivalence class.
For each $n \geq 0$, there is an endofunctor
$\Pi_n:Par_t\Mod \rightarrow Par_t\Mod$
such that
$\Pi_n \Delta(\kappa^{(m)}) = 0$ for $m \neq n,n+1$, and
moreover  there exist short exact sequences
$0\to \Delta(\kappa^{(n)}) \rightarrow \Pi_n \Delta(\kappa^{(n)})
\rightarrow \Delta(\kappa^{(n+1)})\rightarrow 0$
$0\to \Delta(\kappa^{(n)}) \rightarrow \Pi_n \Delta(\kappa^{(n+1)})
\rightarrow \Delta(\kappa^{(n+1)})\rightarrow 0$. The functor
$\Pi_n$ is a
composition of the special projective functors $D_{b|a}\:(a,b \in \Z)$, hence, it
is a projective functor.

In view of <ref>, it suffices to prove
the lemma in the special case that $\kappa =(t)$,
when $S = \{\varnothing, (t+1), (t+1,1),(t+1,1^2),\dots\}$ as in <ref>.
Then we take $\Pi_0 := D_{0|t}\circ \cdots \circ D_{t-1|1} \circ D_{t|0}$
and $\Pi_n := D_{-n|-n}$ for $n > 0$.
Now it is just a matter of applying <ref> to see that these functors have the stated properties.

The situation for $\Pi_0$ is the most interesting.
To understand this,
let $u := \lceil\frac{t}{2}\rceil$
and $v := \lfloor\frac{t}{2}\rfloor$.
Then one checks
that $D_{v+1|u-1} \circ \cdots \circ D_{t-1|1}\circ D_{t|0}
(\Delta(\varnothing)) \cong \Delta((u))$; each of these functors adds a single
node to the first row of the Young diagram.
After that we apply $D_{v|u}$ to get a
module with a two step $\Delta$-flag, with
a copy of $\Delta((u+1))$ at the top
and a copy of $\Delta((v))$ at the bottom.
Note this is obtained from <ref> in a slightly different way according to whether $u=v$ (i.e., $t$ is even) or $u  =v+1$ (i.e., $t$ is odd).
Also, this is now a module in an atypical block.
Finally we apply $D_{0|t} \circ D_{1|t-1}\circ\cdots D_{v-1|u+1}$ to
end up with the desired two step $\Delta$-flag with a copy of
$\Delta(\kappa^{(1)}) =
\Delta((t+1))$ at the top and $\Delta(\kappa^{(0)}) =
\Delta(\varnothing)$ at the bottom; each of these functors adds a single node to the first row of the Young diagram labelling the module at the top and removes a node from the Young diagram labelling the module at the bottom.
This is what $\Pi_0$ is meant to do to $\Delta(\varnothing)$.
A similar argument shows that
$\Pi_0 \Delta((t+1))$ has a $\Delta$-flag with the same two sections.
It is also easy to check that $\Pi_0 \Delta(\kappa^{(m)}) = 0$ for $m > 1$, indeed, $D_{t|0}$ already annihilates these standard modules.

The functors $\Pi_n = D_{-n|-n}$ for $n > 0$ are easier to analyze.
Noting that $\kappa^{(n)} = \Delta((t+1,1^{n-1}))$,
the module $\Pi_n \Delta(\kappa^{(n)})$ has a two step $\Delta$-flag with
$\Delta(\kappa^{(n+1)}) = \Delta((t+1,1^{n}))$ at the top and
$\Delta(\kappa^{(n)})$ at the bottom; this uses
the $t-|\lambda|=a=b\notin\rem(\lambda)$ case from <ref>.
Similarly, $\Pi_n \Delta(\kappa^{(n+1)})$ has a $\Delta$-flag
with the same two sections.
Finally, one checks that $\Pi_{n} \Delta(\kappa^{(m)}) = 0$ for $m \neq n,n+1$.

In the proof of the next theorem, we will show that the functor $\Pi_n$
from <ref> satisfies
$\Pi_n \Delta(\kappa^{(n)}) \cong \Pi_n \Delta(\kappa^{(n+1)})
\cong \Pi_n L(\kappa^{(n+1)}) \cong P(\kappa^{(n+1)})$
for all $n \geq 0$.

Now we can prove the main result about blocks.
This can also be deduced from
<cit.>, but the proof of that
appealed to results of Martin [27] in order to obtain
the precise submodule structure of the indecomposable projectives,
whereas we are able to establish this by exploiting the
highest weight structure and the Chevalley duality $?^\sigmadual$.

Let $\kappa \in \Par_t$ and $S := \{\kappa^{(0)}, \kappa^{(1)},\dots\}$ be the corresponding $\sim_t$-equivalence class.

For each $n \geq 0$, the standard module $\Delta(\kappa^{(n)})$ is of length
two with head $L(\kappa^{(n)})$ and socle $L(\kappa^{(n+1)})$.

The indecomposable projective module $P(\kappa^{(0)})$
is isomorphic to $\Delta(\kappa^{(0)})$, while for $n \geq 1$
the  module $P(\kappa^{(n)})$
has a two step $\Delta$-flag with top section $\Delta(\kappa^{(n)})$ and bottom section $\Delta(\kappa^{(n-1)})$.

For each $n \geq 1$, $P(\kappa^{(n)})$ is self-dual with irreducible
head and socle isomorphic to
$L(\kappa^{(n)})$ and completely reducible heart $\rad P(\kappa^{(n)}) / \soc P(\kappa^{(n)})\cong L(\kappa^{(n-1)})\oplus L(\kappa^{(n+1)})$.

To improve the readability, we  write simply $P(n), \Delta(n)$ and $L(n)$ in place of $P(\kappa^{(n)}), \Delta(\kappa^{(n)})$ and $L(\kappa^{(n)})$.
For $n \geq 0$, <ref> shows that the module
$P_n := \Pi_{n-1} \circ \cdots \Pi_1 \circ \Pi_0 (\Delta(0))$
has a two step $\Delta$-flag with top section $\Delta(n)$
and bottom section $\Delta(n-1)$.
Since $\Delta(0)$ is projective by the minimality observed in <ref> and each $\Pi_i$ is a projective functor,
$P_n$ is projective. Since $P_n$ has $L(n)$ in its head, it must contain the indecomposable projective $P(n)$ as a summand, so
we either have that $P(n) \cong P_n$ if $P_n$ is indecomposable, or $P(n) \cong \Delta(n)$ otherwise.
In the former case, $(P(n):\Delta(m)) = \delta_{m,n}+\delta_{m,n-1}$, while $(P(n):\Delta(m)) = \delta_{m,n}$ in the latter situation.
Now we apply BGG reciprocity to deduce for any $m \geq 0$  that
$[\Delta(m):L(n)] = \delta_{n,m}+\delta_{n,m+1}$ if $P_n$
is indecomposable and $[\Delta(m):L(n)] = \delta_{n,m}$ otherwise. Hence, for each $m \geq 0$, we either have that $\Delta(m)
\cong L(m)$, or $\Delta(m)$ is of composition
length two with composition factors $L(m)$ and

We claim for any $n \geq 0$ that $\Delta(n) \cong L(n)$ if and only if $\Delta(n+1)\cong L(n+1)$.
Suppose first that $\Delta(n) \cong L(n)$.
Since $\Pi_n$ commutes with duality by <ref>, this implies that
$\Pi_n \Delta(n)$ is self-dual. But
this module has a two step
$\Delta$-flag with top section $\Delta(n+1)$
and bottom section $\Delta(n)\cong L(n)$. The only way such a module can be self-dual is if $\Delta(n+1) \cong L(n+1)$ (and the module must be completely reducible).
Conversely, suppose for a contradiction that $\Delta(n+1) \cong L(n+1)$
but $\Delta(n) \not\cong L(n)$.
Then $\Delta(n)$ is of length two with composition factors
$L(n)$ and $L(n+1)$, so that $P(n+1)$ has a two step $\Delta$-flag
with top section $\Delta(n+1)\cong L(n+1)$
and bottom section
$\Delta(n)$. Since $\Pi_{n+1} \Delta(n) = 0$ according to <ref>
and $\Pi_{n+1}$ is exact,
we must have that $\Pi_{n+1} L(n+1) = 0$. Since $\Delta(n+1) \cong L(n+1)$, this implies that $\Pi_{n+1} \Delta(n+1) = 0$, which contradicts <ref>.

