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      {\sc Proof.} }
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\def\intP{{\stackrel\circ P}}
\author{I.\,A.\,Dynnikov\thanks{This work is supported in part by
RFBI (grant no. 99-01-00090)}}
\title{Finitely presented semigroups and groups in knot theory}
\date{Moscow State University\\
Vorobyovy gory, Moscow 119899\\
e-mail: dynnikov@mech.math.msu.su}
\begin{document}
\maketitle
\abstract{In this paper we construct finitely presented semigroups
whose central elements are in one-to-one correspondence
with isotopy classes of 
%  v russkom variante dobavit': neorientirovannykh
non-oriented 
%
links in $\real^3$. Solving the
word problem for those semigroups is equivalent to solving
the classification problem for links and tangles.
Also, we give a construction of finitely presented groups
containing the braid group as a subgroup.}
\section{Introduction}
By a link in $\real^3$ we shall mean the union of a number
of pairwise non-intersecting simple 
% v ruskkom variante dobavit': zamknutykh
closed 
%
polygonal arcs in
$\real^3$. In this paper we shall consider only
{\it non-oriented\/} links. Two links are called equivalent
if they are ambiently isotopic in $\real^3$. Let us denote
by $L_n$, $n=0,1,2,\dots$, the set of isotopy classes of
$n$-component links. The set $L_0$ consists of exactly one
element, the empty link; $L_1$ is the set of isotopy classes
of all knots in $\real^3$.

Let
$$L=L_0\cup L_1\cup L_2\cup\dots$$
be the set of isotopy classes of all links. $L$ carries natural
structure of an Abelian semigroup with multiplication given
by taking the disjoint union of two links separated by
a plane and with the empty link as the unit. The semigroup
$L$ is not finitely generated. As an Abelian semigroup
it is freely generated by the isotopy classes of
so-called nonsplit links.

In the present paper we construct a series of semigroups
$Y_n$, $n=3,4,\dots,\infty$, possesing the following properties:
\begin{enumerate}
\item\label{pr1}
the center $Z(Y_n)$ of the semigroup $Y_n$ is isomorphic to
the semigroup $L$;
\item\label{pr2}
the group $D_n$ of invertible elements of $Y_n$ contains a subgroup
isomorphic to the braid group of infinitely many strands;
\item
for $n<\infty$ the semigroup $Y_n$ and the group $D_n$ are finitely
presented ones, i.e.\ they can be presented by a finite number
of generators and defining relations.
\end{enumerate}

Constructing a semigroup having only properties~\ref{pr1})~and
\ref{pr2}) is easy. We shall do this in Section~\ref{standard_tangles}
by modifying slightly the standard construction of the category
of non-oriented tangles. A decomposition of an element
of that semigroup into a product of the generators is
given by any generic diagram of the corresponding tangle,
the defining relations express isotopies of diagrams and
the Reidemeister moves. Those relations, more preceisely their
analogues for the category of oriented tangles, have been
used in~\cite{tur89} for constructing link invariants.

The main difference of the so constructed semigroup $T$ from
the semigroups $Y_n$, $n=3,4,\dots$, is in that it can not
be presented by finitely many defining relations.
However, the semigroup $T$ can be embedded in each
of the semigroups $Y_n$, which are finitely presented.

In this way we reduce the topological problem of recognizing
links to a purely algebraic one, the word problem for a
finitely presented semigroup. According to paper
by S.\,V.\,Matveev~\cite{matv}, an algorithm
for recognizing links exists. Moreover, as S.\,V.\,Matveev
told the author, by enhancing that algorithm
,
one can construct an algorithm for recognizing tangles, and hence,
one can solve the word problem for the semigroups $Y_n$.
However, it would be interesting to see a purely algebraic proof
of solvability of the word problem for those semigroups and
 to
obtain in this way an alternative proof of recognizibility of links.

\section{Tangles}\label{standard_tangles}
Recall the construction of the category of (non-oriented)
tangles~\cite{tur89}. A tangle of type $(k,l)$ is an arbitrary
one-dimensional compact PL-submanifold 
% vstavit' v russkom variante:
$t$ 
%
in $\real^3$ lying in the region
\begin{equation}\label{slice}
\slice=\{(x,y,z)\in\real^3|\,0\le x\le1\}
\end{equation}
and having the set $\{(0,i,0)\}_{1\le i\le k}\cup
\{(1,j,0)\}_{1\le j\le l}
% vstavit' v russkom variante:
=t\cap\partial\slice
%
$ as its boundary. Two tangles
are called equivalent if they are ambiently isotopic in
$\slice$ assuming that all the boundary points of
$\slice$ are fixed under isotopy. Objects of the tangle category
$\cal T$ are non-negative integers, morphisms from
the object $l$ to the object $k$ are isotopy classes of
tangles of type $(k,l)$.

The composition $t_1\circ t_2$ of two arbitrary tangles
$t_1$ and $t_2$ of type $(k,l)$ and $(l,m)$ respectively is defined
as the union
$$\psi_1(t_1)\cup\psi_2(t_2),$$
where $\psi_{1,2}$ are the maps from $\slice$ to itself
defined by
$$\psi_1(x,y,z)=(x/2,y,z),\quad\psi_2(x,y,z)=((x+1)/2,y,z).$$
In this way, the composition of morphisms of the
category $\cal T$ is well defined. From the point
of view of this category the semigroup $L$ is the semigroup
of morphisms of the object $0$ to itself.

By modifying definitions slightly, we can obtain
a semigroup, a simpler object than the tangle category.
To this end, let us change the definition of tangle
in the following way. We shall call a tangle any PL-submanifold
$t$ in $\slice$ whose boundary $\partial t$ is the set
$$\{(0,i,0),(1,i,0)\}_{i>0}$$
with $t\cap\partial\slice=\partial t$, and such
that all its components lying far enough from the origin
are straight line segments of the form $[(0,i,0),(0,j,0)]$
(we shall denote the straight line segment connecting
points $A$ and $B$ by $[A,B]$). Under this condition the
difference $j-i$ is constant for large $i$. Objects of
this kind could be called tangles of type $(\infty,\infty)$.
An example of a tangle in the sence of the just given
definition
\begin{figure}\caption{A tangle of type $(\infty,\infty)$}\label{a tangle}
$$\begin{picture}(200,250)
\put(40,20){\circle*2}
\put(40,60){\circle*2}
\put(40,100){\circle*2}
\put(40,140){\circle*2}
\put(40,180){\circle*2}
\put(40,220){\circle*2}
\put(160,20){\circle*2}
\put(160,60){\circle*2}
\put(160,100){\circle*2}
\put(160,140){\circle*2}
\put(160,180){\circle*2}
\put(160,220){\circle*2}
\put(40,20){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,1,0)$ }}}
\put(40,60){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,2,0)$ }}}
\put(40,100){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,3,0)$ }}}
\put(40,140){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,4,0)$ }}}
\put(40,180){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,5,0)$ }}}
\put(40,220){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,6,0)$ }}}
\put(160,20){\raisebox{-0.4ex}{\hbox{ $(1,1,0)$}}}
\put(160,60){\raisebox{-0.4ex}{\hbox{ $(1,2,0)$}}}
\put(160,100){\raisebox{-0.4ex}{\hbox{ $(1,3,0)$}}}
\put(160,140){\raisebox{-0.4ex}{\hbox{ $(1,4,0)$}}}
\put(160,180){\raisebox{-0.4ex}{\hbox{ $(1,5,0)$}}}
\put(160,220){\raisebox{-0.4ex}{\hbox{ $(1,6,0)$}}}
\put(40,20){\line(3,2){86}}
\put(160,100){\line(-3,-2){26}}
\put(160,60){\line(-3,2){60}}
\put(160,140){\line(-3,-2){60}}
\put(40,60){\line(3,-1){35}}
\put(160,20){\line(-3,1){75}}
\put(40,100){\line(3,2){120}}
\put(40,140){\line(3,2){120}}
\put(40,180){\line(3,2){75}}
\put(40,220){\line(3,2){15}}
\put(100,235){\hbox to 0pt{\hss$\vdots$\hss}}
\end{picture}$$
\end{figure}
is shown in fig.~\ref{a tangle}.

Notice that formally speaking, the composition of two tangles
may now not be a tangle, but it is a PL-submanifold ambiently
isotopic to a tanlge in $\slice$. Hence the composition is
well defined on the set $T$ of isotopy classes
of such tangles. Endowed with this operation the set
$T$ becomes a semigroup.

There is an evident functor $F$ from the category $\cal T$ to the semigroup
$T$. It is defined as follows. Let $f$ be a morphism from an
object $l$ to $k$. The morphism $f$ can be presented by a tangle
(in the standard sence) lying by whole in the region
$$y\le k+x(l-k).$$
Let us add to this tangle the segments $[(0,k+i,0),(1,l+i,0)]$,
$i>0$. The isotopy class of the obtained tangle (in the new sence)
is what we count as the image $F(f)$ of the morphism $f$.

