\;\;\;\;\;\;\;\;\;\;\;\;=\iota_{t,s\star}\big(\iota_{t,s}^\star(\pr_1\phi_{t,s})^*(a)\cup (\pr_2\phi_{t,\emptyset})^*(b)\big)
\end{array}
\end{equation}
Let us consider %the finite dimensional vertions $\phi_{t,\emptyset}^N$ and $\phi_{t,s}^N$ of $\phi_{t,\emptyset}$ and $\phi_{t,s}$
the compatibly oriented Cartesian square
\begin{tikzcd}[column sep=42pt]
\BS(t)\ltimes{K}E^N\arrow{r}{\pr_2\phi_{t,\emptyset}^N}\arrow{d}[swap]{\iota_{t,s}\ltimes{K}\id}&K\lq E^N\arrow{d}{\iota_s\ltimes{K}\id}\\
\BS(ts)\ltimes{K}E^N\arrow{r}{\pr_2\phi_{t,s}^N}&\BS(s)\ltimes{K}E^N
\end{tikzcd}
By~(\ref{eq:24}), we get $(\iota_{t,s}\ltimes{K}\id)_*(\pr_2\phi_{t,\emptyset}^N)^*=(\pr_2\phi_{t,s}^N)^*(\iota_s\ltimes{K}\id)_*$. Taking the limit $N\to\infty$,
we get $\iota_{t,s\star}(\pr_2\phi_{t,\emptyset})^*=(\pr_2\phi_{t,s})^*\iota_{s\star}$. Applying this substitution to
the right-hand side of~(\ref{eq:25}), we get
\begin{multline*}
\iota_{t,s\star}\theta_{t,\emptyset}(a\otimes b)=(\pr_1\phi_{t,s})^*(a)\cup(\pr_2\phi_{t,s})^*\iota_{s\star}(b)\\[3pt]
=\phi_{t,s}^*(\pr_1^*(a)\cup\pr_2^*\iota_{s\star}(b))=\theta_{t,s}(a\otimes \iota_{s\star}(b))=\theta_{t,s}(\id\otimes\iota_{s\star})(a\otimes b).
\end{multline*}
This equality proves the commutativity of diagram~(\ref{eq:26}). We prove similarly that the diagram
\begin{tikzcd}[column sep=35pt]
H(t)\otimes_\R H(r)\arrow{r}{\id\otimes\iota_{s\star}}\arrow{d}[swap]{\theta_{t,r}}&H(ts)\otimes_\R H(r)(2)\arrow{d}{\theta_{ts,r}}\\
\end{tikzcd}
is commutative. Computing in coordiates as in the three-dimensional diagram above, we get
\begin{equation}\label{eq:29}
\begin{array}{c}\hspace{-160pt}
\theta_{tsr}^{-1}(-\iota_{t,s,r\star})\theta_{tr}(a_1\otimes\cdots\otimes a_n\otimes d\otimes b_2\otimes\cdots\otimes b_{m+1})\\[6pt]
\hspace{130pt}\displaystyle=a_1\otimes\cdots\otimes a_n\otimes\( d\frac{\alpha_s}2\otimes 1+d\otimes\frac{\alpha_s}2\)\otimes b_2\otimes\cdots\otimes b_{m+1}.
\end{array}
\end{equation}
This morphism is represented by the diagram
\begin{equation}\label{eq:27}\tag{$-\iota_{t,s,r\star}$}
\begin{tikzpicture}[baseline=19pt]
\draw[dashed] (-0.5,0) -- (6.5,0);
\draw[dashed] (-0.5,1.5) -- (6.5,1.5);
\draw[red] (3,1.5) node[anchor=south]{\footnotesize$s$}-- (3,0.75);
\draw[red,fill=red] (3,0.75) circle(0.06);
\draw (2.25,0) node[anchor=north]{\footnotesize$t_n$} -- (2.25,1.5) node[anchor=south]{\footnotesize$t_n$};
\draw (0,0) node[anchor=north]{\footnotesize$t_1$} -- (0,1.5) node[anchor=south]{\footnotesize$t_1$};
\draw (0.75,0) node[anchor=north]{\footnotesize$t_2$} -- (0.75,1.5) node[anchor=south]{\footnotesize$t_2$};
\draw  (1.5,0.7) node{\footnotesize$\cdots$};
\draw (3.75,0) node[anchor=north]{\footnotesize$r_1$} -- (3.75,1.5) node[anchor=south]{\footnotesize$r_1$};
\draw (4.5,0) node[anchor=north]{\footnotesize$r_2$} -- (4.5,1.5) node[anchor=south]{\footnotesize$r_2$};
\draw  (5.25,0.7) node{\footnotesize$\cdots$};
\draw (6,0) node[anchor=north]{\footnotesize$r_m$} -- (6,1.5) node[anchor=south]{\footnotesize$r_m$};
\end{tikzpicture}
\end{equation}