From the claim, we see that if $\Delta(n)$ is irreducible for any one $n \geq 0$, then it is irreducible for all $n \geq 0$.
Since all atypical “blocks” are equivalent by <ref>,
it follows in that case that
the standard modules $\Delta(\lambda)$ for all $\lambda \in \P$
are irreducible. This implies that the minimal ordering $\succeq_t$ from
<ref> is trivial, hence, the blocks are trivial and $Par_t$ is
semisimple, which contradicts <ref>.
Thus, we have proved
that $\Delta(n)$ must be of length two
for every $n \geq 0$, and (i) is proved.

Property (ii) follows immediately from (i) and BGG reciprocity as noted earlier.

It remains to prove (iii).
Take $n \geq 1$. By <ref>,
we have that $\Pi_{n-1} \Delta(n+1) = 0$. Since $\Pi_{n-1}$
is exact and $L(n+1)$ is a composition factor of $\Delta(n+1)$, it follows that
$\Pi_{n-1} L(n+1) = 0$ too. From this, we deduce that
$\Pi_{n-1} \Delta(n) \cong \Pi_{n-1} L(n)$. By <ref> again,
$\Pi_{n-1} \Delta(n-1)$ has the same
composition length as $\Pi_{n-1} \Delta(n) \cong \Pi_{n-1} L(n)$.
Also $\Delta(n-1)$ has $L(n)$ as a constituent. Using the exactness of $\Pi_{n-1}$ again, we must therefore have that
$\Pi_{n-1} \Delta(n-1) \cong \Pi_{n-1} L(n)$.
As observed earlier in the proof, this module is isomorphic to $P(n)$,
so using that $L(n)$ is self-dual and $\Pi_{n-1}$ commutes with duality, we now see that $P(n)$ is self-dual.
We also know that it has length four with irreducible head
$[P(n):L(n)] = 2$ and
$[P(n):L(n-1)] =
[P(n):L(n+1)] = 1$.
The only possible structure is the one claimed.

All “blocks” of $Par_t\Mod$ are indecomposable, hence, they coincide with the blocks.

The minimal ordering $\succeq_t$ from <ref> is
the partial order such that $\kappa^{(m)} \succeq_t \kappa^{(n)}$
for each $\kappa \in \P_t$ and $m  \leq n$,
with all other pairs of partitions being incomparable.

In general, in an upper finite highest weight category, the standard objects can have infinite length. Our final corollary, which is also noted in <cit.>,
shows that this is not the
case in $Par_t\Modlfd$.
Consequently, the full subcategory consisting of all modules of
finite length has enough projectives and injectives, indeed, this subcategory is
an essentially finite highest weight category in the sense of <cit.>.

The locally unital
algebra $Par_t$ is locally Artinian, i.e., the left ideals $Par_t 1_n$ and the right ideals $1_n Par_t$ are of finite length for all $n \geq 0$.

<ref> shows that all indecomposable projective left $Par_t$-modules
are of finite length, hence, all finitely generated projectives are of finite length too. This includes all of the left ideals $Par_t 1_n$.
Since there is a duality $?^\sigmadual$, it also follows that all
fintely cogenerated injective left $Par_t$-modules are of finite length.
This includes all of the duals $(1_n Par_t)^\circledast$, hence, each $1_n Par_t$ is of finite length as a right module.

§.§ Proof of <ref>

It just remains to prove <ref>.
In fact, we will prove the following slightly stronger result, from which <ref> follows easily on applying the functors involved
to the Specht module $S(\lambda)$.
To state this stronger result,
let $j_!:\sym\Modfd \rightarrow Par_t \Modlfd$ be the standardization functor from <ref>,
$E_a$ and $F_b$ be the refined induction and restriction functors from
$D_{b|a}$ be the special projective functor from <ref>,
and $\PR_c:\sym\Modfd \rightarrow \sym\Modfd$ be the functor defined by
multiplication by the identity element of the symmetric group
$S_c$ if $c \in \N$, i.e., it is the projection onto $\kk S_c \Modfd$
followed followed by the inclusion of $\kk S_c \Modfd$ into
$\sym\Modfd$, or the zero functor if $c \in \kk - \N$.

For $a,b \in \kk$,
there is a filtration of the functor
$D_{b|a} \circ j_!:\sym\Modfd \rightarrow Par_t\Modlfd$ by subfunctors
$0 = S_0 \subseteq S_1 \subseteq S_2 \subseteq S_3 \subseteq S_4 = D_{b|a} \circ j_!$ such that

\begin{align*}
S_4 / S_3 &\cong j_! \circ E_a \circ \PR_{t-b},\\
S_3 / S_2 &\cong j_! \circ \PR_{t-a} \circ \PR_{t-b},\\
S_2 / S_1 &\cong j_! \circ E_a \circ F_b,\\
S_1 / S_0 &\cong j_! \circ \PR_{t-a} \circ F_b.
\end{align*}

(Recall that a subfunctor $S$ of a functor
$T:\sym\Modfd \rightarrow Par_t\Modlfd$ is a functor $S:\sym\Modfd \rightarrow Par_t \Modlfd$ such that $S V$ is a submodule of $T V$ for all $V \in \sym\Modfd$
and $S f = Tf |_{SV}$ for all $f\in\Hom_{\sym}(V,V')$; then the quotient  $T / S$ is the obvious functor with $(T / S)(V) := T V / SV$.)

The proof will take up the rest of the subsection.
We begin by constructing a filtration of the functor
$D\circ j_!:\sym\Modfd\rightarrow Par_t\Modlfd$.
Note that
$D \circ j_! \cong M\otimes_{\sym}$ where $M$ is the

\begin{equation}\label{M}
M := 1_{\mid\,\star} Par_t \otimes_{Par^\sharp} \infl^\sharp \sym.
\end{equation}

We also have the $(Par_t,\sym)$-bimodules

\begin{align}
N_4 &= Par_t\otimes_{Par^\sharp} \infl^\sharp (\sym 1_{\mid\,\star}),\label{N4}\\
N_3 &:= Par_t\otimes_{Par^\sharp} \infl^\sharp \sym,\label{N3}\\
N_2 &:= Par_t\otimes_{Par^\sharp} \infl^\sharp (\sym 1_{\mid\,\star}\otimes_{\sym} 1_{\mid\,\star}\sym)\label{N2}\\
N_1 &:= Par_t\otimes_{Par^\sharp} \infl^\sharp (1_{\mid\,\star} \sym).\label{N1}
\end{align}

The functors $\sym\Modfd\rightarrow Par_t\Modlfd$
defined by tensoring with $N_4, N_3, N_2$ and $N_1$ are isomorphic to
$j_! \circ E$, $j_!$, $j_! \circ E \circ F$
and  $j_! \circ F$, respectively.