The proof of the following assertion is easy.

\begin{lemma}
Tangles $t_1$ and $t_2$ of type $(k,l)$ are equivalent
if and only if their images $F(t_1)$ and $F(t_2)$ coincide.
An element of the semigroup $T$ has the form $F(t)$, where
$t$ is a link {\/\rm(}i.e.\ a tangle of type $(0,0)${\rm)},
if and only if it is central.
\end{lemma}

It is easy to present the semigroup $T$ by generators and defining
relations.

\begin{lemma}
The semigroup $T$ is generated by the elements $\alpha_i,\beta_i,\sigma_i,
\sigma_i^{-1}$, $i>0$, shown in fig.~\ref{generators}.
\begin{figure}\caption{Generators of the semigroup $T$}\label{generators}
$$\begin{array}{ll}
\alpha_i:&\beta_i:\\
      \begin{picture}(180,200)
            \put(90,20){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(90,180){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(60,40){\circle*2}
            \put(60,80){\circle*2}
            \put(60,120){\circle*2}
            \put(60,160){\circle*2}
            \put(60,40){\line(1,0){60}}
            \put(60,40){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i-1,0)$ }}}
            \put(60,80){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i,0)$ }}}
            \put(60,120){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+1,0)$ }}}
            \put(60,160){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+2,0)$ }}}
            \put(120,40){\circle*2}
            \put(120,80){\circle*2}
            \put(120,120){\circle*2}
            \put(120,160){\circle*2}
            \put(120,40){\raisebox{-0.4ex}{\hbox{ $(1,i-1,0)$}}}
            \put(120,80){\raisebox{-0.4ex}{\hbox{ $(1,i,0)$}}}
            \put(120,120){\raisebox{-0.4ex}{\hbox{ $(1,i+1,0)$}}}
            \put(120,160){\raisebox{-0.4ex}{\hbox{ $(1,i+2,0)$}}}
            \put(120,80){\line(-1,1){20}}
            \put(120,120){\line(-1,-1){20}}
            \put(60,80){\line(3,4){60}}
            \put(60,120){\line(3,4){40}}
            \put(60,160){\line(3,4){10}}
            \end{picture}\hskip1em
&
      \begin{picture}(180,200)
            \put(90,20){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(90,180){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(60,40){\circle*2}
            \put(60,80){\circle*2}
            \put(60,120){\circle*2}
            \put(60,160){\circle*2}
            \put(60,40){\line(1,0){60}}
            \put(60,40){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i-1,0)$ }}}
            \put(60,80){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i,0)$ }}}
            \put(60,120){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+1,0)$ }}}
            \put(60,160){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+2,0)$ }}}
            \put(120,40){\circle*2}
            \put(120,80){\circle*2}
            \put(120,120){\circle*2}
            \put(120,160){\circle*2}
            \put(120,40){\raisebox{-0.4ex}{\hbox{ $(1,i-1,0)$}}}
            \put(120,80){\raisebox{-0.4ex}{\hbox{ $(1,i,0)$}}}
            \put(120,120){\raisebox{-0.4ex}{\hbox{ $(1,i+1,0)$}}}
            \put(120,160){\raisebox{-0.4ex}{\hbox{ $(1,i+2,0)$}}}
            \put(60,80){\line(1,1){20}}
            \put(60,120){\line(1,-1){20}}
            \put(120,80){\line(-3,4){60}}
            \put(120,120){\line(-3,4){40}}
            \put(120,160){\line(-3,4){10}}
      \end{picture}\medskip\\
\sigma_i:&\sigma_i^{-1}:\\
      \begin{picture}(180,200)
            \put(90,20){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(90,170){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(60,40){\circle*2}
            \put(60,80){\circle*2}
            \put(60,120){\circle*2}
            \put(60,160){\circle*2}
            \put(60,40){\line(1,0){60}}
            \put(60,40){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i-1,0)$ }}}
            \put(60,80){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i,0)$ }}}
            \put(60,120){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+1,0)$ }}}
            \put(60,160){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+2,0)$ }}}
            \put(120,40){\circle*2}
            \put(120,80){\circle*2}
            \put(120,120){\circle*2}
            \put(120,160){\circle*2}
            \put(120,40){\raisebox{-0.4ex}{\hbox{ $(1,i-1,0)$}}}
            \put(120,80){\raisebox{-0.4ex}{\hbox{ $(1,i,0)$}}}
            \put(120,120){\raisebox{-0.4ex}{\hbox{ $(1,i+1,0)$}}}
            \put(120,160){\raisebox{-0.4ex}{\hbox{ $(1,i+2,0)$}}}
            \put(60,160){\line(1,0){60}}
            \put(60,80){\line(1,0){10}}
            \put(60,120){\line(1,0){10}}
            \put(120,80){\line(-1,0){10}}
            \put(120,120){\line(-1,0){10}}
            \put(70,120){\line(1,-1){40}}
            \put(70,80){\line(1,1){15}}
            \put(110,120){\line(-1,-1){15}}
      \end{picture}&
      \begin{picture}(180,200)
            \put(90,20){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(90,170){\hbox to 0pt{\hss$\vdots$\hss}}
            \put(60,40){\circle*2}
            \put(60,80){\circle*2}
            \put(60,120){\circle*2}
            \put(60,160){\circle*2}
            \put(60,40){\line(1,0){60}}
            \put(60,40){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i-1,0)$ }}}
            \put(60,80){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i,0)$ }}}
            \put(60,120){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+1,0)$ }}}
            \put(60,160){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,i+2,0)$ }}}
            \put(120,40){\circle*2}
            \put(120,80){\circle*2}
            \put(120,120){\circle*2}
            \put(120,160){\circle*2}
            \put(120,40){\raisebox{-0.4ex}{\hbox{ $(1,i-1,0)$}}}
            \put(120,80){\raisebox{-0.4ex}{\hbox{ $(1,i,0)$}}}
            \put(120,120){\raisebox{-0.4ex}{\hbox{ $(1,i+1,0)$}}}
            \put(120,160){\raisebox{-0.4ex}{\hbox{ $(1,i+2,0)$}}}
            \put(60,160){\line(1,0){60}}
            \put(60,80){\line(1,0){10}}
            \put(60,120){\line(1,0){10}}
            \put(120,80){\line(-1,0){10}}
            \put(120,120){\line(-1,0){10}}
            \put(70,80){\line(1,1){40}}
            \put(70,120){\line(1,-1){15}}
            \put(110,80){\line(-1,1){15}}
      \end{picture}
\end{array}$$
\end{figure}
The following relations holds in $T$:
\begin{align}
\label{iso1}
\alpha_i\beta_{i+1}&=1,\\
\label{iso3}
\alpha_{i+1}\beta_i&=1,\\
\label{iso4}
\alpha_i\alpha_{i+j+1}&=\alpha_{i+j-1}\alpha_i\\
\label{iso5}
\alpha_i\beta_{i+j+1}&=\beta_{i+j-1}\alpha_i,\\
\label{iso6}
\beta_i\beta_{i+j-1}&=\beta_{i+j+1}\beta_i\\
\label{iso7}
\beta_i\alpha_{i+j-1}&=\alpha_{i+j+1}\beta_{i},\\
\label{iso8}
\alpha_i\sigma_{i+j+1}&=\sigma_{i+j-1}\alpha_i,\\
\label{iso9}
\alpha_{i+j+1}\sigma_i&=\sigma_i\alpha_{i+j+1},\\
\label{iso12}
\beta_i\sigma_{i+j-1}&=\sigma_{i+j+1}\beta_i,\\
\label{iso13}
\beta_{i+j+1}\sigma_i&=\sigma_i\beta_{i+j+1},\\
\label{iso10}
\sigma_i\sigma_{i+j+1}&=\sigma_{i+j+1}\sigma_i,\\
\label{iso11}
\alpha_i\sigma_{i+1}^{\pm1}\beta_{i+2}&=\sigma_i^{\mp1},\\
\label{iso2}
\alpha_{i+2}\sigma_{i+1}^{\pm1}\beta_i&=\sigma_i^{\mp1},\\
\label{r1}
\alpha_i\sigma_{i+1}^{\pm1}\beta_i&=1,\\
\label{r2}
\alpha_{i+1}\sigma_i^{\pm1}\beta_{i+1}&=1,\\
\label{r3}
\sigma_{i}\sigma_i^{-1}=\sigma_i^{-1}\sigma_i&=1,\\
\label{r4}
\sigma_i\sigma_{i+1}\sigma_i&=\sigma_{i+1}\sigma_i\sigma_{i+1},
\end{align}
where $i,j>0$.
These relations can be taken as a set of defining
relations of the semigroup $T$.
\end{lemma}

This assertion claims nothing new compared with a similar
statement about defining relations of the tangle
category (see~\cite{tur89}). Relations~(\ref{iso1})--(\ref{iso2})
correspond to isotopies of tangle diagrams, relations~(\ref{r1}),
(\ref{r2}) to the Reidemeister move of type~I, relations~(\ref{r3})
and (\ref{r4}) to the Reidemeister moves of type II and III respectively.

\vskip1em
\noindent{\bf Remark.}
The notion of tangle could be generalized a little differently,
by adding the points $(0,i,0),(1,i,0)$ with $i\le0$ to the boundary.
In this case, the obtained semigroup of tangles is finitely
generated due to the relations
$$\begin{aligned}
\tau^i\alpha_1\tau^{-i}&=\alpha_{i+1},\\
\tau^i\beta_1\tau^{-i}&=\beta_{i+1},\\
\tau^i\sigma_1^{\pm1}\tau^{-i}&=\sigma_{i+1}^{\pm1},
\end{aligned}$$
where $\tau$ is the tangle shown in fig.~\ref{shift}.
\begin{figure}\caption{The tangle $\tau$}\label{shift}
\centerline{\begin{picture}(180,220)
\put(30,20){\circle*2}
\put(30,60){\circle*2}
\put(30,100){\circle*2}
\put(30,140){\circle*2}
\put(30,180){\circle*2}
\put(150,20){\circle*2}
\put(150,60){\circle*2}
\put(150,100){\circle*2}
\put(150,140){\circle*2}
\put(150,180){\circle*2}
\put(30,20){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,-2,0)$ }}}
\put(30,60){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,-1,0)$ }}}
\put(30,100){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,0,0)$ }}}
\put(30,140){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,1,0)$ }}}
\put(30,180){\raisebox{-0.4ex}{\hbox to 0pt{\hss$(0,2,0)$ }}}
\put(150,20){\raisebox{-0.4ex}{\hbox to 0pt{ $(1,-2,0)$\hss}}}
\put(150,60){\raisebox{-0.4ex}{\hbox to 0pt{ $(1,-1,0)$\hss}}}
\put(150,100){\raisebox{-0.4ex}{\hbox to 0pt{ $(1,0,0)$\hss}}}
\put(150,140){\raisebox{-0.4ex}{\hbox to 0pt{ $(1,1,0)$\hss}}}
\put(150,180){\raisebox{-0.4ex}{\hbox to 0pt{ $(1,2,0)$\hss}}}
\put(30,20){\line(3,-1){70}}
\put(30,60){\line(3,-1){120}}
\put(30,100){\line(3,-1){120}}
\put(30,140){\line(3,-1){120}}
\put(30,180){\line(3,-1){120}}
\put(150,180){\line(-3,1){70}}
\end{picture}}
\end{figure}
But there does not exist a finite presentation of this semigroup.

\section{Tangles embedded in book}\label{yinfty}
Here we shall define a semigroup $Y_\infty$, into which the
semigroup $T$ can be embedded. The semigroup $Y_\infty$ itself
is not finitely presented, but we will need it for the further
construction of finitely presented semigroups which
are subsemigroups of $Y_\infty$.