The split map $\mu_s^\star$ can be extended horizontally in the same way. Consider the diagram
\begin{equation}\tag{$\mu_{t,s,r}^\star$}
\begin{tikzpicture}[baseline=19pt]
\draw[dashed] (-0.5,0) -- (8,0);
\draw[dashed] (-0.5,1.5) -- (8,1.5);
\draw[red] (3,1.5) node[anchor=south]{\footnotesize$s$}-- (3.75,0.75);
\draw[red] (4.5,1.5) node[anchor=south]{\footnotesize$s$}-- (3.75,0.75);
\draw[red] (3.75,0.75)-- (3.75,0)node[anchor=north]{\footnotesize$s$};
%\draw[red,fill=red] (3,0.75) circle(0.06);
\draw (2.25,0) node[anchor=north]{\footnotesize$t_n$} -- (2.25,1.5) node[anchor=south]{\footnotesize$t_n$};
\draw (0,0) node[anchor=north]{\footnotesize$t_1$} -- (0,1.5) node[anchor=south]{\footnotesize$t_1$};
\draw (0.75,0) node[anchor=north]{\footnotesize$t_2$} -- (0.75,1.5) node[anchor=south]{\footnotesize$t_2$};
\draw  (1.5,0.7) node{\footnotesize$\cdots$};
\draw (5.25,0) node[anchor=north]{\footnotesize$r_1$} -- (5.25,1.5) node[anchor=south]{\footnotesize$r_1$};
\draw (6,0) node[anchor=north]{\footnotesize$r_2$} -- (6,1.5) node[anchor=south]{\footnotesize$r_2$};
\draw  (6.75,0.7) node{\footnotesize$\cdots$};
\draw (7.5,0) node[anchor=north]{\footnotesize$r_m$} -- (7.5,1.5) node[anchor=south]{\footnotesize$r_m$};
\end{tikzpicture}
\end{equation}
for sequences of simple reflections $t=(t_1,\ldots,t_n)$ and $r=(r_1,\ldots,r_m)$. Let $\mu_{t,s,r}:\BS(tssr)\to\BS(tsr)$
be the map\footnote{here $tssr=(t_1,\ldots,t_n,s,s,r_1,\ldots,r_m)$.} given by $\mu_{t,s,r}([g_1:\cdots:g_n:g'_1:g_2':g''_1:\cdots:g''_m\rr)=[g_1:\cdots:g_n:g'_1g_2':g''_1:\cdots:g''_m\rr$.
Arguing as for $\iota_{t,s,r}^\star$, we get
\theta_{tssr}^{(-1)}\mu_{t,s,r}^\star\theta_{tsr}(a_1\otimes\cdots\otimes a_{n+1}\otimes b_1\otimes\cdots\otimes b_{m+1})
=a_1\otimes\cdots\otimes a_{n+1}\otimes1\otimes b_1\otimes\cdots\otimes b_{m+1}.

At this point, we can compute the coordinate form of merge diagram~(\ref{eq:27:m}).
We clearly have $\mu_s\iota_{s,s,\emptyset}=\id$. Taking the push-forwards, we get $(-\mu_{s\star})(-\iota_{s,s,\emptyset\star})=\id$.
Expressed diagrammatically, we get our first relation:
%\begin{tikzcd}[row sep=small,column sep=small]
%{}\arrow[dash,dashed]{rrrr}{s}&&\arrow[dash,end anchor={[yshift=-2pt]north},start anchor={[yshift=-1pt]north}]{d}&&{}\\
%{}&\arrow[dash,end anchor={[yshift=-2pt]north},start anchor={[yshift=-1pt]north}]{ur}&&\arrow[dash,end anchor={[yshift=-2pt]north},start anchor={[yshift=-1pt]north}]{ul}&{}\\
\begin{center}
\begin{tikzpicture}[baseline=25pt]
\draw[dashed] (0,0) -- (3,0);
\draw[dashed] (0,2) -- (3,2);
\draw[red] (1.5,2) node[color=black,anchor=south]{\footnotesize$s$} -- (1.5,1.5);
\draw[red] (1.5,1.5) -- (1,1);
\draw[red] (1.5,1.5) -- (2,1);
\draw[red] (1,1) -- (1,0) node[color=black,anchor=north]{\footnotesize$s$};
\draw[red] (2,1) -- (2,0.5);
\draw[red,fill=red] (2,0.5) circle(0.06);
%\draw (3.7,1) node{$=$};
\end{tikzpicture}
\qquad
\qquad
\begin{tikzpicture}[baseline=25pt]
\draw[dashed] (0,0) -- (3,0);
\draw[dashed] (0,2) -- (3,2);
\draw[red] (1.5,2) node[color=black,anchor=south]{\footnotesize$s$} -- (1.5,0) node[color=black,anchor=north]{\footnotesize$s$};
\end{tikzpicture}
\end{center}
\noindent
Note that we have proved this relation without resorting to coordinatization. By~(\ref{eq:29}), we get
\begin{multline*}
1\otimes1=\theta_s^{-1}\mu_{s\star}\iota_{s,s,\emptyset\star}\theta_s(1\otimes 1)=\theta_s^{-1}(-\mu_{s\star})\theta_{(s,s)}\(1\otimes\frac{\alpha_s}2\otimes 1+1\otimes 1\otimes\frac{\alpha_s}2\)\\[8pt]
=\theta_s^{-1}(-\mu_{s\star})\theta_{(s,s)}\(1\otimes\frac{\alpha_s}2\otimes 1\)-\theta_s^{-1}\mu_{s\star}\theta_{(s,s)}\(1\otimes 1\otimes 1\)\frac{\alpha_s}2.
\end{multline*}
As $\theta_s^{-1}\mu_{s\star}\theta_{(s,s)}\(1\otimes 1\otimes 1\)=0$ for the degree reason, we get
\theta_s^{-1}(-\mu_{s\star})\theta_{(s,s)}\(1\otimes\frac{\alpha_s}2\otimes 1\)=1\otimes1.
Let us take any $b\in H(s)$. Applying~(\ref{eq:32}) and the above formula, we get
%%>Написать о нём.
\begin{multline*}
\theta_s^{-1}(-\mu_{s\star})\theta_{(s,s)}(1\otimes b\otimes 1)=\\
-P_s(b)\,\theta_s^{-1}\mu_{s\star}\theta_{(s,s)}(1\otimes 1\otimes 1)
+\partial_s(b)\,\theta_s^{-1}(-\mu_{s\star})\theta_{(s,s)}\(1\otimes\frac{\alpha_s}2\otimes 1\)
=\partial_s(b)\otimes 1.
\end{multline*}
As this map is a homomorphism of bimodules, we get
\theta_s^{-1}(-\mu_{s\star})\theta_{(s,s)}(a\otimes b\otimes c)=a\partial_s(b)\otimes c.
The arguments described in this section, allow us to extend $-\mu_{s\star}$ to the morphism $-\mu_{t,s,r\star}$
represented by the diagram
\begin{equation}\tag{$-\mu_{t,s,r\star}$}
\begin{tikzpicture}[baseline=19pt]
\draw[dashed] (-0.5,0) -- (8,0);
\draw[dashed] (-0.5,1.5) -- (8,1.5);
\draw[red] (3,0) node[anchor=north]{\footnotesize$s$}-- (3.75,0.75);
\draw[red] (4.5,0) node[anchor=north]{\footnotesize$s$}-- (3.75,0.75);
\draw[red] (3.75,0.75)-- (3.75,1.5)node[anchor=south]{\footnotesize$s$};
%\draw[red,fill=red] (3,0.75) circle(0.06);
\draw (2.25,0) node[anchor=north]{\footnotesize$t_n$} -- (2.25,1.5) node[anchor=south]{\footnotesize$t_n$};
\draw (0,0) node[anchor=north]{\footnotesize$t_1$} -- (0,1.5) node[anchor=south]{\footnotesize$t_1$};
\draw (0.75,0) node[anchor=north]{\footnotesize$t_2$} -- (0.75,1.5) node[anchor=south]{\footnotesize$t_2$};
\draw  (1.5,0.7) node{\footnotesize$\cdots$};
\draw (5.25,0) node[anchor=north]{\footnotesize$r_1$} -- (5.25,1.5) node[anchor=south]{\footnotesize$r_1$};
\draw (6,0) node[anchor=north]{\footnotesize$r_2$} -- (6,1.5) node[anchor=south]{\footnotesize$r_2$};
\draw  (6.75,0.7) node{\footnotesize$\cdots$};
\draw (7.5,0) node[anchor=north]{\footnotesize$r_m$} -- (7.5,1.5) node[anchor=south]{\footnotesize$r_m$};
\end{tikzpicture}
\end{equation}
and prove the coordinate formula
\begin{multline*}
\theta_{tsr}^{-1}(-\mu_{t,s,r\star})\theta_{tssr}(a_1\otimes\cdots\otimes a_n\otimes a\otimes b\otimes c\otimes b_2\otimes\cdots\otimes b_{m+1})\\[3pt]
=a_1\otimes\cdots\otimes a_n\otimes a\partial_s(b)\otimes c\otimes b_2\otimes\cdots\otimes b_{m+1}.
\end{multline*}