For $m \geq n \geq 0$, let $B_{m,n}$ be the basis for $1_m Par^- 1_n$
defined by representatives for the equivalence classes
of normally ordered upward partition diagrams.
By <ref>, the vector space $M$
is isomorphic to $1_{\mid\,\star} Par^- \otimes_{\KK} \sym$, hence, it
has basis

\begin{equation}\label{loud}
\left\{f \otimes g\:\big|\:m\geq 0,n \geq 0, m+1 \geq n, f \in B_{m+1,n}, g \in S_n\right\}.
\end{equation}

For any $f\in B_{m+1,n}$,
let $c(f)$ be the connected component of the diagram containing the top
left vertex.
In the language from <ref>, this component
could be a trunk, an upward tree, an upward leaf, or an upward branch.
Then we introduce the following subspaces of $M$:

* Let
$M_1$ be the subspace of $M$ spanned by all
$f \otimes g$ in this basis such that $c(f)$ is a trunk.

* Let
$M_2$ be the subspace spanned by all $f \otimes g$ such that
$c(f)$ is either a trunk or an upward tree.

* Let
$M_3$ be the subspace spanned by all $f \otimes g$ such that
$c(f)$ is either a trunk, an upward tree, or an upward leaf.

* Let $M_0 := 0$ and $M_4 := M$.

The following is a generalization of <ref>.

The subspaces $0 = M_0 \subset M_1 \subset M_2 \subset M_3 \subset M_4 = M$
are sub-bimodules of the $(Par_t,\sym)$-bimodule $M$. Moreover,
there are bimodule isomorphisms $\theta_i:N_i \stackrel{\sim}{\rightarrow} M_i / M_{i-1}$
for each $i=1,\dots,4$.

The fact that each $M_i$ is a sub-bimodule of $M$ is easily checked by
vertically composing a basis vector $f \otimes g$
with an arbitrary partition diagram on the top and with any permutation diagram on the bottom. One just needs to note
that the action on top involves
$\res_{\mid\,\star}$, so that the top left vertex is untouched. This implies that
the type $c(f)$ does not change if it is a trunk or an upward leaf, while if it is an upward tree it can only be changed to another
upward tree or to a trunk.

We show in this paragraph that there is a bimodule isomorphism

\begin{align}\label{theta1}
\theta_1:N_1 &\rightarrow M_1,&
\begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
&\mapsto     \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.6,0.5)--(-0.6,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}  \end{align}

for any $m\geq 0,n > 0$,
$f \in 1_{m} Par_t 1_{n-1}$ and $g \in S_n$.
This is a well-defined bimodule homomorphism.
By <ref>, $N_1$ is isomorphic
as a vector space to $Par^-\otimes_{\KK} 1_{\mid\,\star} \sym$, hence, it has basis

\begin{equation}\label{N1basis}
\left\{f \otimes g\:\big|\:m \geq n-1 \geq 0, f \in B_{m,n-1}, g \in S_n\right\}.
\end{equation}

The vector space $M_1$ has basis given by all $f_1 \otimes g$ for
$m+1\geq n > 0,
f_1 \in B_{m+1,n}$ and $g \in S_n$ such that $c(f_1)$ is a trunk.
As it is normally ordered, any such $f_1$ is of the form

\begin{align*}
f_1= \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.6,0.5)--(-0.6,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\end{align*}

for a unique $f \in B_{m,n-1}$.
Moreover, $f_1 \otimes g = \theta_1(f \otimes g)$ for every $g \in S_n$.
It follows that $\theta_1$ takes a basis for $N_1$ to a basis for $M_1$,
so it is an isomorphism.

Next we show that there is a bimodule isomorphism

\begin{align}\label{theta2}
\theta_2:N_2&\rightarrow M_2 / M_1,&
\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} &\mapsto \begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.4)--(0.5,0.4)--(0.5,-0.2)--(-0.5,-0.2);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.4)--(-0.4,0.6);
\draw[-,thick] (0.4,0.4)--(0.4,0.6);
\draw[-,thick] (-0.5,0.1)--(0.5,0.1);
\draw[-,thick] (-0.6,0.6) to[out=down,in=160,looseness=1] (-0.4,-0.35);
\node at (-0.2,0.45) {$\cdot$};
\node at (-0,0.45) {$\cdot$};
\node at (0.2,0.45) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (0,0.25) {$\scriptstyle{f}$};
\node at (0,-0.08) {$\scriptstyle{g}$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} + M_1
\end{align}

for $m\geq 0, n > 0, f \in 1_{m} Par_t 1_{n}$ and $g, h \in S_n$.
Again, this is a well-defined bimodule homomorphism.
By <ref>, $N_2$
is isomorphic to $Par^- \otimes_{\KK} \sym 1_{\mid\,\star}\otimes_{\sym}
1_{\mid\,\star}\sym$. Also $\kk S_n$ is free as a right $\kk S_{n-1}$-module
with basis given by
$\left\{(i\:\,i\!+\!1\:\,\cdots\:\,n)\:|\:1 \leq i \leq n\right\}$, which is a set of $S_n / S_{n-1}$-cosets.
It follows that
$N_2$ has basis

\begin{equation}\label{N2basis}
\left\{f \otimes (i\:\,i\!+\!1\:\,\cdots\:\,n) \otimes g\:\big|\:
m \geq n > 0, f \in B_{m,n}, 1 \leq i \leq n, g \in S_n\right\}.
\end{equation}

The vector space $M_2 / M_1$ has a basis given by all $f_2 \otimes g + M_1$
for $m +1\geq n > 0, f_2 \in B_{m+1,n}$ and
$g \in S_n$ such that $c(f_2)$ is an upward tree.
Any such $f_2$ is equal to

\begin{align*}
f_2=  \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.7,-0.8)--(-0.7,0.5);
\draw[-,thick] (-0.7,-0.7) to [out=20,in=down,looseness=1] (0,-.2);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.8);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.8);
\node at (-0.3,-0.3) {$\cdot$};
\node at (-.2,-0.3) {$\cdot$};
\node at (.2,-0.3) {$\cdot$};
\node at (0.3,-0.3) {$\cdot$};
\node at (-.2,-0.7) {$\cdot$};
\node at (0,-0.7) {$\cdot$};
\node at (0.2,-0.7) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (0.05,-.35) {$\stringlabel{i}$};
\node at (0.5,-.35) {$\stringlabel{1}$};
\node at (-0.5,-.35) {$\stringlabel{n}$};
\end{tikzpicture}
\end{align*}

for a
unique $f \in B_{m,n}$ and
a unique $1 \leq i \leq n$ (the index of the string at which
the component $c(f_2)$ meets the bottom of $f$).
Moreover, $f_2 \otimes g = \theta_2(f \otimes (i\:\,i\!+\!1\:\,\cdots\:\,n) \otimes g)$ for each $g \in S_n$. It follows that $\theta_2$ takes a basis for $N_2$
to a basis for $M_2 / M_1$, so it is an isomorphism.

The isomorphism $\theta_3$ is defined by

\begin{align}\label{theta3}
\theta_3:N_3 &\rightarrow M_3/M_2,&
\begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.5)--(-0.7,0.1);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (-0.7,0.05) {$\dt$};
\end{tikzpicture}\otimes \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}+M_2
\end{align}

for $m \geq 0$, $n \geq 0$, $f \in 1_{m} Par_t 1_n$ and $g \in S_n$.
This is obviously a well-defined bimodule homomorphism.
It is an isomorphism because it takes the
\begin{equation}\label{N3basis}
\{f \otimes g\:|\:m \geq n \geq 0,
f \in B_{m,n}, g \in S_n\}
\end{equation}

for $N_3$ to the basis
for $M_3/M_2$ consisting of all $f_3 \otimes g + M_2$ for $m+1> n \geq 0,
f_3 \in B_{m+1,n}$ and $g \in S_n$ such that $c(f_3)$ is an upward leaf.