Denote by $P_k$ ($k\in\integer$) the half-plane
\begin{equation}
\{(x,y,z)\in\real^3|\;y\ge0,z=ky\}
\end{equation}
and consider the part of the union of these half-planes
that lies in the region~(\ref{slice}):
\begin{equation}
\book=\slice\cap\Bigl(\bigcup\limits_{k\in\integer}P_k\Bigr).
\end{equation}
It is natural to call the set $\book$ a book and the line $y=z=0$
the binding line. Let us denote by $E^0_{k,l}$ and $E^1_{k,l}$, where
$k,l\in\integer$, $l>0$, the following points lying
at the boundary of the book $\book$:
\begin{equation}
E^0_{k,l}=(0,l,kl),\quad E^1_{k,l}=(1,l,kl).
\end{equation}

By a tangle in the book $\book$ we shall call any PL-submanifold $t$ in
$\slice$, verifying the following conditions:
\begin{enumerate}
\item
$\partial t=t\cap\partial\slice=
\{E^0_{k,l},E^1_{k,l}\}_{k,l\in\integer,\,l>0}$;
\item\label{cond1}
$t$ is contained entirely in $\book$;
\item
$t$ is transversal to the binding line, i.e.\ it has
only a finite number of intersecion points with $l$ which
do not disappear under small variation of the
tangle $t$ without violating condition~\ref{cond1};
\item
the restriction of the $x$-coordinate to any connected component
of the intersection $t\cap P_i$ is a monotonic function for
any $i\in\integer$.
\end{enumerate}
We shall call two tangles of this kind equivalent if they
are ambiently isotopic in $\slice$ (but not necessarily in $\book$\,!).

The composition of tangles in the book $\book$ is defined in the same
manner as in the case of ordinary tangles. The set of
isotopy classes of tangles in $\book$ endowed with
composition is a semigroup with the unit
$$I=\bigcup\limits_{k,l\in\integer,l>0}[E^0_{k,l},E^1_{k,l}].$$
We shall denote this semigroup by $Y_\infty$. Tangles
in the book $\book$ will be called simply tangles in sequel.

For any $n\ge3$ we denote by $Y_n$ the subsemigroup in $Y_\infty$
consisting of all isotopy classes of tangles $t$ such that
$t\cap\intP_k=I\cap\intP_k$ for $k<0$ 
% V russkom variante zamenit' "ili" na "i pri"
and for 
%
$k>n-1$. Here $\intP_k$
denotes the half-plane $P_k$ with its boundary removed:
$\intP_k=\{(x,y,z)\in\real^3|\;y>0,z=ky\}$.

Notice that any tangle $t$ is isotopic to a tangle $t'$ for which
there exist $N_1,N_2\in\integer$ such that $t'\cap\intP_k=I\cap\intP_k$
for $k<N_1$ and for $k>N_2$. In addition to this, the tangle
$t'$ can be chosen so that for sufficiently large
$N_3$ and some integers $n_k$, $k\in\integer$, the segments
$[E^0_{k,n},E^1_{k,n+n_k}]$ are contained by whole in $t'$ for
$n>N_3$
% Vstavit v russkom texte: "i liubogo $k\in\integer$"
and any $k\in\integer$. 
%
It is readily seen that the map
$\varphi:Y_\infty\rightarrow\integer^\integer$
\begin{equation}
\varphi(t)=(\dots,n_{-2},n_{-1},n_0,n_1,n_2,\dots)
\end{equation}
defined in this way is a group homomorphism.

It is natural to define the complexity $m(t)$ of a tangle $t$ in the book
$\book$ as the number of intersection points of $t$ with the binding
line. Let us consider the simplest nontrivial tangles, i.e.~ones having
complexity $1$. If $m(t)=1$, then it is easy to see that for
exactly two values $i\in\integer$ the intersection $t\cap\intP_i$
is not isotopic to the intersection $I\cap\intP_i$. For those two
exeptional values of $i$ we have $\varphi(t)_i=\pm1$. Moreover, if
$\varphi(t)_i=1$, then in the half-plane $P_i$ the tangle
$t$ has the following connected componenets (up to isotopy):
$$t\cap P_i=\Bigl(\bigcup\limits_{l>0}[E^0_{i,l},E^1_{i,l+1}]\Bigr)
\cup[A,E^1_{i,1}],$$
where $A$ is the only intersection point of the tangle $t$ with the
binding line. In the case $\varphi(t)_i=-1$, $m(t)=1$,
we have (up to isotopy in $P_i$):
$$t\cap P_i=\Bigl(\bigcup\limits_{l>0}[E^0_{i,l+1},E^1_{i,l}]\Bigr)
\cup[E^0_{i,1},A].$$

Denote by $a_i,b_i,c_i,d_i$ the elements of $Y_\infty$ that are presented
by tangles of complexity $1$ such that the following holds:
\begin{equation}
\begin{aligned}
\varphi(a_i)_j&=\phantom-\delta_{i-1,j}+\delta_{i,j},\\
\varphi(b_i)_j&=\phantom-\delta_{i-1,j}-\delta_{i,j},\\
\varphi(c_i)_j&=-\delta_{i-1,j}-\delta_{i,j},\\
\varphi(d_i)_j&=-\delta_{i-1,j}+\delta_{i,j}.
\end{aligned}
\end{equation}
Here $\delta_{i,j}$ is the Kroneker delta.

Having drawn corresponding pictures, it is easy to check that
the following relations take place int the semigroup $Y_\infty$:
\begin{align}
\label{rel1}
a_id_{i+1}&=a_{i+1}b_i,\\
\label{rel2}
d_ic_{i+1}&=b_{i+1}c_i,\\
\label{rel4}
a_id_i&=a_{i+1}b_{i+1},\\
\label{rel5}
b_ic_i&=d_{i+1}c_{i+1},\\
\label{rel8}
a_ib_ic_id_i&=1,\\
\label{rel3}
b_id_i=d_ib_i&=1,\\
\label{rel6}
x_iy_j&=y_jx_i,\\
\label{rel7}
x_id_{i-1}d_id_{i+1}&=d_{i-1}d_id_{i+1}x_i
\end{align}
for all $i,j\in\integer$ such that $|i-j|>1$ and $x_i\in\{a_i,b_i,c_i,d_i\}$,
$y_j\in\{a_j,b_j,c_j,d_j\}$.

\begin{theorem}\label{y.infty}
The elements $\{a_i,b_i,c_i,d_i\}_{i\in\integer}$ generate
the semigroup $Y_\infty$. Relations {\rm(\ref{rel1})--(\ref{rel7})\/}
can be taken as a set of defining relations of the
semigroup $Y_\infty$.
\end{theorem}

\proof
To begin with, we prove the simpler part of the theorem,
namely, the assertion that the semigroup $Y_\infty$ is generated by the
mentioned elements. Indeed, any tangle can be cut into tangles
of complexity $1$ by planes of the form $x={\rm const}$. Therefore,
it is enough to show that any tangle of complexity $1$
can be presented up to isotopy by a composition
of our generators. Assume that for the tangle $t$ of complexity $m(t)=1$
we have
$$\varphi(t)_i=\varphi(t)_j=1,$$
where $j>i$. Then it is isotopic to
$$a_{i+1}d_{i+2}d_{i+3}\cdots d_j.$$
If $m(t)=1$ and
$$\varphi(t)_i=-\varphi(t)_j=1,$$
$j>i$, then $t$ is equivalent to the tangle
$$b_jb_{j-1}\cdots b_{i+1}.$$
The other two cases are similar.

The rest of this section is devoted to the proof of
sufficientness of relations~(\ref{rel1})--(\ref{rel7}) for realizing
any isotopy between two tangles in the book $\book$. A sketch of
the proof look like this. According to relation~(\ref{rel3}) the
elements $b_i$ and $d_i$ are each other inverse. Therefore,
equivalence of two words $w_1$ and $w_2$ in the alphabet
$A=\{a_i,b_i,c_i,d_i\}_{i\in\integer}$ modulo
relations~(\ref{rel1})--(\ref{rel7}) takes place whenever so does
equivalence of words $b_iw_1$ and $b_iw_2$ as well as $d_iw_1$ and $d_iw_2$,
$w_1b_i$ and $w_2b_i$, $w_1d_i$ and $w_2d_i$. We shall show that
for any word $w$ one can find such words $u$ and $v$ containing
only letters $b_i$ and $d_i$ with $i\in\integer$
that the word $uwv$ is equivalent modulo relations~(\ref{rel1})--(\ref{rel7})
to a word presenting an element of a special subsemigroup
in $Y_\infty$ which is isomorphic to $T$. We shall prove that
defining relations~(\ref{iso1})--(\ref{r4}) of the semigroup
$T$ can be derived from~(\ref{rel1})--(\ref{rel7}).

For $i\in\integer$ we call a word $w=x_1x_2\dots x_n$, where
$x_1,\dots,x_n\in A$, $i$-{\it balanced} if we have
$$\varphi(w)_i=0$$
and for all $k=1,\dots,n$ the following enequality holds:
$$\varphi(x_1\dots x_k)_i\ge0.$$

In the geometrical language, $i$-balanceness of the word $w$
corresponding to a tangle $t\subset\book$ means that
any connected component of the intersection $t\cap\intP_i$
connects either points of the form $E_{i,m}^0$ and $E_{i,m}^1$ or
two vertices of the tangle $t$ that lie at the binding line.
For any $\varepsilon>0$ each $i$-balanced tangle is ambiently
isotopic in $\book$ to a tangle $t'$ such that $t'$ does not differ
from the identity tangle in the half-plane $P_i\cap\{y\ge\varepsilon\}$.

The following interpretation of of the notion of $i$-balanced word
is also convinient. For a given word $w$ in the alphabet $A$ and
an integer $i$ let us replace each letter $x$ with a left parenthesis
if $\varphi(x)_i=1$, with a right parenthesis if $\varphi(x)_i=-1$,
and with a space if $\varphi(x)_i=0$. The word $w$ is $i$-balanced
if and only if the obtained parenthesis string is balanced.