\subsection{Generalized Bott-Samelson varieties}\label{Generalized Bott-Samelson varieties} For any nonempty set $S$ of simple reflections,
let $G_S$ be the subgroup of $G$ generated by all groups $G_s$, where $s\in S$.
Let $S=(S_1,\ldots,S_n)$ be a sequence of nonempty sets of reflections. Then we define
\BS(S)=G_{S_1}\mathop{\times}\limits_K{G_{S_2}}\mathop{\times}\limits_K\cdots\mathop{\times}\limits_KG_{S_n}/K
Clearly, if each $S_i$ is a singleton, then $\BS(S)=\BS(s)$, where $S_i=\{s_i\}$ and $s=(s_1,\ldots,s_n)$.
Bearing in mind this fact, we will omit brackets for singletons. For example, we abbreviate
$\BS(\{s\},\{t,r\},\{p\},\{q\})$ to $\BS(s,\{t,r\},p,q)$. We denote the equivariant cohomologies by
$H(S)=H_K^\bullet(\BS(S),\k)$, use the similar abbreviation and draw $H(S)$ and the identical morphisms
between them similarly to the cohomologies and morphisms for the usual Bott-Samelson varieties. For example,
the identity map $H(s,\{t,r\},p,q)\to H(s,\{t,r\},p,q)$ is depicted as

\vspace{-6pt}

\begin{center}
\begin{tikzpicture}
\draw[dashed] (0,1.5) -- (4,1.5);
\draw[dashed] (0,0) -- (4,0);
\draw[red] (0.5,0)  node[anchor=north]{\footnotesize$s$};
\draw(1.5,0)  node[anchor=north]{\footnotesize$\color{black}\{\color{blue}t\color{black},\color{green}r\color{black}\}$};
\draw[orange] (2.5,0)  node[anchor=north]{\footnotesize$p$};
\draw[violet] (3.5,0)  node[anchor=north]{\footnotesize$q$};

%\draw (2.15,0) node[anchor=north]{\footnotesize$\cdots$};
\draw[red] (0.5,1.5)  node[anchor=south]{\footnotesize$s$};
\draw(1.5,1.5)  node[anchor=south]{\footnotesize$\color{black}\{\color{blue}t\color{black},\color{green}r\color{black}\}$};
%\draw[green] (3,1.5)  node[anchor=south]{\footnotesize$s_n$};
\draw[orange] (2.5,1.5)  node[anchor=south]{\footnotesize$p$};
\draw[violet] (3.5,1.5)  node[anchor=south]{\footnotesize$q$};
%\draw (2.15,1.5) node[anchor=south]{\footnotesize$\cdots$};
\draw[red] (0.5,0) -- (0.5,1.5);
\draw[blue,dash pattern= on 4pt off 4pt] (1.5,0) -- (1.5,1.5);
\draw[green,dash pattern= on 4pt off 4pt,dash phase=4pt] (1.5,0) -- (1.5,1.5);

\draw[orange] (2.5,0) -- (2.5,1.5);
\draw[violet] (3.5,0) -- (3.5,1.5);
%\draw (2.15,0.7) node{\footnotesize$\cdots$};
\end{tikzpicture}
\end{center}

\vspace{-4pt}

We will consider here only the case where each set $S_i$ is a singleton with exactly one exception
for which this set consists of two distinct simple reflections (as in the example above).
These varieties also have complex structures coming from algebraic realization similar to~(\ref{eq:BSalg}).

\subsection{Two-color morphisms}\label{Two-color morphisms}
%First we restrict our attention to the Bott-Samelson varieties $\BS(s,t)$
%for sequences $s$ of simple reflections.
For the simplicity of exposition, we will omit the coordinatization maps $\theta_s$,
thus thinking of them as identity maps. We denote
x_s=\frac{\alpha_s}2,\quad c_s=x_s\otimes 1+1\otimes x_s
for any simple reflection $s$.

If we consider $H(s)$ as a left $\R$-module, then it has the following $\R$-basis:
\{1\otimes x_{s_1}^{\epsilon_1}\otimes\cdots\otimes x_{s_n}^{\epsilon_n}\suchthat \epsilon_1,\ldots,\epsilon_n\in\{0,1\}\}.
It consists of $2^n$ elements, where $n$ is the length of $s$.
We denote by $H(s)^{<}$ the left $\R$-sub\-module of $H(s)$ generated by all
above basis elements such that $\epsilon_i=0$ for some $i$.
The remaining basis element $1\otimes x_{s_1}\otimes\cdots\otimes x_{s_n}$
will be called the {\it normal element} of $H(s)$, see~\cite[D\'efinition 4.6]{Libedinsky}.

