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\title{Triangulations with Regular Structures}
\author{V.A.Malyshev}
\date{}

\begin{document}
\maketitle

%\begin{abstract}
%ø{\small\it  $\!\!\!\!\!\!\!\!\!\!\!\!$ }
%\end{abstract}

\begin{abstract}
ø{\small\it  $\!\!\!\!\!\!\!\!\!\!\!\!$ The direct product of two simplices and the
intersection of a simplex with a half-space have triangulations with regular structure.
This gives natural triangulations for convex polytopes and permits to construct some
nontrivial homeomorphisms.}
\end{abstract}

\begin{center}
 {\bf Contents}
\end{center}



\noindent {Introduction}
\medskip

\noindent {1. Associated triagulations}

\noindent {2. Direct products}

\noindent {3. Regular intersections}

\noindent {4. Singular intersections}

\noindent {5. Cylinders of simplicial mappings}

\medskip
\noindent {Bibliography}


%%%%%%%%%%%%%%%%%%%
\section*{Introduction}


In the paper we consider polytopes with regular triangulations. The main examples are the
direct product of two simplices and the intersection of a simplex with a half-space. In
particular, the last triangulation permits to obtain a natural triangulation for any
convex polytope. Indeed, let a convex polytope belong to a simplex. Since the polytope
may be represented as an intersection of half-spaces, the triangulation of the polytope
is constructed step by step. At each step we use the triangulation of the intersection of
a simplex with a half-space.

The triangulations with regular structure permit to construct some homeomorphisms. Let
$$
ø\Delta^{n}=\{\lambda\in \mathbb{R}^{n+1}:\;\lambda_{0}\ge0,\ldots,\lambda_{n}\ge0,\,
\lambda_{0}+\cdots+\lambda_{n}=1\}
$$
be the unite $n$-dimensional simplex. We assume that $\Delta^{-1}=\emptyset$. Consider a
simplicial surjection
$$
øF:\;\Delta^{n}\longrightarrow\Delta^{k}.
$$
Then the cylinder ${\rm Cyl}F$ of the mapping $F$ is homeomorphic to the ball
$$D^{n+1}\approx {\rm cyl}\Delta^{k}*\Delta^{n-k-1}$$ and has a natural triangulation with
$k+1$ simplices of dimension $n+1$.

\bigskip
\begin{center}
 \begin{picture}(400,20)
 \put(-40,30){\line(1,0){70}}
 \put(-25,15){{\footnotesize{\it 2000 Mathematics Subject Classification.} Primary 52-01, 52-55,
 52B11, 52B55.}}
 \put(-25,0){{\footnotesize{\it Key words and phrases.} Triangulations, polytopes.}}
\end{picture}
\end{center}
\newpage


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. Associated  triagulations}


Let $A_{0},A_{1},\ldots,A_{n}$ be a set of points in the space $\mathbb{R}^{m}$. By
$$
ø\Delta={\rm conv}\{A_{0},A_{1},\ldots,A_{n}\}
$$
we denote the convex hull of the set $A_{0},A_{1},\ldots,A_{n}$. The set
$A_{0},A_{1},\ldots,A_{n}$ is called nonsingular if
$$
ø\lambda_{0}=\lambda_{1}=\cdots=\lambda_{n}=0
$$
whenever
$$
ø\lambda_{0}+\lambda_{1}+\cdots+\lambda_{n}=0
$$
and
$$
ø\lambda_{0}A_{0}+\lambda_{1}A_{1}+\cdots+\lambda_{n}A_{n}=0.
$$
If $A_{0},A_{1},\ldots,A_{n}$ is nonsingular, then $\Delta$ is a simplex of dimension
$n$.

Let $T=\{\Delta\}$ be a finite collection of simplices $\Delta\subset \mathbb{R}^{m}$.
The collection $T$ is called a triangulation~\cite{Rokh1} if the following conditions
hold:

$1^{0}$. If $\Gamma$ is a face of $\Delta$ and $\Delta\in T$, then $\Gamma\in T$.

$2^{0}$. If $\Gamma,\Delta\in T$ and $\Gamma\subset \Delta$, then $\Gamma$ is a face of
$\Delta$.

$3^{0}$. If $\Gamma,\Delta\in T$, then $\Gamma\cap \Delta\in T$ .

\noindent Example: The collection of all faces of a simplex $\Delta$ is a triangulation.

Let $G$ be an oriented graph without closed paths. Let each vertex $v\in G$ be associated
with a point $A_{v}\in\mathbb{R}^{m}$, and let the set $A_{v_0},A_{v_1},\ldots,A_{v_n}$
be nonsingular for any path $v_0,v_1,\ldots,v_n$ on the graph $G$. If simplices
$$
\Delta={\rm conv}\{A_{v_0},A_{v_1},\ldots,A_{v_n}\}
$$
with their faces form a triangulation, then we say that the triangulation is associated
with the graph $G$. Obviously, for construction of the associated triangulation we may
use only maximal paths on the graph $G$.