Let $\widetilde T$ be the subsemigroup of $Y_\infty$ consisting
of all the isotopy classes of tangles that are $i$-balanced for
all $i\ne0$.

\begin{lemma}
The semigroup $\widetilde T$ is isomorphic to the semigroup $T$ that
was introduced in Section~{\rm\ref{standard_tangles}}.
\end{lemma}
\proof
It is easy to see that if a tangle $t$ is $i$-balanced for all
$i\ne0$, then it can be transformed by an ambient isotopy in $\book$
to the form such that
$$t\cap\{|z|\ge1/2\}=I\cap\{|z|\ge1/2\}.$$
Forgetting the ``trivial'' part of the tangle $t$, which lies in
the region $|z|\ge1/2$, we obtain a tangle with boundary
$\{E^0_{0,k},E^1_{0,k}\}_{k>0}$, which defines an element from $T$.
So, we get a map $\xi:\widetilde T\rightarrow T$. Clearly,
this map is injective. We shall construct the inverse map.
To this end we provide the images of the elements
$\alpha_i,\beta_i,\sigma_i^{\pm1}$, $i>0$,
generating the semigroup $T$, under the map $\xi^{-1}$:
\begin{align}
\label{image1}
\alpha_i&\mapsto d_1^{i-1}a_1b_1^i\\
\label{image2}
\beta_i&\mapsto d_1^ic_1b_1^{i-1}\\
\label{image3}
\sigma_i&\mapsto d_1^{i-1}b_0d_1d_0b_1^i\\
\label{image4}
\sigma_i^{-1}&\mapsto d_1^ib_0b_1d_0b_1^{i-1}.
\end{align}
The fact that the tangles in the right hand side of
these formulas are send under the map
$\xi$ into the generators $\alpha_i,\beta_i,\sigma_i^{\pm1}$
shown in fig.~\ref{generators} can be checked straightforwardly.
In what follows we identify the semigroups $\widetilde T$ and $T$
,
considering $T$ as a subsemigroup in $Y_\infty$.

\begin{lemma}\label{razlozh}
If a word $w$ in the alphabet $A$ is $i$-balanced for
all $i\ne0$, then it is equivalent up to
relations~{\rm(\ref{rel1})--(\ref{rel7})\/} to a word of the form
\begin{equation}\label{product}
x_1\dots x_m,
\end{equation}
where $x_1,\dots,x_m\in\{d_1^{i-1}a_1b_1^i,d_1^ic_1b_1^{i-1},
d_1^{i-1}b_0d_1d_0b_1^i,d_1^ib_0b_1d_0b_1^{i-1}\}_{i>0}$.
\end{lemma}

\proof
To begin with, we check that, applying relations~(\ref{rel1})--(\ref{rel7})
to the word $w$, we can get rid of all entries of letters
$a_i,b_i,c_i,d_i$ with $i\ne 0,1$ in $w$. Denote by
$A_i$ the set $\{a_i,b_i,c_i,d_i\}$. Let $N$
be the maximal $i$ such that the word $w$ contains letters from $A_i$,
let $N>1$. Relations~(\ref{rel4}), (\ref{rel5}) and (\ref{rel3}) implies:
$$\begin{aligned}
a_N&=a_Nb_Nd_N=a_{N-1}d_{N-1}d_N\\
c_N&=b_Nd_Nc_N=b_Nb_{N-1}c_{N-1}.
\end{aligned}$$
So, by applying those relations one can get rid of letters $a_N,c_N$
in the word $w$. In the process, the word $w$ will remain
$i$-blanced for $i\ne0$. Indeed, on the ``bracket'' language
the mentioned replacements look like follows:
$$\begin{aligned}
i&=N-2:\qquad&&{\lefteqn\varnothing\phantom(}\quad\rightarrow\quad
{\lefteqn\varnothing
\phantom(\ \phantom(\ \phantom(}\quad\rightarrow\quad
(\ )\\
i&=N-1:\qquad&&{(}\quad\rightarrow\quad
{(\ (\ )}\quad\rightarrow\quad(\ (\ )\quad,\\
i&=N:\qquad&&
{(}\quad\rightarrow\quad{(\ )\ (}\quad\rightarrow\quad
\phantom(\ \phantom(\ (
\end{aligned}$$
$$\begin{aligned}
i&=N-2:\qquad&&{\lefteqn\varnothing\phantom(}\quad\rightarrow\quad
{\lefteqn\varnothing
\phantom(\ \phantom(\ \phantom(}\quad\rightarrow\quad
\phantom(\ (\ )\\
i&=N-1:\qquad&&{)}\quad\rightarrow\quad
{(\ )\ )}\quad\rightarrow\quad(\ )\ )\quad.\\
i&=N:\qquad&&
{)}\quad\rightarrow\quad{)\ (\ )}\quad\rightarrow\quad{)}
\end{aligned}$$
Now we shall get rid of entries of letters $b_N,d_N$. Due to
$N$-balanceness of the word $w$, its last letter cannot be $d_N$
,
and the number of $b_N$ and $d_N$ entries coincide. Let us find
the last letter $d_N$ in $w$. Let $x$ be the letter following after $d_N$.
There are the following three possibilities.

1) $x=b_N$. In this case we remove the subword $d_Nb_N$.

2) $x\in A_{N-1}$. Make the following replacements:
$$d_Nx\mapsto b_{N-1}b_{N-2}d_{N-2}d_{N-1}d_Nx\mapsto
b_{N-1}b_{N-2}xd_{N-2}d_{N-1}d_N.$$

3) $x\in A_i$, where $i\le N-2$. Interchange $d_N$ and $x$.
\\
In all the three cases $i$-balanceness of the word $w$ for $i\ne0$ is
preserved. In case~1) the number of $d_N$ (as well as $b_N$)
entries in $w$ is decreased by~$1$. In cases~2) and 3) the number
of $d_N$ and $w$ entries is preserved but the number of
letters following after the last $d_N$ is decreased. Hence
after a finite number of such replacements we get
a word not envolving letters $b_N,d_N$.

So, applying the described procedure we can change the
word $w$ so that it contains only letters from $A_i$, where $i\le1$.
Similarly, using transforms
$$\begin{aligned}
a_N&\mapsto a_Nd_Nb_N\mapsto a_{N+1}b_{N+1}b_N,\\
c_N&\mapsto c_Nb_Nd_N\mapsto c_{N+1}d_{N+1}d_N,\\
b_Nd_N&\mapsto\varnothing,\\
xd_N&\mapsto xd_Nd_{N+1}d_{N+2}b_{N+2}b_{N+1}\mapsto
d_Nd_{N+1}d_{N+2}xb_{N+2}b_{N+1},\quad x\in A_{N+1},\\
xd_N&\mapsto d_Nx,\quad x\in A_i,\quad i\ge N+2,
\end{aligned}$$
where $N<0$, one can make the word $w$ does not contain letters
from $A_i$ with $i<0$. During this procedure we insert only
only letters $b_0,d_0,b_1,d_1$ in the word $w$.
By means of replacements
$$\begin{aligned}
a_0&\mapsto a_0d_0b_0\mapsto a_1b_1b_0,\\
c_0&\mapsto c_0b_0d_0\mapsto c_1d_1d_0
\end{aligned}$$
we obtain a word envolving only letters $b_0,d_0,a_1,b_1,c_1,d_1$
and still being $i$-balanced for all $i\ne0$.

Thus, it suffices to prove the lemma in the case when
the word $w$ contains only letters $b_0,d_0,a_1,b_1,c_1,d_1$
and is $(-1)$-balanced and $1$-balanced. By definition,
$(-1)$-balanceness means that one obtain a well balanced
parenthesis string by replacing all $b_0$'s
with a left bracket and all $d_0$'s with a right bracket in
the word $w$. We shall call the {\it depth\/} of the word $w$
the maximal value of $\varphi(u)_{-1}$ taken over all possible
decompositions of the word $w$ into the product $w=uv$ of two subwords.
In other words, the depth of $w$ is the depth of the bracket hierarchy
obtained by replacing each $b_0$ with a left bracket and each $d_0$ with
a right one.