We leave the proof of the following result to the reader.

\begin{proposition}\label{proposition:4}
$c_{s_1}\cdots c_{s_n}=1\otimes x_{s_1}\otimes\cdots\otimes x_{s_n}+h$ for some $h\in H(s)^{<}$.
\end{proposition}

For any distinct simple reflections $s$ and $t$, we denote by $m_{s,t}$
the order of the product $st$. We have $m_{s,t}=m_{t,s}\in\{2,3,4,6\}$.
% and this number can take one of the following values:
We define the following sequence
\begin{array}{ll}
(s,t),&\text{ if }m_{s,t}=2,\\
(s,t,s),&\text{ if }m_{s,t}=3,\\
(s,t,s,t)&\text{ if }m_{s,t}=4,\\
(s,t,s,t,s,t)&\text{ if }m_{s,t}=6.
\end{array}
\right.
%We have
%st=ts,\quad sts=tst,\quad stst=tsts,\quad ststst=tststs,
The product of the reflections of $[s,t]$ is denoted by $w_{s,t}$. The braid relation for this pair reads as $w_{s,t}=w_{t,s}$.
We will use the notation $\BS(s,t,\ldots)=\BS([s,t])$ and $H(s,t,\ldots)=H([s,t])$
and denote by $1\otimes x_s\otimes x_t\otimes\cdots$ the normal element of $H(s,t,\ldots)$.
Similarly $c_sc_t\cdots$ denotes one of the product $c_sc_t$, $c_sc_tc_s$, $c_sc_tc_sc_t$, $c_sc_tc_sc_tc_sc_t$
depending on $m_{s,t}$.

The image of the twisting map $A_{[s,t]}$, see~(\ref{eq:30}),
is actually contained in the Schubert variety
$X_{s,t}=\overline{K_{w_{s,t}}/K}=\BS(\{s,t\})$. Thus we get the map
$\eta_{s,t}:\BS(s,t,\ldots)\to X_{s,t}$. The twisting maps for both sides are $A_{[s,t]}$
and the natural inclusion to $G/K$, respectively. Note also that both these spaces
are smooth manifolds having real dimension $2m_{s,t}$. Therefore, we get the following two morphisms of $\R$-$\R$-bimodules:
\eta_{s,t}^\star:H_K^\bullet(X_{s,t},\k)\to H(s,t,\ldots),\quad
\eta_{s,t\star}:H(s,t,\ldots)\to H_K^\bullet(X_{s,t},\k)
presented by the diagrams
%\draw[dashed] (0,0) -- (3,0);
%\draw[dashed] (0,1.5) -- (3,1.5);
%\draw[red] (1.5,1.5) node[anchor=south]{\footnotesize$s$} -- (1.5,0.75);
%\draw[red] (1.5,0.75) -- (0.75,1.5) node[anchor=south]{\footnotesize$s$};
%\draw[red] (1.5,0.75) -- (2.25,1.5) node[anchor=south]{\footnotesize$s$};
%Composing these maps, we will get the desired maps.

\vspace{-5pt}
\begin{tikzpicture}[baseline=19pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red] (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$} -- (0,1.5) node[anchor=south]{\footnotesize$s$};
\draw[blue] (1.5,0) -- (0.75,1.5) node[anchor=south]{\footnotesize$t$};
\draw[purple] (1.5,0) -- (3,1.5);
\draw  (1.75,1) node{\footnotesize$\cdots$};
\end{tikzpicture}
\hspace{60pt}
\begin{tikzpicture}[baseline=19pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red] (1.5,1.5) node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$} -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,1.5) -- (0.75,0) node[anchor=north]{\footnotesize$t$};
\draw[purple] (1.5,1.5) -- (3,0);
\draw  (1.75,0.5) node{\footnotesize$\cdots$};
\end{tikzpicture}
\hfill(\eta_{s,t\star})
\vspace{1pt}

\noindent
Here the purple line can be either red or blue, depending on the parity of $m_{s,t}$.

First, we calculate the following composition.
\begin{lemma}\label{lemma:10}
\end{lemma}
\begin{proof}
Applying the projection formula, we get
\eta_{s,t\star}\eta_{s,t}^\star(h)=\eta_{s,t\star}(1\cup\eta_{s,t}^\star(h))=\eta_{s,t\star}(1)\cup h
for any $h\in H_K^\bullet(X_{s,t},\k)$. Therefore, it suffices to prove that $\eta_{s,t\star}(1)=1$.
There is an open subset $V\subset X_{s,t}$ such that
$\eta_{s,t}$ induces the orientation preserving diffeomorphism $\eta'_{s,t}:U\ito V$, where $U=\eta_{s,t}^{-1}(V)$.
Consider the following Cartesian square
\begin{tikzcd}
\end{tikzcd}
By (the equariant version of)~(\ref{eq:24}), we get
As $X_{s,t}$ is connected, the above calculation proves that $\eta_{s,t\star}(1)=1$.
\end{proof}
Diagramatically expresed this lemma looks as follows:
\vspace{-5pt}
\begin{equation*}
\begin{tikzpicture}[baseline=29pt,y=40pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw  (1.75,0.75) node{\footnotesize$\cdots$};
\draw [red] plot [smooth,tension=1.2] coordinates {(1.5,0) (0.25,0.75) (1.5,1.5)};
\draw [blue] plot [smooth,tension=1.2] coordinates {(1.5,0) (1,0.75) (1.5,1.5)};
\draw [purple] plot [smooth,tension=1.2] coordinates {(1.5,0) (2.75,0.75) (1.5,1.5)};
\draw (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\draw (1.5,1.5) node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\end{tikzpicture}
\qquad
\qquad
%\draw[dashed] (-0.5,0) -- (3.5,0);
\begin{tikzpicture}[baseline=29pt,y=40pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red,dash pattern= on 4pt off 4pt] (1.5,0) -- (1.5,1.5);
\draw[blue,dash pattern= on 4pt off 4pt,dash phase=4pt] (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$}-- (1.5,1.5)node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\end{tikzpicture}
\end{equation*}
\vspace{1pt}