Finally, we construct the isomorphism $\theta_4$.
The vector space $N_4$ has basis

\begin{equation}\label{N4basis}
\{f \otimes g\:|\:m-1\geq n \geq 0, f \in B_{m,n+1}, g \in S_{n+1}\}.
\end{equation}

We define the linear map

\begin{align}\label{theta4}
\theta_4:N_4 &\rightarrow M_4 / M_3,&
\begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-.9,.5) to [out=down,in=180,looseness=1] (-0.45,-0.6) to [out=0,in=down,looseness=1] (0,-.2);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.8);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.8);
\node at (-0.25,-0.3) {$\cdot$};
\node at (-.15,-0.3) {$\cdot$};
\node at (.2,-0.3) {$\cdot$};
\node at (0.3,-0.3) {$\cdot$};
\node at (-.2,-0.7) {$\cdot$};
\node at (0,-0.7) {$\cdot$};
\node at (0.2,-0.7) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (0.05,-.35) {$\stringlabel{i}$};
\node at (0.5,-.35) {$\stringlabel{1}$};
\node at (-0.6,-.35) {$\stringlabel{n\!+\!1}$};
\end{tikzpicture}\otimes \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g'}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} + M_3,
\end{align}

where $f \otimes g$ is a vector from the basis for $N_4$ just displayed,
and $g' \in S_n$ and $1 \leq i \leq n+1$ are defined from the equation
$g = (i\:\,i\!+\!1\:\,\cdots\:\,n\!+\!1) g'$.
To see that this linear map is actually a bimodule isomorphism, we construct a
bimodule homorphism in the other direction and show that it is a two-sided inverse of $\theta_4$. Consider the map

\begin{align}\label{phi}
\phi: M &\rightarrow N_4,
\begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}&\mapsto \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)to(-0.4,0.3)to[out=up,in=up,looseness=2](-0.7,0.3)to(-0.7,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,-0.5)--(-0.7,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\end{align}

for $m \geq 0, n \geq 0, f \in 1_{m+1} Par_t 1_n$ and $g \in S_n$.
It is easy to show that this is a well-defined bimodule homomorphism.
Moreover, $M_3\subseteq \ker\phi$
since applying $\phi$ to any basis vector
$f \otimes g \in M_3$ produces a downward leaf, a
cap or a downward tree which can be pushed across the tensor to act
as zero on $\infl^\sharp \sym$.
Hence, $\phi$ induces a homomorphism $\bar\phi:M_4 / M_3\to N_4$.
It remains to check that $\bar \phi \circ \theta_4$ and $\theta_4 \circ \bar \phi$ are both identity morphisms, which is straightforward.

In the next two lemmas, we
finally need to make some explicit calculations with the relations involving the left and right dots in the affine partition category. However, we are working now with $\Par_t$, not with $\APar$, so all
string diagrams from now on
should be interpreted as the canonical images of these morphisms in $\APar$
under the functor $p_t:\APar \rightarrow Par_t$ from <ref>.
We will also use the notation from <ref> for an open dot on the interior of a string, meaning the canonical image of this morphism in $\ASym$ under the functor
$p:\ASym \rightarrow \Sym$ from <ref>. This is quite different from an open dot at the end of a string!

Suppose that $m \geq 0, n \geq 0, f \in 1_{m} Par_t 1_n$ and $g \in S_n$.

(i) The following holds in the bimodule $M = 1_{\mid\,\star}Par_t\otimes_{Par^\sharp} \infl^\sharp \sym $ for $i=1,\dots,n$:

\begin{equation*}
\begin{tikzpicture}[anchorbase]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (-0.05,0) to[out=down,in=up] (-.2,-.65) to[out=down,in=left] (-.05,-.8) to[out=right,in=down] (.1,-.65) to[out=up,in=down,looseness=1] (-1.2,.2) to (-1.2,.6);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (.1,-0.63)--(-.08,-0.63);
\closeddot{-0.08,-.63};
\draw[-,thick] (-0.05,-.8)to(-.05,-1.1);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.4,-0.7) {$\cdot$};
\node at (-.55,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.3,-0.7) {$\cdot$};
\node at (.5,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.05,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} \equiv
\begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.1)--(-0.7,0.5);
\opendot{-.7,.1};
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\otimes \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} \pmod{M_2}.
\end{equation*}

(ii) The following holds in the bimodule $M$ for $i=0,1,\dots,n$
(the case $i=0$ is when there are no strings to the right of the dangling dots):

\begin{equation*}
\begin{tikzpicture}[anchorbase]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-1.2,.6) to (-1.2,.2) to[out=down,in=up,looseness=1.7] (-.25,-.9);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\opendot{-.25,-.9};
\draw[-,thick] (-.07,-0.7)--(-.27,-0.7);
\closeddot{-0.07,-.7};
\draw[-,thick] (0.1,0)to(0.1,-1.1);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.4,-0.7) {$\cdot$};
\node at (-.55,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.3,-0.7) {$\cdot$};
\node at (.5,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (.1,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} \equiv
\begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.1)--(-0.7,0.5);
\opendot{-.7,.1};
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\otimes \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture} \pmod{M_2}.
\end{equation*}

(i) We proceed by induction on $i=1,\dots,n$. The base case $i=1$ follows
from <ref>.
For the induction step, we take $i > 1$ and assume the result has been proved for $i-1$. Then we apply <ref> to commute the left dot past the string to its right.
This produces a sum of five terms.
Ordering these terms in the same way as they appear on the right hand side of <ref>, the induction hypothesis can be applied to the first term, to produce the right hand side that we are after. It remains to show that the other four terms lies in $M_2$.
These terms are as follows:

\begin{align*}
\begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-1.2,.6) to (-1.2,.2) to [out=down,in=120] (0,-.63) to (.2,-1.1);
\draw[-,thick] (-.15,-1.1) to (-.15,-.9) to (0,-.63) to[out=60,in=down] (.2,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (-.15,-.9) to [out=120,in=down] (-.1,0);
\draw[-,thick] (0,-0.63)--(.2,-0.63);
\closeddot{0.2,-.63};
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.4,-0.7) {$\cdot$};
\node at (-.55,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.35,-0.7) {$\cdot$};
\node at (.5,-0.7) {$\cdot$};
\node at (0.65,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.2,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
+  \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (-.15,0) to (-.15,-1.1);
\draw[-,thick] (.2,0) to (.2,-1.1);
\draw[-,thick] (-1.2,.6) to (-1.2,.2) to [out=down,in=140] (0.2,-.5);
\draw[-,thick] (.2,-.77) to (-.15,-1);
\draw[-,thick] (0,-0.63)--(.2,-0.63);
\closeddot{0,-.63};
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.3,-0.7) {$\cdot$};
\node at (-.5,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.35,-0.7) {$\cdot$};
\node at (.5,-0.7) {$\cdot$};
\node at (0.65,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.15,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\ -  \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (0.3,0) to[out=down,in=up] (-.02,-.65) to[out=down,in=left] (.145,-.8) to[out=right,in=down] (.3,-.65) to[out=up,in=down,looseness=1] (-1.2,.2) to (-1.2,.6);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (.3,-0.63)--(.12,-0.63);
\closeddot{0.12,-.63};
\draw[-,thick] (0.13,-.8)to(.13,-.9) to (.3,-1.1);
\draw[-,thick] (.13,-.9) to (-.05,-1.1);
\draw[-,thick] (-.15,0) to [out=-90,in=120,looseness=1] (0.02,-1.02);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.25,-0.7) {$\cdot$};
\node at (-.45,-0.7) {$\cdot$};
\node at (-0.65,-0.7) {$\cdot$};
\node at (0.4,-0.7) {$\cdot$};
\node at (.55,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.05,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\ -  \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (-0.2,0) to [out=down,in=130,looseness=1.2] (-0.08,-1.05);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (.25,-0.7)--(.05,-0.7);
\closeddot{0.05,-.7};
\draw[-,thick] (-1.2,.6) to (-1.2,.2) to [out=down,in=135] (.075,-.4) to (.3,0);
\draw[-,thick] (-.15,-1.1) to[out=30,in=down,looseness=1] (.25,-.75) to [out=up,in=0,looseness=1.5] (.075,-.52) to [out=left,in=up,looseness=1.5] (-.1,-.75) to [out=down,in=150,looseness=1] (.3,-1.1);
\draw[-,thick] (.075,-.52) to (.075,-.4);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.35,-0.7) {$\cdot$};
\node at (-.525,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.4,-0.7) {$\cdot$};
\node at (.55,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.15,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\ .
\end{align*}

The second and third terms here are zero already in $M$ because,
in both of them, the diagram to the left of the tensor is
equivalent to a diagram with a downward tree at the bottom.
It remains to show that the first and the fourth terms lie in $M_2$.
For the fourth term, we note that

\begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (-0.2,0) to [out=down,in=130,looseness=1.2] (-0.08,-1.05);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (.25,-0.7)--(.05,-0.7);
\closeddot{0.05,-.7};
\draw[-,thick] (-1.2,.6) to (-1.2,.2) to [out=down,in=135] (.075,-.4) to (.3,0);
\draw[-,thick] (-.15,-1.1) to[out=30,in=down,looseness=1] (.25,-.75) to [out=up,in=0,looseness=1.5] (.075,-.52) to [out=left,in=up,looseness=1.5] (-.1,-.75) to [out=down,in=150,looseness=1] (.3,-1.1);
\draw[-,thick] (.075,-.52) to (.075,-.4);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.35,-0.7) {$\cdot$};
\node at (-.525,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.4,-0.7) {$\cdot$};
\node at (.55,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.15,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}
\begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (-0.2,0) to [out=down,in=130,looseness=1.2] (-0.08,-.83);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (.25,-0.5)--(.05,-0.5);
\closeddot{0.05,-.5};
\draw[-,thick] (-1.2,.6) to (-1.2,.2) to [out=down,in=up] (-.2,-1.1);
\draw[-,thick] (-.21,-1) to [out=50,in=-120] (-.15,-.9) to[out=60,in=down,looseness=1] (.25,-.55) to [out=up,in=0,looseness=1.5] (.075,-.32) to [out=left,in=up,looseness=1.5] (-.1,-.55) to [out=down,in=120,looseness=1] (.3,-.9) to [out=-60,in=up,looseness=1] (.3,-1.1);
\draw[-,thick] (.075,-.32) to (.075,0);
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.35,-0.9) {$\cdot$};
\node at (-.525,-0.9) {$\cdot$};
\node at (-0.7,-0.9) {$\cdot$};
\node at (0.4,-0.7) {$\cdot$};
\node at (.55,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (-.15,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\ .

The left dot can now be absorbed into the morphism $f$, changing it to some other morphism $f'$. The result is a linear combination of morphisms in all of which the top left vertex is connected to the bottom edge, so that the connected component containing this vertex is either a tree or a trunk, and it belongs to the sub-bimodule $M_2$.
The reason the first term lies in $M_2$ is very similar, one just needs to rewrite
the right crossing using <ref>, and then it is easy to see that the
top left vertex is again connected to the bottom edge.

Again we proceed by induction.
The  base case $i=0$ follows
from <ref> using that $T = t1_\one$.
For the induction step, we consider some $i > 0$. Then we apply <ref> to commute the right dot past the string to its right. This produces a sum of five terms.
This time, the induction hypothesis can be applied to the first term, to
produce the vector that we are after but scaled by $(t-i+1)$ rather than
the desired $(t-i)$. The remaining four terms are as follows:

\begin{align*}
\begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (0,-1.1) to (0,-.5) to[out=135,in=down,looseness=1] (-1.2,.2) to (-1.2,.6);
\draw[-,thick] (0,-.5) to [out=45,in=down,looseness=1] (0.15,0);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (0.15,-0.23)--(-.05,-0.23);
\closeddot{-0.05,-.23};
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.4,-0.7) {$\cdot$};
\node at (-.55,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.3,-0.7) {$\cdot$};
\node at (.5,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (0,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}+
\begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (0,-1.1) to (0,-.5) to[out=135,in=down,looseness=1] (-1.2,.2) to (-1.2,.6);
\draw[-,thick] (0,-.5) to [out=45,in=down,looseness=1] (0.15,0);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (0,-0.7)--(-.2,-0.7);
\closeddot{-0.2,-.7};
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.4,-0.7) {$\cdot$};
\node at (-.55,-0.7) {$\cdot$};
\node at (-0.7,-0.7) {$\cdot$};
\node at (0.3,-0.7) {$\cdot$};
\node at (.5,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (0,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}-
\begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (0,-1.1) to (0,-.5) to[out=135,in=down,looseness=1] (-1.2,.2) to (-1.2,.6);
\draw[-,thick] (0,-.5) to [out=45,in=down,looseness=1] (0.15,0);
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (0,-0.7)--(.2,-0.7);
\closeddot{0.2,-.7};
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.25,-0.7) {$\cdot$};
\node at (-.45,-0.7) {$\cdot$};
\node at (-0.65,-0.7) {$\cdot$};
\node at (0.4,-0.7) {$\cdot$};
\node at (.55,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (0,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
-        \begin{tikzpicture}[anchorbase,scale=.9]\draw[-] (-.9,0)--(-.9,0.4)--(.9,0.4)--(.9,0)--(-.9,0);
\draw[-,thick] (-0.8,0)--(-0.8,-1.1);
\draw[-,thick] (.2,-1.1) to (0,-.5) to[out=110,in=down,looseness=1] (-1.2,.2) to [out=up,in=down] (-1.2,.6);
\draw[-,thick] (-.15,-.9) to[out=70,in=-130,looseness=0] (0,-.5) to [out=70,in=down,looseness=1] (0.13,0);
\opendot{-.15,-.9};
\draw[-,thick] (-0.8,0.4)--(-0.8,0.6);
\draw[-,thick] (0.8,0)--(0.8,-1.1);
\draw[-,thick] (0.8,0.4)--(0.8,0.6);
\draw[-,thick] (0,-0.5)--(-.25,-0.5);
\closeddot{-0.25,-.5};
\node at (-0.3,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.3,0.5) {$\cdot$};
\node at (-0.25,-0.7) {$\cdot$};
\node at (-.45,-0.7) {$\cdot$};
\node at (-0.65,-0.7) {$\cdot$};
\node at (0.4,-0.7) {$\cdot$};
\node at (.55,-0.7) {$\cdot$};
\node at (0.7,-0.7) {$\cdot$};
\node at (0,0.2) {$\scriptstyle{f}$};
\node at (0.2,-1.2) {$\stringlabel{i}$};
\node at (.8,-1.2) {$\stringlabel{1}$};
\node at (-.8,-1.2) {$\stringlabel{n}$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\ .
\end{align*}