Our next goal is to make our word has depth $1$ or $0$
by applying relations (\ref{rel1})--(\ref{rel7}).
Suppose that the depth of the word $w$ is greater than $1$.
Then for all $x,y\in A_1$ we shall implement the following
replacements in $w$ as long as possible:
$$\begin{aligned}
b_0xy&\mapsto b_0xd_0b_0y,\\
b_0^2xd_0b_0&\mapsto b_0^2xd_0^2b_0^2,\\
b_0^2xd_0y&\mapsto b_0^2xd_0^2b_0y.
\end{aligned}$$
Under these replacements, both $(-1)$-balanceness and the depth
of $w$ are preserved. It easy to see that since the depth of $w$
is greater than $1$ by assumption, after a finite
number of such replacements, subwords of the form
\begin{equation}\label{b2xd2}
b_0^2xd_0^2,\quad\mbox{where }x\in A_1
,
\end{equation}
will necessarily appear. With each of those subwords we perform the
following transforms, which are obtained by applying one ore more
relations (\ref{rel1})--(\ref{rel7}):
$$\begin{aligned}
b_0^2a_1d_0^2&\mapsto b_0\bigl(b_0a_1(d_0d_1d_2)b_2b_1\bigr)d_0\\
&\mapsto b_0\bigl(b_0(d_0d_1d_2)a_1b_2b_1)d_0\\
&\mapsto b_0\bigl(d_1d_2a_1b_2b_1\bigr)d_0\\
&\mapsto b_0b_{-1}d_{-1}\bigl(d_1d_2a_1b_2b_1\bigr)d_0\\
&\mapsto b_0b_{-1}\bigl(d_1d_2a_1b_2b_1\bigr)d_{-1}d_0\\
&\mapsto b_0b_{-1}\bigl(b_0a_1d_0\bigr)d_{-1}d_0\\
&\mapsto b_0b_{-1}b_{-2}d_{-2}\bigl(b_0a_1d_0\bigr)d_{-1}d_0\\
&\mapsto b_0b_{-1}b_{-2}b_0a_1d_0d_{-2}d_{-1}d_0\\
&\mapsto b_0b_{-1}b_{-2}b_0a_0d_0d_1d_0d_{-2}d_{-1}d_0\\
&\mapsto b_0b_{-1}b_{-2}b_0a_0b_{-1}(d_{-1}d_0d_1)d_0d_{-2}d_{-1}d_0\\
&\mapsto b_0b_{-1}b_{-2}b_0a_0b_{-1}d_0(d_{-1}d_0d_1)(d_{-2}d_{-1})d_0\\
&\mapsto b_0b_{-1}b_{-2}b_0a_0b_{-1}d_0d_{-1}d_0(d_{-2}d_{-1})d_1d_0\\
&\mapsto b_0b_{-1}b_{-2}b_0a_0b_{-1}d_0d_{-1}d_0(d_{-2}d_{-1}d_0d_1)b_1b_0d_1d_0\\
&\mapsto b_0b_{-1}b_{-2}(d_{-2}d_{-1}d_0d_1)b_0a_0b_{-1}d_0d_{-1}d_0b_1b_0d_1d_0\\
&\mapsto d_1b_0a_0b_{-1}d_0d_{-1}d_0b_1b_0d_1d_0\\
&\mapsto d_1b_0a_0b_{-1}d_0(d_{-1}d_0d_1)b_1b_1b_0d_1d_0\\
&\mapsto d_1b_0a_0b_{-1}(d_{-1}d_0d_1)d_0b_1b_1b_0d_1d_0\\
&\mapsto d_1b_0(a_0d_0d_1)d_0b_1b_1b_0d_1d_0\\
&\mapsto d_1b_0a_1d_0b_1^2b_0d_1d_0;
\end{aligned}$$
by replacing in these formulas $a_1$ with $d_1$ and $a_0$ with $b_0$,
we obtain a chain of tranforms that yields
$$b_0^2d_1d_0^2\mapsto d_1b_0d_1d_0b_1^2b_0d_1d_0;$$
by reversing the letter order in all the words and replacing
$a_i,b_i,c_i,d_i$ with $c_i,d_i,a_i,b_i$ respectively, we get tranforms
$$\begin{aligned}
b_0^2c_1d_0^2&\mapsto
b_0b_1d_0d_1^2b_0c_1d_0b_1,\\
b_0^2b_1d_0^2&\mapsto
b_0b_1d_0d_1^2b_0b_1d_0b_1,
\end{aligned}$$
which also are obtained by applying relations (\ref{rel1})--(\ref{rel7}).
All the mentioned tranforms preserve $i$-balanceness for $i\ne0$.
Having applied those transforms to all the subwords of the
form~(\ref{b2xd2}), we get a word of a smaller depth. Repeating
the procedure sufficiently many times we get a word of depth $1$,
i.e.\ a word decomposable into the product of subwords from
the following lists:
$$a_1,\quad b_1,\quad c_1,\quad d_1,\quad
b_0a_1d_0,\quad b_0b_1d_0,\quad b_0c_1d_0,\quad b_0d_1d_0.$$
By using tranforms
$$\begin{aligned}
b_0a_1d_0&\mapsto b_0(a_1d_2)b_2d_0\\
&\mapsto b_0(a_2b_1)b_2d_0\\
&\mapsto a_2b_0b_1b_2d_0\\
&\mapsto (a_2b_1)d_1b_0b_1b_2d_0\\
&\mapsto (a_1d_2)d_1b_0b_1b_2d_0\\
&\mapsto (a_1b_1)b_0(d_0d_1d_2)d_1b_0b_1b_2d_0\\
&\mapsto (a_0d_0)b_0d_1(d_0d_1d_2)b_0b_1b_2d_0\\
&\mapsto (a_0d_1)d_0d_1b_0d_2b_1b_2d_0\\
&\mapsto (a_1b_0)d_0d_1b_0b_1b_0(d_0d_1d_2)b_1b_2d_0\\
&\mapsto a_1d_1b_0b_1b_0b_1(d_0d_1d_2)b_2d_0\\
&\mapsto a_1d_1b_0b_1b_0b_1d_0d_1d_0\\
&\mapsto a_1d_1b_0b_1d_0(b_0^2b_1d_0^2)b_0d_1d_0\\
&\mapsto a_1d_1b_0b_1d_0(b_0b_1d_0d_1^2b_0b_1d_0b_1)b_0d_1d_0\\
&=a_1d_1(b_0b_1d_0)(b_0b_1d_0)d_1^2(b_0b_1d_0)b_1(b_0d_1d_0)
\end{aligned}$$
and a similar chain resulting in the transform
$$b_0^2c_1d_0^2\mapsto
(b_0b_1d_0)d_1(b_0d_1b_0)b_1^2(b_0d_1d_0)(b_0d_1d_0)b_1c_1,$$
applied to all subwords of the form $b_0a_1d_0$ and $b_0c_1d_0$,
we get a word which decomposes into the product of subwords from
a smaler list:
$$a_1,\quad b_1,\quad c_1,\quad d_1,\quad
b_0b_1d_0,\quad b_0d_1d_0.$$
We keep notation $w$ for the obtained word. Let
$$w=y_1\dots y_m,$$
where $y_1,\dots,y_m\in\{a_1,b_1,c_1,d_1,b_0b_1d_0,b_0d_1d_0\}$,
be the corresponding decomposition.

Under all transforms discussed above the word $w$ was remaining
$1$-balanced. This means that by replacing all the subwords of
the form $a_1,d_1,b_0d_1d_0$ with a left bracket and all the subwords
of the form $b_1,c_1,b_0c_1d_0$ with a right bracket, one gets
a well balanced bracket string. This implies that there exist such
$i_1,\dots,i_{m-1}\ge0$ that all the words
$$\begin{aligned}
x_1&=y_1b_1^{i_1},\\
x_2&=d_1^{i_1}y_2b_1^{i_2},\\
&\;\;\vdots\\
x_{m-1}&=d_1^{i_{m-2}}y_{m-1}b_1^{i_{m-1}},\\
x_m&=d_1^{i_{m-1}}y_m
\end{aligned}$$
are $1$-balanced. Additionally, the word $x_1\dots x_m$
is equivalent to the word $w$ modulo relations (\ref{rel3}).
Having removed all subwords of the form $d_1^kb_1^k$ in each $x_j$,
we get words from the required set
$$\{d_1^{i-1}a_1b_1^i,d_1^ic_1b_1^{i-1},
d_1^{i-1}b_0d_1d_0b_1^i,d_1^ib_0b_1d_0b_1^{i-1}\}_{i>0}$$
and some number of words $\varnothing$.
This completes the proof of Lemma~\ref{razlozh}.\proved

Now we return to the proof of Theorem~\ref{y.infty}. Let two words
$w_1,w_2$ in the alphabet $A$ correspond to isotopic tangles.
We wish to show that they can be obtained from each other by applying
relations~(\ref{rel1})--(\ref{rel7}). Clearly we have
$\varphi(w_1)=\varphi(w_2)$. Also clear that one can
find a word $u$, consisting only of the letters $\{b_i,d_i\}$,
such that $\varphi(u)_i=-\varphi(w_1)_i$ for all $i\ne0$.
We have:
$$\varphi(w_1u)_i=\varphi(w_2u)
% vstavit' v russkom variante:
_i
%
=0,\quad\mbox{for all }i\ne0.$$
It easily follows from this that for a sufficiently large $N$ the words
$$w_j'=
d_1^{N^2}\dots d_{N-1}^{2N}d_N^Nb_0^{N^2}\dots b_{2-N}^{2N}b_{1-N}^N
w_jud_{1-N}^Nd_{2-N}^{2N}\dots d_0^{N^2}b_N^Nb_{N-1}^{2N}\dots b_1^{N^2},$$
where $j=1,2$, are $i$-balanced for all $i\ne0$.

Due to relations~(\ref{rel3}), the words $w_1$ and $w_2$ are
equivalent modulo relations~(\ref{rel1})--(\ref{rel7}) if and only if
the same is true about $w_1'$ and $w_2'$.

Thus it is enough to proof the assertion in the case when the words
$w_1$ and $w_2$ are $i$-balanced for all $i\ne0$. In this case, according
to Lemma~\ref{razlozh}, by using relations~(\ref{rel1})--(\ref{rel7})
both words can be decomposed into the product of the
generators~(\ref{image1})--(\ref{image4}) of the semigroup $T\in Y_\infty$.
Therefore, it suffices to show that relations~(\ref{iso1})--(\ref{r4}),
which defines the semigroup $T$, follows from
relations~(\ref{rel1})--(\ref{rel7}) under
indentifications~(\ref{image1})--(\ref{image4}). We devote the next
section to checking this.

\section{Checking relations}\label{checking}
So, here we shall check that relations~(\ref{iso1})--(\ref{r4})
for the elements~(\ref{image1})--(\ref{image4}) are satisfied by virtue
of~(\ref{rel1})--(\ref{rel7}).

Relation~(\ref{iso1}):
$$\begin{aligned}
\alpha_i\beta_{i+1}&=d_1^{i-1}a_1b_1^id_1^{i+1}c_1b_1^i\\
&=d_1^{i-1}(a_1d_1c_1b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1(b_0c_0)b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1d_2b_2(b_0c_0)b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1d_2(b_0c_0)b_2b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1d_2(d_1c_1)b_2b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1b_1b_0(d_0d_1d_2)d_1c_1b_2b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1b_1b_0d_1c_1(d_0d_1d_2)b_2b_1)b_1^{i-1}\\
&=d_1^{i-1}((a_1b_1)b_0(d_1c_1)d_0)b_1^{i-1}\\
&=d_1^{i-1}((a_0d_0)b_0(b_0c_0)d_0)b_1^{i-1}\\
&=d_1^{i-1}(a_0b_0c_0d_0)b_1^{i-1}\\
&=d_1^{i-1}b_1^{i-1}\\
&=1.
\end{aligned}$$
Note that simaltaniously we have proven the relation
$$b_ia_id_ic_i=1,$$
which is symmetric to~(\ref{rel8}).

Relation~(\ref{iso3}):
$$\begin{aligned}
\alpha_{i+1}\beta_i&=d_1^ia_1b_1^{i+1}d_1^ic_1b_1^{i-1}\\
&=d_1^ia_1b_1c_1b_1^{i-1}\\
&=d_1^i(a_1b_1c_1d_1)b_1^i\\
&=d_1^ib_1^i\\
&=1.
\end{aligned}$$
Let us prove the following auxiliary relations:
\begin{align}
\label{aux1}
d_1^{i-1}a_1b_1^i&=b_0^{i-1}a_0d_0^i,\\
\label{aux2}
d_1^ic_1b_1^{i-1}&=b_0^ic_0d_0^{i-1},
\end{align}
$i>0$. To this end, we apply induction. For $i=1$ these relations
coincide with~(\ref{rel4}), (\ref{rel5}). For $i>1$ we have:
$$\begin{aligned}
d_1^{i-1}a_1b_1^i&=d_1(d_1^{i-2}a_1b_1^{i-1})b_1\\
&=d_1(d_1^{i-2}a_1b_1^{i-1})b_1\\
&=d_1(b_0^{i-2}a_0d_0^{i-1})b_1\\
&=b_0b_{-1}(d_{-1}d_0d_1)(b_0^{i-2}a_0d_0^{i-1})b_1\\
&=b_0b_{-1}(b_0^{i-2}a_0d_0^{i-1})(d_{-1}d_0d_1)b_1\\
&=b_0b_{-1}(b_0^{i-2}a_0d_0^{i-1})d_{-1}d_0\\
&=b_0b_{-1}(d_1^{i-2}a_1b_1^{i-1})d_{-1}d_0\\
&=b_0(d_1^{i-2}a_1b_1^{i-1})b_{-1}d_{-1}d_0\\
&=b_0(b_0^{i-2}a_0d_0^{i-1})d_0\\
&=b_0^{i-1}a_0d_0^i.
\end{aligned}$$
Similarly, one proves~(\ref{aux2}).