Example: The standard triagulation of the simplex
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is associated with the graph
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. Direct products}

All graphs that we use for construction of associated triangulations belong to the unit
lattice. In particular, vertices and edges of our graphs are vertices and edges of the
lattice. Since edges of the lattice are oriented in positive directions, we do not write
arrows on the edges of our graphs.

In the section we consider the standard triangulation of the direct product of two
simplices. Let $A_0,\ldots,A_p\in \mathbb{R}^{m}$ and $B_0,\ldots,B_q\in \mathbb{R}^{l}$
be nonsingular sets. Consider simplices
$$
X_{p}={\rm conv}\{A_0,\ldots,A_p\}
$$
and
$$
Y_{q}={\rm conv}\{B_0,\ldots,B_q\}.
$$
Let $S_{pq}$ be the graph
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We see that $S_{pq}$ has $C^p_{p+q}=C^q_{p+q}$ maximal paths connecting $0\times 0$ and
$p\times q$.

Assume that the vertex $i\times j\in S_{pq}$ is associated with the point $A_i\times
B_j\in X_{p}\times Y_{q}$. We claim that the triangulation associated with the graph
$S_{pq}$ is the triangulation of the direct product $X_{p}\times Y_{q}$. In particular,
the triangulation has $C^p_{p+q}=C^q_{p+q}$ simplices of maximal dimension $n=p+q$.
\medskip

\noindent{\bf Examples.} If $p=1$ and $q=1$, then $S_{1\,1}$ has two maximal paths
$$
ø\begin{array}{ccc}
  ø0\times0 & ø1\times0 & ø1\times1 \\
  ø0\times0 & ø0\times1 & ø1\times1
\end{array}
$$
Hence the triangulation of $X_{1}\times Y_{1}$ has two maximal simplices with vertices
$$
ø\begin{array}{ccc}
  øA_{0}\times B_{0}&A_{1}\times B_{0}&A_{1}\times B_{1}\medskip \\
  øA_{0}\times B_{0}&A_{0}\times B_{1}&A_{1}\times B_{1} \\
\end{array}
$$
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If $p=2$ and $q=1$, then $S_{2\,1}$ has three maximal paths
$$
\begin{array}{cccc}
  ø0\times0 & ø1\times0 & ø2\times0 & ø2\times1 \\
  ø0\times0 & ø1\times0 & ø1\times1 & ø2\times1 \\
  ø0\times0 & ø0\times1 & ø1\times1 & ø2\times1 \\
\end{array}
$$
Hence the triangulation of $X_{2}\times Y_{1}$ has three maximal simplices with vertices
$$
ø\begin{array}{cccc}
  øA_{0}\times B_{0}&A_{1}\times B_{0}&A_{2}\times B_{0}&A_{2}\times B_{1}\medskip \\
  øA_{0}\times B_{0}&A_{1}\times B_{0}&A_{1}\times B_{1}&A_{2}\times B_{1}\medskip \\
  øA_{0}\times B_{0}&A_{0}\times B_{1}&A_{1}\times B_{1}&A_{2}\times B_{1}
\end{array}
$$
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\medskip

\begin{theorem}
\label{simpro} The triangulation associated with $S_{pq}$ is the triangulation of
$X_{p}\times Y_{q}$.
\end{theorem}
{\bf Proof. Part 1.} Let
$$
i_0\times j_0,\;i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
be a maximal path. We claim that the set
$$
A_{i_0}\times B_{j_0},A_{i_1}\times B_{j_1},\ldots,A_{i_n}\times B_{j_n}
$$
is nonsingular.


We see that
$$
\begin{array}{l}
ø 0=i_0\le i_1\le\cdots\le i_n=p\medskip\\
  0=j_0\le j_1\le\cdots\le j_n=q
\end{array}
$$
and
$$
i_k+j_k=k
$$
for all $0\le k\le n$. Put
$$
\begin{array}{lll}
 I(i)&=&\{k:\; i_k=i\},\medskip\\
 J(j)&=&\{k:\; j_k=j\},\\
\end{array}
$$
where $0\le i\le p$ and $0\le j\le q$.
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Assume that
$$
\lambda_0+\lambda_1+\cdots+\lambda_n=0
$$
and
$$
\lambda_0A_{i_0}\times B_{j_0}+ \lambda_1A_{i_1}\times B_{j_1}+\cdots+
\lambda_nA_{i_n}\times B_{j_n}=0.
$$
Since $A_0,A_1,\ldots,A_p$ and $B_0,B_1,\ldots,B_q$ are nonsingular, we have
$$
\sum_{k\in I(i)}\lambda_k=0
$$
for $0\le i\le p$ and
$$
\sum_{k\in J(j)}\lambda_k=0
$$
for $0\le j\le q$.