\subsection{The $2m_{s,t}$-valent vertex} Here, we are going to consider
the composition $\eta_{t,s}^\star\eta_{s,t\star}:H(s,t,\ldots)\to H(t,s,\ldots)$,
which is a morphism of $\R$-$\R$-bimodules. It is depicted as follows:

\vspace{-5pt}
\begin{equation}\tag{$\eta_{t,s}^\star\eta_{s,t\star}$}
\begin{tikzpicture}[baseline=29pt,y=40pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,0.75) -- (0.75,0) node[anchor=north]{\footnotesize$t$};
\draw[purple] (1.5,0.75) -- (3,0);
\draw  (1.75,0.25) node{\footnotesize$\cdots$};
\draw[blue] (1.5,0.75) -- (0,1.5) node[anchor=south]{\footnotesize$t$};
\draw[red] (1.5,0.75) -- (0.75,1.5) node[anchor=south]{\footnotesize$s$};
\draw[purple] (1.5,0.75) -- (3,1.5);
\draw  (1.75,1.2) node{\footnotesize$\cdots$};
\end{tikzpicture}
\end{equation}

Note that all such morphisms
are proportional~\cite[Proposition 4.3]{Libedinsky}.

%Supposing the converse $\eta_{t,s}^\star\eta_{s,t\star}=0$, we get by Lemma~\ref{lemma:10} the contradiction

\begin{lemma}\label{lemma:12}
$\ker\mu_{s,t\star}\subset H(s,t,\ldots)^{<}$.
\end{lemma}
\begin{proof}
Let $h\in\ker\mu_{s,t\star}$. We have $h=\alpha\otimes x_s\otimes x_t\otimes\cdots+h'$ for some $h'\in H(s,t,\ldots)^{<}$.
For the degree reason, we get $\eta_{t,s}^\star\eta_{s,t\star}(h')\in H(t,s,\ldots)^{<}$.
Therefore, it follows from $\eta_{t,s}^\star\eta_{s,t\star}(h)=0$ that
\alpha\,\eta_{t,s}^\star\eta_{s,t\star}(1\otimes x_s\otimes x_t\otimes\cdots)\in H(t,s,\ldots)^{<}.
By~\cite[Proposition 4.3 and Lemme 4.7]{Libedinsky}, this is only possible if $\alpha=0$.
\end{proof}

Our next aim is to prove the following normalization condition.

\begin{lemma}\label{lemma:13}
$\eta_{t,s}^\star\eta_{s,t\star}(1\otimes x_s\otimes x_t\otimes\cdots)=1\otimes x_t\otimes x_s\otimes\cdots+h$
for some $h\in H(t,s,\ldots)^{<}$.
\end{lemma}
\begin{proof}
Let us consider the following chain of inclusions:
\begin{tikzcd}
\pt\arrow{r}{\iota_{\emptyset,s,\emptyset}}&\BS(s)\arrow{r}{\iota_{(s),t,\emptyset}}&[9pt]\BS(s,t)\arrow{r}{\iota_{(s,t),s,\emptyset}}&[12pt]\cdots\arrow{r}{\iota_{(s,t,\ldots),r,\emptyset}}&[20pt]\BS(s,t,\ldots),
\end{tikzcd}
where $r=s$ or $r=t$ depending on the parity of $m_{s,t}$.
We denote the resulting composition by $\iota$. Its image is $[1:1:\cdots:1\rr$.
Taking the equivariant push-forward

\vspace{-10pt}
\begin{equation*}
\begin{tikzpicture}[baseline=29pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
%\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
%\draw[blue] (1.5,0.75) -- (0.75,0) node[anchor=north]{\footnotesize$t$};
%\draw[purple] (1.5,0.75) -- (3,0);
%\draw  (1.75,0.25) node{\footnotesize$\cdots$};
%\draw[blue] (1.5,0.75) -- (0,1.5) node[anchor=south]{\footnotesize$t$};
%\draw[red] (1.5,0.75) -- (0.75,1.5) node[anchor=south]{\footnotesize$s$};
%\draw[purple] (1.5,0.75) -- (3,1.5);
\draw  (1.75,1) node{\footnotesize$\cdots$};
\draw[red] (0,0.2) -- (0,1.5) node[anchor=south]{\footnotesize$s$};
\draw[blue] (0.75,0.5) -- (0.75,1.5) node[anchor=south]{\footnotesize$t$};
\draw[purple] (3,1.2) -- (3,1.5) node[anchor=south]{\footnotesize$r$};
\draw[red,fill=red] (0,0.2) circle(0.06);
\draw[blue,fill=blue] (0.75,0.5) circle(0.06);
\draw[purple,fill=purple] (3,1.2) circle(0.06);
\end{tikzpicture}
\end{equation*}

\bigskip

\noindent
we get by~(\ref{eq:29}) that the image of $(-1)^{m_{s,t}}$
is equal to $c_sc_t\cdots$. Hence $\eta_{s,t\star}(c_sc_t\cdots)=(\eta_{s,t}\iota)_\star((-1)^{m_{s,t}})$.
Arguing similarly, we get $\eta_{t,s\star}(c_tc_s\cdots)=(\eta_{t,s}\iota)_\star((-1)^{m_{s,t}})$.
As $\eta_{s,t}\iota=\eta_{t,s}\iota$, we get by Lemmas~\ref{lemma:10} and~\ref{lemma:12} that
\eta_{t,s}^\star\eta_{s,t\star}(c_sc_t\cdots)-c_tc_s\cdots\in\ker\eta_{t,s\star}\subset H(t,s,\ldots)^{<}.
To conclude the proof, it suffices to apply Proposition~\ref{proposition:4} and the fact
that $\eta_{t,s}^\star\eta_{s,t\star}$, being a degree preseving homomorphism $H(s,t,\ldots)\to H(t,s,\ldots)$
of left $\R$-modules, maps $H(s,t,\ldots)^{<}$ to $H(t,s,\ldots)^{<}$.
\end{proof}

Thus the composition $\eta_{t,s}^\star\eta_{s,t\star}$ is just the map $f_{s,t}$ as in~\cite[Lemme 4.7]{Libedinsky}.