In the first term here, the left dot is some morphism in $Par_t$, which has the effect of changing $f$ to some other morphism $f'$.
After doing that, it is clear that the top left vertex is still connected to the bottom edge, so the first term lies in $M_2$. For the second and third terms, the left and right dots can be commuted across the tensor using <ref>, then again we see that these morphisms lie in $M_2$ since the top left vertex
is connected to the bottom edge again.
For the final term, we note

\begin{tikzpicture}[anchorbase]
\draw[-,thick](-0.125,-0.25)--(0.25,0.5);
\draw[-,thick](0.25,-0.5)--(-0.25,0.5);
\draw[-,thick](0,0)--(-0.25,0);
\opendot{-.125,-.25};
\closeddot{-0.25,0};
\end{tikzpicture}
\stackrel{\cref{june}}{=} \begin{tikzpicture}[anchorbase]
\draw[-,thick] (-.25,.5) to [out=-90,in=90] (.25,0.15) to [out=-90,in=60] (-.175,-.35);
\draw[-,thick] (.25,.5) to [out=-90,in=90] (-.25,0.15) to [out=-90,in=90] (.3,-.5);
\draw[-,thick] (0.02,-.165) to (.25,-.165);
\opendot{-.175,-.35};
\closeddot{0.25,-.165};
\end{tikzpicture}
\stackrel{\cref{crazycrossing}}{=}
\begin{tikzpicture}[anchorbase]
\draw[-,thick] (-.25,.5) to [out=-90,in=90] (.25,0.15) to [out=-90,in=60] (-.175,-.35);
\draw[-,thick] (.25,.5) to [out=-90,in=90] (-.25,0.15) to [out=-90,in=90] (.3,-.5);
\draw[-,thick] (0.43,-.165) to (.2,-.165);
\draw[-,thick] (0.24,0.05) to[out=-30,in=30,looseness=1.75] (0.26,-.38);
\opendot{-.175,-.35};
\closeddot{0.2,-.165};
\end{tikzpicture}
\,\: \substack{\cref{relssym2}\\{\displaystyle =}\\\cref{relsfrob2}}\:\,
\begin{tikzpicture}[anchorbase]
\draw[-,thick] (-.25,0.5) to[out=-80,in=up] (.3,0) to [out=down,in=0,looseness=1] (0,-0.25) to [out=180,in=down,looseness=1] (-.3,0) to [out=up,in=-100] (.25,.5);
\draw[-,thick] (.3,0)--(0.03,0);
\draw[-,thick] (0,-.5)--(0,-.25);
\closeddot{0.03,0};
\end{tikzpicture}\ .

Making this substitution in the middle of the picture reveals that the final term is exactly the expression studied in (i). On applying the conclusion of (i), we deduce that it  contributes exactly the needed correction to complete the proof.

Consider the bimodule endomorphisms $\rho$ and $\lambda$
of $M$ defined on
$1_{m+1} Par_t 1_n \otimes \kk S_n$ by the left action of
$x_{m+1}^R \otimes 1_n$ and $x_{m+1}^L\otimes 1_n$, respectively,
for each $m,n \geq 0$.
These endomorphisms preserve each of the sub-bimodules $M_i\:(i=1,2,3,4)$,
hence, $\rho$ and $\lambda$ induce endomorphisms also denoted $\rho$ and $\lambda$
of each of the subquotients
$M_i / M_{i-1}$.
Moreover, for each $i$, the isomorphism $\theta_i$
from <ref> satisfies

\begin{align}\label{maincommutationfact}
\theta_i\circ \rho_i &= \rho \circ \theta_i,&
\theta_i\circ \lambda_i &= \lambda \circ \theta_i,
\end{align}

where $\rho_i, \lambda_i:N_i \rightarrow N_i$ are defined as follows:

$\rho_1$ and $\lambda_1$ are the bimodule endomorphisms of $N_1$
defined on the subspace $1_{m} Par_t 1_{n-1} \otimes \kk S_n$
by the left actions of $(t-n+1) 1_{m} \otimes 1_n$
and  $1_{m} \otimes x_n$,
respectively, for each $m\geq 0, n > 0$.

(ii) $\rho_2$ and $\lambda_2$ are the bimodule endomorphisms of $N_2$
defined on $1_{m} Par_t 1_n \otimes \kk S_n \otimes \kk S_n$
by the right action of
$1_{n} \otimes x_n\otimes 1_n$ and the left action of
$1_{m} \otimes 1_n \otimes x_n$, respectively.

(iii) $\rho_3$ and $\lambda_3$ are both equal to the bimodule endomorphism
of $N_3$
defined on $1_{m} Par_t 1_n \otimes \kk S_n$
by multiplication by $(t-n)$.

(iv) $\rho_4$ and $\lambda_4$ are the bimodule endomorphisms of $N_4$
defined on $1_{m} Par_t 1_{n+1} \otimes \kk S_{n+1}$
by the right actions of  $1_{n+1} \otimes x_{n+1}$ and $(t-n) 1_{n+1} \otimes 1_{n+1}$,

(i)  Recall the definition of $\theta_1$ from <ref>.
Take a vector $f \otimes g$ in the basis for $N_1$ from <ref>.
By <ref>,
we have that $x_n^L \equiv x_n\pmod{K_n}$ and $x_n^R \equiv (t-n+1) 1_n
\pmod{K_n}$
where $K_n$ is the two sided ideal of $1_n Par_t 1_n$ from <ref>.
Since the strictly downward partition diagrams which generate $K^+$
are zero on $\infl^\sharp \sym$, it follows that

\begin{align*}
\rho(\theta_1(f\otimes g)) &=
\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.8,0.5)--(-0.8,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.8,0.4)to(-0.6,0.4);
\closeddot{-0.6,0.4};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}=
\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.8,0.5)--(-0.8,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.8,-0.35)to(-0.6,-0.35);
\closeddot{-0.6,-0.35};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\\ &= (t-n+1)\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.5)--(-0.7,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}= \theta_1(\rho_1(f \otimes g)),\\
\lambda(\theta_1(f\otimes g)) &= \!\!\!\!\!
\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.5)--(-0.7,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.7,0.35)to(-0.9,0.35);
\closeddot{-0.9,0.35};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}=
\!\!\!\!\!\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.5)--(-0.7,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.7,-0.35)to(-0.9,-0.35);
\closeddot{-0.9,-0.35};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\\&= \begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,0.5)--(-0.7,-0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\opendot{-0.4,0.35};
\end{tikzpicture} = \theta_1(\lambda_1(f \otimes g)).
\end{align*}

This shows as the same time that $\rho$ and $\lambda$ both leave $M_1$ invariant.