Let us prove~(\ref{iso4}).
$$\begin{aligned}
\alpha_i\alpha_{i+j+1}&=d_1^{i-1}a_1b_1^id_1^{i+j}a_1b_1^{i+j+1}\\
&=d_1^{i-1}(a_1d_1)(d_1^{j-1}a_1b_1^j)b_1^{i+1}\\
&=d_1^{i-1}(a_2b_2)(b_0^{j-1}a_0d_0^j)b_1^{i+1}\\
&=d_1^{i-1}(b_0^{j-1}a_0d_0^j)(a_2b_2)b_1^{i+1}\\
&=d_1^{i-1}(d_1^{j-1}a_1b_1^j)(a_1d_1)b_1^{i+1}\\
&=d_1^{i+j-2}a_1b_1^ja_1b_1^i\\
&=(d_1^{i+j-2}a_1b_1^{i+j-1})(d_1^{i-1}a_1b_1^i)\\
&=\alpha_{i+j-1}\alpha_i.
\end{aligned}$$
Relation~(\ref{iso5}) is proven similarly, this time by using~(\ref{aux2}).
As for~(\ref{iso6}), we have:
$$\begin{aligned}
\beta_i\beta_{i+j-1}&=d_1^ic_1b_1^{i-1}d_1^{i+j-1}c_1b_1^{i+j-2}\\
&=d_1^{i+1}(b_1c_1)(d_1^jc_1b_1^{j-1})b_1^{i-1}\\
&=d_1^{i+1}(d_2c_2)(b_0^jc_0d_0^{j-1})b_1^{i-1}\\
&=d_1^{i+1}(b_0^jc_0d_0^{j-1})(d_2c_2)b_1^{i-1}\\
&=d_1^{i+1}(d_1^jc_1b_1^{j-1})(b_1c_1)b_1^{i-1}\\
&=(d_1^{i+j+1}c_1b_{i+j})(d_1^ic_1b_1^{i-1})\\
&=\beta_{i+j+1}\beta_i.
\end{aligned}$$
One proves~(\ref{iso7}) in a similar way.

In order to prove~(\ref{iso8})--(\ref{iso10}) we shall make use of
the following relations (for $i>0$):
\begin{align}
\label{sigma1}
b_0d_1d_0b_1=b_0^2b_{-1}d_0d_{-1}d_0=d_1d_2d_1b_2b_1^2,\\
\label{sigmai}
d_1^{i-1}(b_0d_1d_0b_1)b_1^{i-1}=b_0^{i-1}(b_0d_1d_0b_1)d_0^{i-1}.
\end{align}
Let us prove~(\ref{sigma1}):
$$\begin{aligned}
b_0^2b_{-1}d_0d_{-1}d_0&=b_0^2b_{-1}d_0(d_{-1}d_0d_1)b_1\\
&=b_0^2b_{-1}(d_{-1}d_0d_1)d_0b_1\\
&=b_0d_1d_0b_1\\
&=b_0d_1(d_0d_1d_2)b_2b_1^2\\
&=b_0(d_0d_1d_2)d_1b_2b_1^2\\
&=d_1d_2d_1b_2b_1^2.
\end{aligned}$$
We shall prove relation~(\ref{sigmai}) by induction. For $i=1$ it is
trivial. Next, we have:
$$\begin{aligned}
d_1^{i-1}(b_0d_1d_0b_1)b_1^{i-1}&=d_1(d_1^{i-2}(b_0d_1d_0b_1)b_1^{i-2})b_1\\
&=d_1(b_0^{i-2}(b_0d_1d_0b_1)d_0^{i-2})b_1\\
&=d_1(b_0^{i-2}(b_0^2b_{-1}d_0d_{-1}d_0)d_0^{i-2})b_1\\
&=b_0b_{-1}b_{-2}(d_{-2}d_{-1}d_0d_1)
(b_0^{i-2}(b_0^2b_{-1}d_0d_{-1}d_0)d_0^{i-2})b_1\\
&=b_0b_{-1}b_{-2}(b_0^{i-2}(b_0^2b_{-1}d_0d_{-1}d_0)d_0^{i-2})
(d_{-2}d_{-1}d_0d_1)b_1\\
&=b_0b_{-1}b_{-2}(b_0^{i-2}(b_0^2b_{-1}d_0d_{-1}d_0)d_0^{i-2})
d_{-2}d_{-1}d_0\\
&=b_0b_{-1}b_{-2}(d_1^{i-2}(b_0d_1d_0b_1)b_1^{i-2})d_{-2}d_{-1}d_0\\
&=b_0b_{-1}(d_1^{i-2}(b_0d_1d_0b_1)b_1^{i-2})d_{-1}d_0\\
&=b_0(d_1^{i-2}b_{-1}(b_0d_1d_0b_1)d_{-1}b_1^{i-2})d_0\\
&=b_0(d_1^{i-2}b_{-1}(d_1d_2d_1b_2b_1^2)d_{-1}b_1^{i-2})d_0\\
&=b_0(d_1^{i-2}(d_1d_2d_1b_2b_1^2)b_1^{i-2})d_0\\
&=b_0(d_1^{i-2}(b_0d_1d_0b_1)b_1^{i-2})d_0\\
&=b_0(b_0^{i-2}(b_0d_1d_0b_1)d_0^{i-2})d_0\\
&=b_0^{i-1}(b_0d_1d_0b_1)d_0^{i-1}
\end{aligned}$$

Relation (\ref{iso8}):
$$\begin{aligned}
\alpha_i\sigma_{i+j+1}&=(d_1^{i-1}a_1b_1^i)(d_1^{i+j}b_0d_1d_0b_1^{i+j+1})\\
&=d_1^{i-1}(a_1d_1)(d_1^{j-1}(b_0d_1d_0b_1)b_1^{j-1})b_1^{i+1}\\
&=d_1^{i-1}(a_2b_2)(b_0^{j-1}(b_0d_1d_0b_1)d_0^{j-1})b_1^{i+1}\\
&=d_1^{i-1}(a_2b_2)(b_0^{j-1}(b_0^2b_{-1}d_0d_{-1}d_0)d_0^{j-1})b_1^{i+1}\\
&=d_1^{i-1}(b_0^{j-1}(b_0^2b_{-1}d_0d_{-1}d_0)d_0^{j-1})(a_2b_2)b_1^{i+1}\\
&=d_1^{i-1}(d_1^{j-1}(b_0d_1d_0b_1)b_1^{j-1})(a_1d_1)b_1^{i+1}\\
&=\sigma_{i+j-1}\alpha_i.
\end{aligned}$$
Relation (\ref{iso9}):
$$\begin{aligned}
\alpha_{i+j+1}\sigma_i&=(d_1^{i+j}a_1b_1^{i+j+1})(d_1^{i-1}b_0d_1d_0b_1^i)\\
&=d_1^{i-1}(d_1^{j+2}a_1b_1^{j+2})b_0d_1d_0b_1^i\\
&=d_1^{i-1}(b_0^{j+1}a_0d_0^{j+2})b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0(b_0^ja_0d_0^{j+1})d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1(d_1^{j-1}a_1b_1^j)d_0b_1^i\\
&=d_1^{i-1}b_0d_1(b_0^{j-1}a_0d_0^j)d_0b_1^i\\
&=d_1^{i-1}b_0d_1d_0(b_0^ja_0d_0^{j+1})b_1^i\\
&=d_1^{i-1}b_0d_1d_0(d_1^ja_1b_1^{j+1})b_1^i\\
&=\sigma_i\alpha_{i+j+1}.
\end{aligned}$$
Relations (\ref{iso12}) and (\ref{iso13}) are proven in a similar way.
Let us prove (\ref{iso10}):
$$\begin{aligned}
\sigma_i\sigma_{i+j+1}
&=(d_1^{i-1}b_0d_1d_0b_1^i)(d_1^{i+j}b_0d_1d_0b_1^{i+j+1})\\
&=d_1^{i-1}b_0d_1d_0(d_1^{j}(b_0d_1d_0b_1)b_1^j)b_1^i\\
&=d_1^{i-1}b_0d_1d_0(b_0^j(b_0d_1d_0b_1)d_0^j)b_1^i\\
&=d_1^{i-1}b_0d_1(b_0^{j-1}(b_0d_1d_0b_1)d_0^{j-1})d_0b_1^i\\
&=d_1^{i-1}b_0d_1(d_1^{j-1}(b_0d_1d_0b_1)b_1^{j-1})d_0b_1^i\\
&=d_1^{i-1}b_0(d_1^j(b_0d_1d_0b_1)b_1^j)d_1d_0b_1^i\\
&=d_1^{i-1}b_0(b_0^j(b_0d_1d_0b_1)d_0^j)d_1d_0b_1^i\\
&=d_1^{i-1}(b_0^{j+1}(b_0d_1d_0b_1)d_0^{j+1})b_0d_1d_0b_1^i\\
&=d_1^{i-1}(d_1^{j+1}(b_0d_1d_0b_1)b_1^{j+1})b_0d_1d_0b_1^i\\
&=\sigma_{i+j+1}\sigma_i.
\end{aligned}$$
Relations (\ref{iso11}):
$$\begin{aligned}
\alpha_i\sigma_{i+1}\beta_{i+2}
&=(d_1^{i-1}a_1b_1^i)(d_1^ib_0d_1d_0b_1^{i+1})(d_1^{i+2}c_1b_1^{i+1})\\
&=d_1^{i-1}(a_1b_0d_1d_0d_1c_1b_1)b_1^{i-1}\\
&=d_1^{i-1}(a_1b_0d_1(d_0d_1d_2)b_2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(a_1b_0(d_0d_1d_2)d_1b_2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(a_1d_1d_2d_1b_2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(a_2b_2d_2d_1b_2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(a_2d_1b_2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(a_2(d_1d_2d_3)b_3b_2^2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}((d_1d_2d_3)a_2b_3b_2^2c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0d_0(d_2d_3a_2b_3b_2^2)c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0(d_2d_3a_2b_3b_2^2)d_0c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(d_1d_2d_3)a_2b_3b_2^2d_0c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_2(d_1d_2d_3)b_3b_2^2d_0c_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_2d_1b_2(d_0c_1)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(a_2b_2)b_1b_0(d_0d_1d_2)d_1b_2(d_0c_1)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(a_1d_1)b_1b_0d_1(d_0d_1d_2)b_2(b_1c_0)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_1b_0d_1(d_0c_0)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_1b_0d_1(b_{-1}c_{-1})b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_1b_0(b_{-1}c_{-1})d_1b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_1b_0(d_0c_0)b_1)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_1c_0b_1)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(a_1b_1)b_0b_{-1}(d_{-1}d_0d_1)c_0b_1)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(a_0d_0)b_0b_{-1}c_0(d_{-1}d_0d_1)b_1)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(a_0b_{-1})c_0d_{-1}d_0)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1a_{-1}(d_0c_0)d_{-1}d_0)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1(a_{-1}b_{-1}c_{-1}d_{-1})d_0)b_1^{i-1}\\
&=d_1^{i-1}(d_1b_0b_1d_0)b_1^{i-1}\\
&=\sigma_i^{-1},
\end{aligned}$$
$$\begin{aligned}
\alpha_i\sigma_{i+1}^{-1}\beta_{i+2}
&=(d_1^{i-1}a_1b_1^i)(d_1^{i+1}b_0b_1d_0b_1^i)(d_1^{i+2}c_1b_1^{i+1}\\
&=d_1^{i-1}((a_1d_1)b_0b_1d_0d_1(d_1c_1)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}((a_2b_2)b_0b_1d_0d_1(b_0c_0)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(b_0(a_2b_2)b_1d_0d_1d_2b_2(b_0c_0)b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(b_0(a_1d_1)b_1d_0d_1d_2(b_0c_0)b_2b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(b_0a_1(d_0d_1d_2)(d_1c_1)b_2b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(b_0a_1(d_1c_1)(d_0d_1d_2)b_2b_1^2)b_1^{i-1}\\
&=d_1^{i-1}(b_0(a_1d_1c_1b_1)d_1d_0b_1)b_1^{i-1}\\
&=d_1^{i-1}(b_0d_1d_0b_1)b_1^{i-1}\\
&=\sigma_i.
\end{aligned}$$
Relations~(\ref{iso2}) are symmetric to relations~(\ref{iso11}),
their proofs can be received from the proofs of relations~(\ref{iso11})
by reversing the letter order in all the words and replacings
$a_i\leftrightarrow c_i$, $b_i\leftrightarrow d_i$.