\subsection{The Jones-Wenzl projector}\label{The Jones-Wenzl projector} This map is given by the composition $\eta_{s,t}^\star\eta_{s,t\star}$
and is represented by the diagram:

\vspace{-5pt}
\begin{equation}\tag{$\eta_{s,t}^\star\eta_{s,t\star}$}
\begin{tikzpicture}[baseline=29pt,y=40pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,0.75) -- (0.75,0) node[anchor=north]{\footnotesize$t$};
\draw[purple] (1.5,0.75) -- (3,0);
\draw  (1.75,0.25) node{\footnotesize$\cdots$};
\draw[red] (1.5,0.75) -- (0,1.5) node[anchor=south]{\footnotesize$s$};
\draw[blue] (1.5,0.75) -- (0.75,1.5) node[anchor=south]{\footnotesize$t$};
\draw[purple] (1.5,0.75) -- (3,1.5);
\draw  (1.75,1.2) node{\footnotesize$\cdots$};
\end{tikzpicture}
\end{equation}
\vspace{1pt}

Form Lemma~\ref{lemma:10}, we immediately get the relations

\vspace{-5pt}
\begin{equation*}
\begin{tikzpicture}[baseline=19pt,y=40pt]
\draw[dashed] (-0.5,-0.5) -- (3.5,-0.5);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw  (1.75,0.5) node{\footnotesize$\cdots$};
\draw [red] plot [smooth,tension=1.2] coordinates {(1.5,0) (0.25,0.5) (1.5,1)};
\draw [blue] plot [smooth,tension=1.2] coordinates {(1.5,0) (1,0.5) (1.5,1)};
\draw [purple] plot [smooth,tension=1.2] coordinates {(1.5,0) (2.75,0.5) (1.5,1)};
%\draw (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
%\draw (1.5,1) node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\draw [red]  (1.5,1) -- (0.25,1.5) node[anchor=south]{\footnotesize$s$};
\draw [blue] (1.5,1) -- (1,1.5) node[anchor=south]{\footnotesize$t$};
\draw [purple] (1.5,1) -- (2.75,1.5);
\draw [red]  (1.5,0) -- (0.25,-0.5) node[anchor=north]{\footnotesize$s$};
\draw [blue] (1.5,0) -- (1,-0.5) node[anchor=north]{\footnotesize$t$};
\draw [purple] (1.5,0) -- (2.75,-0.5);
\draw  (1.75,1.3) node{\footnotesize$\cdots$};
\draw  (1.75,-0.3) node{\footnotesize$\cdots$};
\end{tikzpicture}
\;\;=\;\;
\begin{tikzpicture}[baseline=19pt,y=40pt]
\draw[dashed] (-0.5,-0.5) -- (3.5,-0.5);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
%\draw  (1.75,0.5) node{\footnotesize$\cdots$};
%\draw [red] plot [smooth,tension=1.2] coordinates {(1.5,0) (0.25,0.5) (1.5,1)};
%\draw [blue] plot [smooth,tension=1.2] coordinates {(1.5,0) (1,0.5) (1.5,1)};
%\draw [purple] plot [smooth,tension=1.2] coordinates {(1.5,0) (2.75,0.5) (1.5,1)};
%\draw (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
%\draw (1.5,1) node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\draw[red,dash pattern= on 4pt off 4pt] (1.5,0) -- (1.5,1);
\draw[blue,dash pattern= on 4pt off 4pt,dash phase=4pt] (1.5,0) -- (1.5,1);
\draw [red]  (1.5,1) -- (0.25,1.5) node[anchor=south]{\footnotesize$s$};
\draw [blue] (1.5,1) -- (1,1.5) node[anchor=south]{\footnotesize$t$};
\draw [purple] (1.5,1) -- (2.75,1.5);
\draw [red]  (1.5,0) -- (0.25,-0.5) node[anchor=north]{\footnotesize$s$};
\draw [blue] (1.5,0) -- (1,-0.5) node[anchor=north]{\footnotesize$t$};
\draw [purple] (1.5,0) -- (2.75,-0.5);
\draw  (1.75,1.3) node{\footnotesize$\cdots$};
\draw  (1.75,-0.3) node{\footnotesize$\cdots$};
\end{tikzpicture}
\;\;=\;\;
\begin{tikzpicture}[baseline=29pt,y=40pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,0.75) -- (0.75,0) node[anchor=north]{\footnotesize$t$};
\draw[purple] (1.5,0.75) -- (3,0);
\draw  (1.75,0.25) node{\footnotesize$\cdots$};
\draw[red] (1.5,0.75) -- (0,1.5) node[anchor=south]{\footnotesize$s$};
\draw[blue] (1.5,0.75) -- (0.75,1.5) node[anchor=south]{\footnotesize$t$};
\draw[purple] (1.5,0.75) -- (3,1.5);
\draw  (1.75,1.2) node{\footnotesize$\cdots$};
\end{tikzpicture}
\end{equation*}