Recall the definition of $\theta_2$ from <ref>.
The argument for $\lambda$ is similar to in (i). It follows from the calculation

\begin{align*}
\lambda(\theta_2(f\otimes g \otimes h))
\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.4)--(0.5,0.4)--(0.5,-0.2)--(-0.5,-0.2);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.4)--(-0.4,0.6);
\draw[-,thick] (0.4,0.4)--(0.4,0.6);
\draw[-,thick] (-0.5,0.1)--(0.5,0.1);
\draw[-,thick] (-0.6,.5) to (-.8,.5);
\closeddot{-.8,.5};
\draw[-,thick] (-0.6,0.6) to[out=down,in=160,looseness=1] (-0.4,-0.35);
\node at (-0.2,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.2,0.5) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (0,0.25) {$\scriptstyle{f}$};
\node at (0,-0.08) {$\scriptstyle{g}$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.4)--(0.5,0.4)--(0.5,-0.2)--(-0.5,-0.2);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.4)--(-0.4,0.6);
\draw[-,thick] (0.4,0.4)--(0.4,0.6);
\draw[-,thick] (-0.5,0.1)--(0.5,0.1);
\draw[-,thick] (-0.4,-.45) to (-.6,-.45);
\closeddot{-.6,-.45};
\draw[-,thick] (-0.6,0.6) to[out=down,in=160,looseness=1] (-0.4,-0.35);
\node at (-0.2,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.2,0.5) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (0,0.25) {$\scriptstyle{f}$};
\node at (0,-0.08) {$\scriptstyle{g}$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}
\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.4)--(0.5,0.4)--(0.5,-0.2)--(-0.5,-0.2);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.4)--(-0.4,0.6);
\draw[-,thick] (0.4,0.4)--(0.4,0.6);
\draw[-,thick] (-0.5,0.1)--(0.5,0.1);
\draw[-,thick] (-0.6,0.6) to[out=down,in=160,looseness=1] (-0.4,-0.35);
\node at (-0.2,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.2,0.5) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (0,0.25) {$\scriptstyle{f}$};
\node at (0,-0.08) {$\scriptstyle{g}$};
\end{tikzpicture}\otimes\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\node at (-.4,.35) {$\dt$};
\end{tikzpicture}
+M_1=\theta_2\left(\begin{tikzpicture}[scale=1.1,anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[baseline=-1.02mm,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{g}$};
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (0.4,-.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (-0.4,-.2)--(-0.4,-0.5);
\node at (-0.2,.35) {$\cdot$};
\node at (-0,.35) {$\cdot$};
\node at (0.2,.35) {$\cdot$};
\node at (-0.2,-.35) {$\cdot$};
\node at (-0,-.35) {$\cdot$};
\node at (0.2,-.35) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\opendot{-0.4,.35};
\end{tikzpicture}\right)
\\&=\theta_2(\lambda_2(f\otimes g \otimes h)),
\end{align*}

where $f \otimes g \otimes h$ is one of the basis vectors for $N_2$ from <ref>.
For $\rho$, we instead have that

\begin{align*}
\rho(\theta_2(f\otimes g \otimes h)) &=
\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.4)--(0.5,0.4)--(0.5,-0.2)--(-0.5,-0.2);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.7);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.7);
\draw[-,thick] (-0.4,0.4)--(-0.4,0.6);
\draw[-,thick] (0.4,0.4)--(0.4,0.6);
\draw[-,thick] (-0.5,0.1)--(0.5,0.1);
\draw[-,thick] (-0.9,.5) to (-.7,.5);
\closeddot{-.7,.5};
\draw[-,thick] (-0.9,0.6) to[out=down,in=160,looseness=1] (-0.4,-0.55);
\node at (-0.2,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.2,0.5) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\node at (0,0.25) {$\scriptstyle{f}$};
\node at (0,-0.08) {$\scriptstyle{g}$};
\end{tikzpicture}\otimes\begin{tikzpicture}[baseline=-1.02mm,scale=1.1]\draw[-] (-0.5,-0.25)--(-0.5,0.15)--(0.5,0.15)--(0.5,-0.25)--(-0.5,-0.25);
\node at (-0,-0.05) {$\scriptstyle{h}$};
\draw[-,thick] (-0.45,-0.25)--(-0.45,-0.7);
\draw[-,thick] (0.45,-0.25)--(0.45,-0.7);
\draw[-,thick] (-0.45,0.15)--(-0.45,0.6);
\draw[-,thick] (0.45,0.15)--(0.45,0.6);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\end{tikzpicture}+M_1=
\begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.4)--(0.5,0.4)--(0.5,-0.2)--(-0.5,-0.2);
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.7);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.7);
\draw[-,thick] (-0.4,0.4)--(-0.4,0.6);
\draw[-,thick] (0.4,0.4)--(0.4,0.6);
\draw[-,thick] (-0.5,0.1)--(0.5,0.1);
\draw[-,thick] (-0.55,-.35) to (-.4,-.35);
\closeddot{-.55,-.35};
\draw[-,thick] (-0.9,0.6) to[out=down,in=160,looseness=1] (-0.4,-0.55);
\node at (-0.2,0.5) {$\cdot$};
\node at (-0,0.5) {$\cdot$};
\node at (0.2,0.5) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\node at (0,0.25) {$\scriptstyle{f}$};
\node at (0,-0.08) {$\scriptstyle{g}$};
\end{tikzpicture}\otimes\begin{tikzpicture}[baseline=-1.02mm,scale=1.1]\draw[-] (-0.5,-0.25)--(-0.5,0.15)--(0.5,0.15)--(0.5,-0.25)--(-0.5,-0.25);
\node at (-0,-0.05) {$\scriptstyle{h}$};
\draw[-,thick] (-0.45,-0.25)--(-0.45,-0.7);
\draw[-,thick] (0.45,-0.25)--(0.45,-0.7);
\draw[-,thick] (-0.45,0.15)--(-0.45,0.6);
\draw[-,thick] (0.45,0.15)--(0.45,0.6);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\end{tikzpicture}+M_1\\
&= \theta_2\left(\begin{tikzpicture}[scale=1.1,anchorbase]\draw[-] (-0.5,0)--(-0.5,0.3)--(0.5,0.3)--(0.5,0)--(-0.5,0)--(-.5,-.3)--(0.5,-.3)--(.5,0);
\node at (0,.15) {$\scriptstyle{f}$};
\node at (0,-.15) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.3)--(-0.4,-0.6);
\draw[-,thick] (0.4,-0.3)--(0.4,-0.6);
\draw[-,thick] (-0.4,0.3)--(-0.4,0.6);
\draw[-,thick] (0.4,0.3)--(0.4,0.6);
\draw[-,thick] (-0.4,-0.45)--(-0.65,-0.45);
\closeddot{-0.65,-0.45};
\node at (-0.2,0.45) {$\cdot$};
\node at (-0,0.45) {$\cdot$};
\node at (0.2,0.45) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-,thick] (-0.4,0.6)--(-0.4,-0.6);
\draw[-,thick] (0.4,-.6)--(0.4,0.6);
\node at (-0.2,0) {$\cdot$};
\node at (-0,0) {$\cdot$};
\node at (0.2,0) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[scale=1.1,anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.6);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.6);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.6);
\draw[-,thick] (0.4,0.2)--(0.4,0.6);
\node at (-0.2,0.4) {$\cdot$};
\node at (-0,0.4) {$\cdot$};
\node at (0.2,0.4) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\end{tikzpicture}\right)
= \theta_2\left(\begin{tikzpicture}[scale=1.1,anchorbase]\draw[-] (-0.5,0)--(-0.5,0.3)--(0.5,0.3)--(0.5,0)--(-0.5,0)--(-.5,-.3)--(0.5,-.3)--(.5,0);
\node at (0,.15) {$\scriptstyle{f}$};
\node at (0,-.15) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.3)--(-0.4,-0.6);
\draw[-,thick] (0.4,-0.3)--(0.4,-0.6);
\draw[-,thick] (-0.4,0.3)--(-0.4,0.6);
\draw[-,thick] (0.4,0.3)--(0.4,0.6);
\opendot{-0.4,-0.45};
\node at (-0.2,0.45) {$\cdot$};
\node at (-0,0.45) {$\cdot$};
\node at (0.2,0.45) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[anchorbase,scale=1.1]
\draw[-,thick] (-0.4,0.6)--(-0.4,-0.6);
\draw[-,thick] (0.4,-.6)--(0.4,0.6);
\node at (-0.2,0) {$\cdot$};
\node at (-0,0) {$\cdot$};
\node at (0.2,0) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[scale=1.1,anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.6);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.6);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.6);
\draw[-,thick] (0.4,0.2)--(0.4,0.6);
\node at (-0.2,0.4) {$\cdot$};
\node at (-0,0.4) {$\cdot$};
\node at (0.2,0.4) {$\cdot$};
\node at (-0.2,-0.45) {$\cdot$};
\node at (-0,-0.45) {$\cdot$};
\node at (0.2,-0.45) {$\cdot$};
\end{tikzpicture}\right)\\
\theta_2\left(\begin{tikzpicture}[scale=1.1,anchorbase]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[baseline=-1.02mm,scale=1.1]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{g}$};
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (0.4,-.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (-0.4,-.2)--(-0.4,-0.5);
\node at (-0.2,.35) {$\cdot$};
\node at (-0,.35) {$\cdot$};
\node at (0.2,.35) {$\cdot$};
\node at (-0.2,-.35) {$\cdot$};
\node at (-0,-.35) {$\cdot$};
\node at (0.2,-.35) {$\cdot$};
\opendot{-0.4,-.35};
\end{tikzpicture}\ \otimes\ \begin{tikzpicture}[anchorbase,scale=1.1]\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (0,0) {$\scriptstyle{h}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\right)
\theta_2(\rho_2(f\otimes g\otimes h)).
\end{align*}