Relations~(\ref{r1}) are symmetric to each other. Let us prove
one of them:
$$\begin{aligned}
\alpha_i\sigma_{i+1}\beta_i
&=(d_1^{i-1}a_1b_1^i)(d_1^ib_0d_1d_0b_1^{i+1})(d_1^ic_1b_1^{i-1})\\
&=d_1^{i-1}(a_1b_0d_1d_0(b_1c_1))b_1^{i-1}\\
&=d_1^{i-1}(a_1b_0d_1d_0(d_2c_2)b_1^{i-1}\\
&=d_1^{i-1}(a_1b_0d_1(b_1c_1)d_0)b_1^{i-1}\\
&=d_1^{i-1}((a_1b_0)c_1d_0)b_1^{i-1}\\
&=d_1^{i-1}(a_0(d_1c_1)d_0)b_1^{i-1}\\
&=d_1^{i-1}(a_0(b_0c_0)d_0)b_1^{i-1}\\
&=1.
\end{aligned}$$
Relations~(\ref{r2}) also are symmetric to each other. We prove one of them:
$$\begin{aligned}
\alpha_{i+1}\sigma_i\beta_{i+1}
&=(d_1^ia_1b_1^{i+1})(d_1^{i-1}b_0d_1d_0b_1^i)(d_1^{i+1}c_1b_1^i)\\
&=d_1^i((a_1b_1)b_1b_0d_1d_0d_1c_1)b_1^i\\
&=d_1^i((a_0d_0)b_1b_0d_1d_0d_1c_1)b_1^i\\
&=d_1^i(d_2(a_0d_0)(b_2b_1b_0)d_1d_0d_1c_1)b_1^i\\
&=d_1^i(d_2(a_1b_1)(b_2b_1b_0)d_1d_0d_1c_1)b_1^i\\
&=d_1^i(d_2(b_2b_1b_0)(a_1b_1)d_1d_0d_1c_1)b_1^i\\
&=d_1^i(b_1b_0a_1(d_0d_1d_2)b_2c_1)b_1^i\\
&=d_1^i(b_1b_0(d_0d_1d_2)a_1b_2c_1)b_1^i\\
&=d_1^i(d_2a_1(b_2c_1))b_1^i\\
&=d_1^i(d_2(a_1d_1)c_2)b_1^i\\
&=d_1^i(d_2(a_2b_2)c_2)b_1^i\\
&=1.
\end{aligned}$$
Relations~(\ref{r3}) trivially follows from~(\ref{rel3}). It remains to
check relation~(\ref{r4}):
$$\begin{aligned}
\sigma_i\sigma_{i+1}\sigma_i
&=(d_1^{i-1}b_0d_1d_0b_1^i)(d_1^ib_0d_1d_0b_1^{i+1})
(d_1^{i-1}b_0d_1d_0b_1^i)\\
&=d_1^{i-1}b_0d_1^2d_0b_1^2b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1b_0b_{-1}(d_{-1}d_0d_1)d_0b_1^2b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1b_0b_{-1}d_0(d_{-1}d_0d_1)b_1^2b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1(d_2b_2)(b_0b_{-1}d_0d_{-1}d_0)b_1b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1d_2(b_0b_{-1}d_0d_{-1}d_0)b_2b_1b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1d_2b_0b_{-1}d_0(d_{-1}d_0d_1)b_1b_2b_1b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0d_1d_2b_0b_{-1}(d_{-1}d_0d_1)d_0b_1b_2b_1b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0^2b_{-1}(d_{-1}d_0d_1d_2)(d_1d_0b_1)b_2b_1b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0^2b_{-1}(d_1d_0b_1)(d_{-1}d_0d_1d_2)b_2b_1b_0d_1d_0b_1^i\\
&=d_1^{i-1}b_0^2b_{-1}d_1d_0b_1d_{-1}d_1d_0b_1^i\\
&=d_1^{i-1}b_0^2b_{-1}d_1d_0b_1d_1d_{-1}d_0b_1^i\\
&=d_1^{i-1}b_0^2b_{-1}d_1d_0(d_{-1}d_0d_1)b_1^{i+1}\\
&=d_1^{i-1}b_0^2b_{-1}d_1(d_{-1}d_0d_1)d_0b_1^{i+1}\\
&=d_1^{i-1}b_0^2d_1d_0d_1d_0b_1^{i+1}\\
&=d_1^{i-1}b_0^2d_1(d_0d_1d_2)b_2d_0b_1^{i+1}\\
&=d_1^{i-1}b_0^2(d_0d_1d_2)d_1b_2d_0b_1^{i+1}\\
&=d_1^{i-1}b_0d_1d_2d_1b_2(b_{-1}d_{-1})d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}d_1d_2d_1b_2d_{-1}d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}b_0(d_0d_1d_2)d_1b_2d_{-1}d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}b_0d_1(d_0d_1d_2)b_2d_{-1}d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}b_0d_1d_0d_1d_{-1}d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}b_0d_1d_0d_{-1}d_1d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}(b_0d_1d_0)(d_{-1}d_0d_1d_2)b_2b_1b_0d_1d_0b_1^{i+1}\\
&=d_1^{i-1}b_0b_{-1}(d_{-1}d_0d_1d_2)(b_0d_1d_0)b_2b_1b_0d_1d_0b_1^{i+1}\\
&=d_1^id_2b_0d_1d_0b_2b_1b_0d_1d_0b_1^{i+1}\\
&=d_1^id_2b_0^2b_{-1}(d_{-1}d_0d_1)d_0b_2b_1b_0d_1d_0b_1^{i+1}\\
&=d_1^id_2b_0^2b_{-1}d_0(d_{-1}d_0d_1)(b_2b_1b_0)d_1d_0b_1^{i+1}\\
&=d_1^id_2(b_0^2b_{-1}d_0d_{-1}d_0)(b_2b_1b_0)d_1^2d_0b_1^{i+1}\\
&=d_1^i(b_0^2b_{-1}d_0d_{-1}d_0)b_1b_0d_1^2d_0b_1^{i+1}\\
&=d_1^ib_0^2b_{-1}d_0(d_{-1}d_0d_1)b_1^2b_0d_1^2d_0b_1^{i+1}\\
&=d_1^ib_0^2b_{-1}(d_{-1}d_0d_1)d_0b_1^2b_0d_1^2d_0b_1^{i+1}\\
&=(d_1^ib_0d_1d_0b_1^{i+1})(d_1^{i-1}b_0d_1d_0b_1^i)
(d_1^ib_0d_1d_0b_1^{i+1})\\
&=\sigma_{i+1}\sigma_i\sigma_{i+1}.
\end{aligned}$$

\section{Semigroups $Y_n$ and groups $D_n$ in the case $n<\infty$.}
In Section~\ref{yinfty} we defined subsemigroups $Y_n\subset Y_\infty$,
where $n=3,4,\dots$. Clearly for any $n\ge3$ the semigroup $Y_n$
coincides with the semigroup in $Y_\infty$ generated by the set
$A_1\cup A_2\cup\dots\cup A_{n-1}$, where $A_i=\{a_i,b_i,c_i,d_i\}$.

\begin{theorem}
For $n\ge7$ the semigroup $Y_n$ is isomorphic to the semigroup
defined by the generating set $A_1\cup A_2\cup\dots\cup A_{n-1}$
and those relations~\/{\rm(\ref{rel1})--(\ref{rel7})} that
involve only generators from that set.
\end{theorem}