\vspace{-5pt}
\begin{equation*}
\begin{tikzpicture}[baseline=19pt,y=40pt]
\draw[dashed] (-0.5,-0.5) -- (3.5,-0.5);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw  (1.75,0.5) node{\footnotesize$\cdots$};
\draw [red] plot [smooth,tension=1.2] coordinates {(1.5,0) (0.25,0.5) (1.5,1)};
\draw [blue] plot [smooth,tension=1.2] coordinates {(1.5,0) (1,0.5) (1.5,1)};
\draw [purple] plot [smooth,tension=1.2] coordinates {(1.5,0) (2.75,0.5) (1.5,1)};
%\draw (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
%\draw (1.5,1) node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\draw [blue] (1.5,1) -- (0.25,1.5) node[anchor=south]{\footnotesize$t$};
\draw [red]  (1.5,1) -- (1,1.5) node[anchor=south]{\footnotesize$s$};
\draw [purple] (1.5,1) -- (2.75,1.5);
\draw [red]  (1.5,0) -- (0.25,-0.5) node[anchor=north]{\footnotesize$s$};
\draw [blue] (1.5,0) -- (1,-0.5) node[anchor=north]{\footnotesize$t$};
\draw [purple] (1.5,0) -- (2.75,-0.5);
\draw  (1.75,1.3) node{\footnotesize$\cdots$};
\draw  (1.75,-0.3) node{\footnotesize$\cdots$};
\end{tikzpicture}
\;\;=\;\;
\begin{tikzpicture}[baseline=19pt,y=40pt]
\draw[dashed] (-0.5,-0.5) -- (3.5,-0.5);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
%\draw  (1.75,0.5) node{\footnotesize$\cdots$};
%\draw [red] plot [smooth,tension=1.2] coordinates {(1.5,0) (0.25,0.5) (1.5,1)};
%\draw [blue] plot [smooth,tension=1.2] coordinates {(1.5,0) (1,0.5) (1.5,1)};
%\draw [purple] plot [smooth,tension=1.2] coordinates {(1.5,0) (2.75,0.5) (1.5,1)};
%\draw (1.5,0) node[anchor=north]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
%\draw (1.5,1) node[anchor=south]{\footnotesize$\color{black}\{\color{red}s\color{black},\color{blue}t\color{black}\}$};
\draw[red,dash pattern= on 4pt off 4pt] (1.5,0) -- (1.5,1);
\draw[blue,dash pattern= on 4pt off 4pt,dash phase=4pt] (1.5,0) -- (1.5,1);
\draw [blue] (1.5,1) -- (0.25,1.5) node[anchor=south]{\footnotesize$t$};
\draw [red]  (1.5,1) -- (1,1.5) node[anchor=south]{\footnotesize$s$};
\draw [purple] (1.5,1) -- (2.75,1.5);
\draw [red]  (1.5,0) -- (0.25,-0.5) node[anchor=north]{\footnotesize$s$};
\draw [blue] (1.5,0) -- (1,-0.5) node[anchor=north]{\footnotesize$t$};
\draw [purple] (1.5,0) -- (2.75,-0.5);
\draw  (1.75,1.3) node{\footnotesize$\cdots$};
\draw  (1.75,-0.3) node{\footnotesize$\cdots$};
\end{tikzpicture}
\;\;=\;\;
\begin{tikzpicture}[baseline=29pt,y=40pt]
\draw[dashed] (-0.5,0) -- (3.5,0);
\draw[dashed] (-0.5,1.5) -- (3.5,1.5);
\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,0.75) -- (0.75,0) node[anchor=north]{\footnotesize$t$};
\draw[purple] (1.5,0.75) -- (3,0);
\draw  (1.75,0.25) node{\footnotesize$\cdots$};
\draw[blue] (1.5,0.75) -- (0 ,1.5) node[anchor=south]{\footnotesize$t$};
\draw[red]  (1.5,0.75) -- (0.75,1.5) node[anchor=south]{\footnotesize$s$};
\draw[purple] (1.5,0.75) -- (3,1.5);
\draw  (1.75,1.2) node{\footnotesize$\cdots$};
\end{tikzpicture}
\end{equation*}

\subsection{Two-color dot contraction}\label{Two-color dot contraction} As an example for $m_{s,t}=3$, let us prove the following relation

\vspace{-5pt}
\begin{equation*}
\begin{tikzpicture}[baseline=25pt]
\draw[dashed] (0,0) -- (3,0);
\draw[dashed] (0,2) -- (3,2);
%\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[red] (1.5,1) -- (0.5,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (1.5,0) node[anchor=north]{\footnotesize$t$};
\draw[red] (1.5,1) -- (2.5,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (0.5,2) node[anchor=south]{\footnotesize$t$};
\draw[red] (1.5,1) -- (1.5,2) node[anchor=south]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (2.2,1.7);
\draw[blue,fill=blue] (2.2,1.7) circle(0.06);
\end{tikzpicture}
\qquad
\qquad
\begin{tikzpicture}[baseline=25pt]
\draw[dashed] (0,0) -- (3,0);
\draw[dashed] (0,2) -- (3,2);
%\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[red] (1.5,1) -- (0.5,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (1.5,0) node[anchor=north]{\footnotesize$t$};
\draw[red] (1.5,1) -- (2.5,0) node[anchor=north]{\footnotesize$s$};
\draw[red] (1.5,1) -- (0.8,1.7);
\draw[blue] (1.5,1) -- (1.5,2) node[anchor=south]{\footnotesize$t$};
\draw[red] (1.5,1) -- (2.5,2) node[anchor=south]{\footnotesize$s$};
\draw[red,fill=red] (0.8,1.7) circle(0.06);
\end{tikzpicture}
\end{equation*}
\vspace{1pt}
\noindent
Indeed, the left-hand side and the right-hand side are equal to
respectively. Hence it suffices to prove that $\eta_{t,s}i_{(t,s),t,\emptyset}=\eta_{s,t}i_{\emptyset,s,(t,s)}$.
This follows from the following calculation:
\begin{tikzcd}
{}&{}[1:g:g'\rr\arrow[mapsto,swap]{ur}{\eta_{s,t}}& \end{tikzcd}
For the other values of $m_{s,t}$ the proofs are similar.

\subsection{Horizontal extensions of two-color morphisms}\label{Horizontal extensions of two-color morphisms}
Finally, it remains to extend the morphisms $\eta_{s,t}^\star$ and $\eta_{s,t\star}$ horizontally.
It can be done similarly to Section~\ref{Horizontal_extensions}.
First, let $p=(p_1,\ldots,p_n)$ be a sequence of simple reflections. Then we define the map
\eta_{p,s,t}:\BS(p_1,\ldots,p_n,s,t,\ldots)\to\BS(p_1,\ldots,p_n,\{s,t\})
by $[g_1:\cdots:g_n:g'_1:\cdots:g'_{m_{s,t}}\rr\mapsto[g_1:\cdots:g_n:g'_1\cdots g'_{m_{s,t}}\rr$.
We have the following commutative diagram
\begin{tikzcd}
H(p)\otimes_\R H_K^\bullet(X_{s,t},\k)\arrow{d}[swap]{\theta_{p,s,t}}{\wr}\arrow{r}{\id\otimes\eta_{s,t}^\star}&H(p)\otimes_\R H(s,t,\ldots)\arrow{d}{\theta_{p,[s,t]}}[swap]{\wr}\\
\end{tikzcd}
where $\theta_{p,s,t}$ corresponds to
\phi_{p,s,t}:\BS(p_1,\ldots,p_n,\{s,t\})\,{}_K\!\!\times E\to(\BS(p)\,{}_K\!\!\times E)\times(X_{s,t}\,{}_K\!\!\times E)
as in Section~\ref{The_embedding} for the following set of data:
%following case of Theorem~\ref{theorem:1}:
X=G_{p_1}\mathop{\times}\limits_K{G_{p_2}}\mathop{\times}\limits_K\cdots\mathop{\times}\limits_KG_{p_n},\quad Y=X_{s,t},\quad L=R=P=Q=K,
and $\alpha:X\to G$ defined by $[g_1:\cdots:g_n]\mapsto g_1\cdots g_n$.