Recall the definition of $\theta_3$ from <ref>.
Note that $\rho = \lambda$ by the third relation from <ref>. For $\rho$,
we need to show that $\rho(\theta_3(f \otimes g)) = (t-n)\theta_3(f \otimes g)$
for any
basis vector $f \otimes g \in 1_m Par_t 1_n \otimes \kk S_n \subset N_3$
from <ref>. This follows from <ref>(ii) taking $i=n$.

Instead of working with $\theta_4$ from <ref>,
it is easier to use the inverse map $\bar\phi$ induced by the homomorphism $\phi:M\rightarrow N_4$ from <ref>.
We need to show that $\phi \circ \rho =\rho_4 \circ \phi$.
This follows from the following calculations for $f \otimes g \in 1_{m+1} Par_t 1_n \otimes \kk S_n$ and $m, n\geq 0$:

\begin{align*}
\phi(\rho(f \otimes g)) &= \phi\left(\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\draw[-,thick] (-0.4,0.35)--(-0.25,0.35);
\closeddot{-0.25,0.35};
\end{tikzpicture}\otimes\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\end{tikzpicture}\right)\ =\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.7,0.3)to(-0.7,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\draw[-,thick] (-0.4,0.3)--(-0.25,0.3);
\closeddot{-0.25,0.3};
\end{tikzpicture}\otimes\!\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,-0.5)--(-0.7,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\end{tikzpicture}= \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.7,0.3)to(-0.7,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\draw[-,thick] (-0.7,-0.3)--(-0.85,-0.3);
\closeddot{-0.85,-0.3};
\end{tikzpicture}\otimes\!\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,-0.5)--(-0.7,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\end{tikzpicture} \\&= \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.7,0.3)to(-0.7,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,-0.5)--(-0.7,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\opendot{-0.7,.3};
\end{tikzpicture}= \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.7,0.3)to(-0.7,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\end{tikzpicture}\otimes\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.7,-0.5)--(-0.7,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.35) {$\cdot$};
\node at (-0,-0.35) {$\cdot$};
\node at (0.2,-0.35) {$\cdot$};
\opendot{-0.7,-.35};
\end{tikzpicture} = \rho_4(\phi(f\otimes g)),\\
\phi(\lambda(f\otimes g))&=\phi\left(\!\!\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.4,0.35)--(-0.6,0.35);
\closeddot{-0.6,0.35};
\end{tikzpicture}\otimes\begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\right)\ =\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.8,0.3)to(-0.8,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.4,0.3)--(-0.6,0.3);
\closeddot{-0.6,0.3};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.8,-0.5)--(-0.8,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\ =\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.8,0.3)to(-0.8,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\draw[-,thick] (-0.8,-0.35)--(-0.6,-0.35);
\closeddot{-0.6,-0.35};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.8,-0.5)--(-0.8,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\\
&=(t-n)\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{f}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.3)to[out=up,in=up,looseness=2](-0.8,0.3)to(-0.8,-0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\node at (-0,0.35) {$\cdot$};
\node at (-0.2,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}\otimes\ \begin{tikzpicture}[scale=1.1,anchorbase]
\draw[-] (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--(-0.5,-0.2);
\node at (-0,-0) {$\scriptstyle{g}$};
\draw[-,thick] (-0.4,-0.2)--(-0.4,-0.5);
\draw[-,thick] (0.4,-0.2)--(0.4,-0.5);
\draw[-,thick] (-0.4,0.2)--(-0.4,0.5);
\draw[-,thick] (0.4,0.2)--(0.4,0.5);
\draw[-,thick] (-0.8,-0.5)--(-0.8,0.5);
\node at (-0.2,0.35) {$\cdot$};
\node at (-0,0.35) {$\cdot$};
\node at (0.2,0.35) {$\cdot$};
\node at (-0.2,-0.4) {$\cdot$};
\node at (-0,-0.4) {$\cdot$};
\node at (0.2,-0.4) {$\cdot$};
\end{tikzpicture}= \lambda_4(\phi(f \otimes g)).
\end{align*}

The functor $D_{b|a} \circ j_!$ is defined by tensoring with the bimodule
$\overline{M}$ that is the simultaneous generalized $a$ eigenspace of
the endomorphism $\rho$ and generalized $b$ eigenspace of the endomorphism $\lambda$
defined in <ref>.
<ref> defines a filtration of $M$ with sections
$M_i / M_{i-1} \cong N_i$ for $i=1,\dots,4$.
Then <ref> shows that the endomorphisms
$\rho$ and $\lambda$ preserve this filtration, hence, the filtration of $M$
induces a filtration of the summand $\overline{M}$. Moreover, for each $i$,
$\overline{M}_i / \overline{M}_{i-1}$ is isomorphic to
the summand $\overline{N}_i$ of $N$ defined by the simultaneous generalized $a$-eigenspace of the endomorphism $\rho_i$ and generalized $b$ eigenspace of the endomorphism $\lambda_i$. By the descriptions of $\rho_i$ and $\lambda_i$, it follows that
$\overline{N}_i \otimes_{\sym}$ is isomorphic to the functor
$j_! \circ E_a \circ \PR_{t-b}$, $j_! \circ \PR_{t-a} \circ \PR_{t-b}$,
$j_! \circ E_a \circ F_b$ or $j_! \circ \PR_{t-a} \circ F_b$
for $i=4,3,2,1$, respectively.
It remains to observe that $\sym$ is semisimple, so every $\sym$-module is flat. This means that
the filtration of $\overline{M}$ induces a filtration $0 = S_0 \subseteq S_1 \subseteq S_2 \subseteq S_3 \subseteq S_4 = D_{b|a} \circ j_!$ such that
$S_i \cong \overline{M}_i / \overline{M}_{i-1} \otimes_{\sym}
\cong \overline{N}_i \otimes_{\sym}$.

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