\proof
Clearly the semigroup $Y_n$ is isomorphic to the semigroup $Y'_n$
in $Y_\infty$ generated by the set $A_{-2}\cup A_{-1}\cup\dots
\cup A_{n-5}\cup A_{n-4}$. The proof of our assertion for the semigroup
$Y'_n$ with $n\ge7$ almost verbally coincides with the proof of
Theorem~\ref{y.infty}. Indeed, the elements~(\ref{image1})--(\ref{image4})
belong to $Y'_n$. All the transforms that were used in the proof
of Lemma~\ref{razlozh} preserve the set of words in the alphabet
$A_{-2}\cup A_{-1}\cup\dots\cup A_{n-5}\cup A_{n-4}$ for $n\ge7$.
In the case of a finite $n$, one can simply omit all the letters
$b_i,d_i$ with $i>n-4$ and $i<-2$ in the definition of
the words $w_{1,2}'$ in the end of Section~\ref{yinfty}.
Finally, during the checking relations in Section~\ref{checking},
only generators from $A_{-2}\cup\dots\cup A_3$ appeared.
\proved

For $5\le n<\infty$ the semigroups $Y_n$ can be defined in a different
way. By adding to the generating set the tangles of complexity $1$
acting nontrivially in $P_{n-1}$ and $P_0$, we get the following
presentation.

\begin{theorem}
If $n\ge5$ the semigroup $Y_n$ is isomorphic to the semigroup presented
by the generating set $\{a_i,b_i,c_i,d_i\}_{i\in\integer_n}$ and
the defining relations
\begin{equation}\label{d_1d_n}
d_1d_2\dots d_n=1,
\end{equation}
$$a_id_{i+1}=a_{i+1}b_i,\eqno(\ref{rel1})$$
$$d_ic_{i+1}=b_{i+1}c_i,\eqno(\ref{rel2})$$
$$a_id_i=a_{i+1}b_{i+1},\eqno(\ref{rel4})$$
$$b_ic_i=d_{i+1}c_{i+1},\eqno(\ref{rel5})$$
$$a_ib_ic_id_i=1,\eqno(\ref{rel8})$$
$$b_id_i=d_ib_i=1,\eqno(\ref{rel3})$$
$$x_iy_j=y_jx_i,\eqno(\ref{rel6})$$
for all $i,j\in\integer_n$ such that $i-j\notin\{-1,0,1\}$
and $x_i\in\{a_i,b_i,c_i,d_i\}$, $y_j\in\{a_j,b_j,c_j,d_j\}$.
All the indices are regarded modulo $n$.
\end{theorem}

Again, the proof is almost the same as that of Theorem~\ref{y.infty}.
Here the generators $a_i,b_i,c_i,d_i$ with $1\le 1\le n-1$ are the
same as above. The new generators $a_n,b_n,c_n,d_n$ acting in
the half-planes $P_{n-1}$ and $P_0$ are expressed in terms of
the otheres due to the relations given above:
$$\begin{aligned}
a_n&=a_1d_2\dots d_{n-1},\\
b_n&=d_1\dots d_{n-1},\\
c_n&=d_1\dots d_{n-2}c_{n-1},\\
d_n&=b_{n-1}\dots b_1.
\end{aligned}$$
Relations~(\ref{rel7}), which are omitted this time, easily follow
from~(\ref{rel6})~and (\ref{d_1d_n}).

If $n=3,4$ the proof fails because it uses comutativity
of the elements from $A_i$ with the elements from $A_{i+2}$
and from $A_{i+3}$, which does not take place in the case $n=3,4$.
However, the semigroups $Y_3$ and $Y_4$ are also finitely presented.
For $Y_3$ this fact was established in papers~\cite{dyn1}, \cite{dyn2}.
For $Y_4$ the proof is similar. We give here only the formulation
of the corresponding claim.

\begin{theorem}
The semigroup $Y_3$ is isomorphic to the semigroup presented by the
genertors $\{a_i,b_i,c_i,d_i\}_{i\in\integer_3}$ and the defining relations
$$\begin{aligned}
a_i=a_{i+1}d_{i-1},\quad
b_i=a_{i-1}c_{i+1},\quad
&c_i=b_{i-1}c_{i+1},\quad
d_i=a_{i+1}c_{i-1},\\
d_1d_2d_3&=1,\\
b_id_i=d_ib_i&=1,\\
x_iy_i&=y_ix_i,
\end{aligned}$$
where $x_i\in\{a_{i-1}b_{i-1},\;d_{i-1}c_{i-1},
\;b_{i+1}d_{i-1}d_{i+1}b_{i-1}\}$,
$y_i\in\{a_i,\;b_i,\;c_i,\;b_{i-1}d_id_{i-1}\},$ $i\in\integer_3$.

The semigroup $Y_4$ is isomorphic to the semigroup presented by
the generators $\{a_i,b_i,c_i,d_i\}_{i\in\integer_4}$ and the defining
reltions
$$\begin{aligned}
d_1d_2d_3d_4&=1,\\
a_id_{i+1}&=a_{i+1}b_i,\\
d_ic_{i+1}&=b_{i+1}c_i,\\
a_id_i&=a_{i+1}b_{i+1},\\
b_ic_i&=d_{i+1}c_{i+1},\\
a_ib_ic_id_i&=1,\\
b_id_i=d_ib_i&=1,\\
x_iy_{i+2}&=y_{i+2}x_i,
\end{aligned}$$
where $x_i\in\{a_i,b_i,c_i,d_i\}$, $y_i\in\{a_i,b_i,c_i,d_i,
b_{i-1}a_id_{i-1},b_{i-1}b_id_{i-1},b_{i-1}c_id_{i-1}\}$,
$i\in\integer_4$.
\end{theorem}

\begin{theorem}
If $n\ge3$ the center $Z(Y_n)$ of the semigroup $Y_n$ is isomorphic
to the semigroup $L$ of isotopy classes of non-oriented links in $\real^3$.
\end{theorem}

\proof
% Vstavit frazu v russkom variante: "Budem govorit', chto slovo
% $w$ sbalansirovano, jesli ono $i$-sbalansirovano dlia vsekh $i\in\integer$.
We say that a word $w$ is {\it balanced\/} if it is balanced for
all $i\in\integer$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Clearly the semigroup $L$ is embedded $Y_\infty$ as the subsemigroup
consisting of all the elements that {\it can\/} be presented
by a balanced word. Also clear that this subsemigroup is contained
in all the semigroups $T,Y_3,Y_4,\dots$ and that all its elements
commute with all elements from $Y_\infty$. Let us prove the
converse statement, i.e.\ if a word $w$ presents a central element
$x$ of the semigroup $Y_n$, then $x$ belongs to
the semigroup $L\subset Y_n\subset Y_\infty$.

Indeed, it is not hard to show that if $x\in Z(Y_n)$, then
$\varphi(x)=0$. Hence for some $N$ and $i_1,i_2,\dots,i_N\in
\{1,2,\dots,n-1\}$ the word $a_{i_1}\dots a_{i_N}wc_{i_N}\dots c_{i_1}$
is balanced. Then for the corresponding tangle $t$, there exists
a sphere embedded in $\real^3$ that separates the tangle
into a link and a tangle presenting the unit of the semigroup
$Y_n$. Therefore, such a sphere exists also for an equivalent
tangle defined by the word $a_{i_1}\dots a_{i_N}c_{i_N}\dots
c_{i_1}w$, and hence for the tangle defined by the word $w$,
since the word $a_{i_1}\dots a_{i_N}c_{i_N}\dots c_{i_1}$
is balanced.\proved

Consider now the group $D_n$ of invertible elements of
the semigroup $Y_n$, $n\ge3$.

\begin{theorem}
If $n\ge3$ the group $D_n$ is finitely presented and it can
be presented by the generators $\{d_i\}_{i\in\integer_n}$ and the following
set of defining relations:

for $n\ge5$
$$d_id_j=d_jd_i,\quad\mbox{if }i-j\ne\pm1,$$
$$d_1d_2\dots d_n=1,$$
where all the indeces are regarded modulo $n$;

for $n=4$
$$\begin{aligned}
d_1d_2d_3d_4&=1,\\
d_id_{i+2}&=d_{i+2}d_i,\\
d_{i+2}(d_{i-1}^{-1}d_id_{i-1})&=
(d_{i-1}^{-1}d_id_{i-1})d_{i+2},
\end{aligned}$$
$i\in\integer_4$;

for $n=3$
$$\begin{aligned}
d_1d_2d_3&=1,\\
d_iu_i&=u_id_i,\\
u_iu_{i+1}&=u_{i+1}u_i,
\end{aligned}$$
where $u_i=d_{i+1}^{-1}d_{i-1}d_{i+1}d_{i-1}^{-1}$, $i\in\integer_3$.

The commutator subgroup of the group $D_n$ is isomorphic to the braid
group $B_\infty$ on the infinite number of strands.
\end{theorem}

\proof
It is known that the group of invertible elements of the semgigroup
$T$  is the braid group $B_\infty$, which is presented by
generators $\sigma_i$ (fig.~\ref{generators}) and relation~(\ref{iso10})
and (\ref{r4}). The proof of the first assertion of the theorem can be
obtained from the proofs of similar statements for the semigroups
$Y_n$, $n=3,4,\dots,\infty$, by replacing $T$ with $B_\infty$ and
deleting all relations and calculations that use the generators
$a_i,c_i$ of the semigroups $Y_n$ and the generators $\alpha_i,\beta_i$
of the semigroup $T$. In particular, in this way we conclude that
the group of the invertible elements of the semigroup
$Y_\infty$ is presented by the generatros $d_i$, $i\in\integer$
and the relations:
$$\begin{aligned}
d_id_j&=d_jd_i,\quad\rlap{if $|i-j|>1$},\\
d_i(d_{i-1}d_id_{i+1})&=(d_{i-1}d_id_{i+1})d_i.
\end{aligned}$$

Also clear that the kernel of the restriction of the homomorphism
$\varphi$ to the group $D_n$ contains the commutator subgroup of
the group $D_n$ and at the same time, it is a subgroup isomorphic
to the braid group $B_\infty$. So, it suffices to show that the
commutator subgroup is the whole subgroup, not a non-trivial part of it.
This follows easily from the fact that the generators
(\ref{image3}), which twist a coule of strands, lie in
the commutator subgroup of $D_\infty$.\proved

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\end{thebibliography}
\end{document}