Now let $q=(q_1,\ldots,q_m)$ be another sequence of simple reflections. We define the map
\eta_{p,s,t,q}:\BS(p_1,\ldots,p_n,s,t,\ldots,q_1,\ldots,q_m)\to\BS(p_1,\ldots,p_n,\{s,t\},q_1,\ldots,q_m)
by $[g_1:\cdots:g_n:g'_1:\cdots:g'_{m_{s,t}}:g''_1:\cdots:g''_m\rr\mapsto[g_1:\cdots:g_n:g'_1\cdots g'_{m_{s,t}}:g''_1:\cdots:g''_m\rr$.
We have the following commutative diagram
\begin{tikzcd}[column sep=35pt]
H(p_1,\ldots,p_n,\{s,t\})\otimes_\R H(q)\arrow{d}[swap]{\theta_{p,s,t,q}}{\wr}\arrow{r}{\eta_{p,s,t}^\star\otimes\id}&H(p_1,\ldots,p_n,s,t,\ldots)\otimes H(q)\arrow{d}{\theta_{p[s,t],q}}[swap]{\wr}\\
\end{tikzcd}
where $\theta_{p,s,t,q}$ is defined similarly to $\theta_{p,s,t}$. We leave it to the reader
to write down similar diagrams for push-forwards, to draw the correspondind diagrams and to compose them
(see, the examples of diagrams in the introduction).

\def\sep{\\[-5pt]}

\begin{thebibliography}{00}

[1] A.\,Borel, Sur la cohomolgie des espaces fibr\'es principaux et des espaces homog\'enes de group de Lie compact, {\it Ann. Math}, 1953, 57, 115--207.\sep
[2] M.\,Brion, Equivariant cohomology and equivariant intersection theory, In: Broer, A., Daigneault, A., Sabidussi, G. (eds) Representation Theories and Algebraic Geometry. Nato ASI Series, vol 514. Springer, Dordrecht, 1998, 1--37.\sep
[3] R.\,Bott, H.\,Samelson,\, Application of the theory of Morse to the symmetric spaces, {\it Am. J. Math.}, {\bf 80} (1958) 964--1029.\sep
[4] W.\,Fulton, Equivariant cohomlgy in algebrac topology, lectures at Columbia university, notes by D. Anderson, Appendix A.\sep
[5] A.\,Hatcher, Algebraic topology, {\it Cambridge Univ. Press}, Cambridge, 2002.\sep
%[6] W.\,Y.\,Hsiang, Cohomology Theory of Topological Transformation Groups, Springer, New York, 1975.
[7] J. C.\,Jantzen, Moment graphs and representations, {\it Inst. Fourier, Geometric Methods in Representation Theory, Summer School Proceedings}, 2008.\sep
[8] N.\,Libedinsky, Sur la cat\'egorie des bimodules de Soergel, {\it J. Algebra}, 2008, 320(7):2675--2694.\sep
[9] J.\,Rotman, An introduction to homological algebra, {\it Academic Press}, New York, 1979.\sep
[10] V.\,Shchigolev, Nested fibre bundles in Bott-Samelson varieties, {\it J. Algebra}, {\bf 605} (2022), 296--338.\sep
[11] R.\,Steinberg, Lectures on Chevalley Groups, New Haven, 1968.\sep
[12] B.\,Webster, G.\,Williamson, A geometric model for Hochschild homology of Soergel bimodules, {\it Geometry \& Topology}, 12(2), (2008), 1243--1263.\sep
[13] B.\,Elias, G.\,Williamson, Soergel calculus, {\it Represent. Theory}, {\bf 20}, 295--374 (2016).
\end{thebibliography}

\bigskip

\noindent
Financial University under the Government of the Russian Federation, \\ Leningradsky Prospekt 49, Moscow, 125993, Russia,

\end{document}

\hspace{-50pt}
\begin{tikzcd}
\R\otimes_{H^\bullet_{G_s}(\pt,\k)}\R\arrow{r}{\theta_{(s)}}&
\end{tikzcd}

\vspace{-20pt}

\begin{tikzcd}
\arrow{r}{\theta_{(s,s)}^{-1}}&\R\otimes_{H^\bullet_{G_s}(\pt,\k)}\R\otimes_{H^\bullet_{G_s}(\pt,\k)}\R.
\end{tikzcd}

\vspace{-5pt}
\begin{equation*}
\begin{tikzpicture}[baseline=19pt]
\draw[dashed] (0,0) -- (3,0);
\draw[dashed] (0,2) -- (3,2);
%\draw[red] (1.5,0.75) -- (0,0) node[anchor=north]{\footnotesize$s$};
\draw[red] (1.5,1) -- (0.5,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (1.5,0) node[anchor=north]{\footnotesize$t$};
\draw[red] (1.5,1) -- (2.5,0) node[anchor=north]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (0.5,2) node[anchor=south]{\footnotesize$t$};
\draw[red] (1.5,1) -- (1.5,2) node[anchor=south]{\footnotesize$s$};
\draw[blue] (1.5,1) -- (2.5,2) node[anchor=south]{\footnotesize$t$};
\end{tikzpicture}
\end{equation*}
\vspace{1pt}

\label{theorem:5}