It is easy to check that
$$
\max I(i)\ne\max J(j)
$$
for $0\le i\le p-1$ and $0\le j\le q$, that
$$
\max I(i')<\max I(i'')
$$
for $0\le i'<i''\le p$, and that
$$
\max J(j')<\max J(j'')
$$
for $0\le j'<j''\le q$. This means that $\lambda_0,\lambda_1,\ldots,\lambda_n$ satisfy a
linear system with a matrix of rank $n+1$. Therefore,
$$
\lambda_0=\lambda_1=\ldots=\lambda_n=0.
$$
By definition, the set
$$
A_{i_0}\times B_{j_0},A_{i_1}\times B_{j_1},\ldots,A_{i_n}\times B_{j_n}
$$
is nonsingular.

{\bf Part 2.} We claim that for any $x\times y\in X_{p}\times Y_{q}$ there exists a
maximal path
$$
i_0\times j_0,\;i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
such that
$$
x\times y\in {\rm conv}\{A_{i_0}\times B_{j_0}, A_{i_1}\times
B_{j_1},\ldots,A_{i_n}\times B_{j_n}\}.
$$
Let
$$
x=\lambda_0A_0+\cdots+\lambda_pA_p
$$
and
$$
y=\mu_0B_0+\cdots+\mu_qB_q.
$$
Put
$$
\begin{array}{lll}
\alpha_0&=&\lambda_0\\ \alpha_1&=&\lambda_0+\lambda_1\\ &\cdots&\\
\alpha_{p-1}&=&\lambda_0+\lambda_1+\cdots+\lambda_{p-1}\medskip\\ \beta_0&=&\mu_0\\
\beta_1&=&\mu_0+\mu_1\\ &\cdots&\\ \beta_{q-1}&=&\mu_0+\mu_1+\cdots+\mu_{q-1}\\
\end{array}
$$
and place the sequence
$$
ø\alpha_{0},\alpha_{1},\ldots,\alpha_{p-1},\;\beta_{0},\beta_{1},\ldots,\beta_{q-1}
$$
in ascending order
$$
\gamma_0\le\gamma_1\le\cdots\le\gamma_{n-1}.
$$
There exist sequences
$$
0=i_0\le i_1\le\cdots\le i_n=p
$$
and
$$
0=j_0\le j_1\le\cdots\le j_n=q
$$
such that
$
i_k+j_k=k
$
for $0\le k\le n$ and
$$
\gamma_k=(i_{k+1}-i_k)\alpha_{i_k}+(j_{k+1}-j_k)\beta_{j_k}
$$
for $0\le k\le n-1$.

Let us check that
$$
x\times y=\sum_{k=0}^n(\gamma_k-\gamma_{k-1})A_{i_k}\times B_{j_k},
$$
where $\gamma_{-1}=0$ and $\gamma_n=1$. Indeed,
$$
\sum_{k=0}^n(\gamma_k-\gamma_{k-1})A_{i_k}= \sum_{i=0}^p\sum_{k\in
I(i)}(\gamma_k-\gamma_{k-1})A_i.
$$
Let $0\le i\le p$, $s=\min I(i)$, and $r=\max I(i)$.
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We see that
$$
\begin{array}{lllll}
\gamma_{s-1}&=&1\alpha_{i-1}&+&0\beta_{j_{s-1}}\medskip\\ \gamma_s    &=&0\alpha_i
&+&1\beta_{j_s}\medskip\\ &\cdots&&&\medskip\\ \gamma_{r-1}&=&0\alpha_i
&+&1\beta_{j_{r-1}}\medskip\\ \gamma_r&=&1\alpha_i        &+&0\beta_{j_r}\medskip\\
\end{array}
$$
Hence
$$
\sum_{k\in I(i)}(\gamma_k-\gamma_{k-1})= \sum_{k=s}^r(\gamma_k-\gamma_{k-1})=
\alpha_i-\alpha_{i-1}=\lambda_i.
$$
Therefore
$$
x=\sum_{k=0}^n(\gamma_k-\gamma_{k-1})A_{i_k}.
$$
The identity
$$
y=\sum_{k=0}^n(\gamma_k-\gamma_{k-1})B_{j_k}
$$
is proved analogously.

{\bf Part $3$.} Consider two maximal paths
$$
i_0\times j_0,\;i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
and
$$
u_0\times v_0,\;u_1\times v_1,\;\ldots,\;u_n\times v_n.
$$
Let
$$
\sum_{k=0}^n\lambda_kA_{i_k}\times B_{j_k}= \sum_{k=0}^n\mu_kA_{u_k}\times B_{v_k},
$$
where
$$
\lambda_{0}\ge 0,\ldots,\lambda_{n}\ge 0,\; \mu_{0}\ge 0,\ldots,\mu_{n}\ge 0
$$
and
$$
ø\lambda_{0}+\cdots+\lambda_{n}=\mu_{0}+\cdots+\mu_{n}=1.
$$
We claim that
$$
\lambda_k=\mu_k
$$
in the case $i_k\times j_k=u_k\times v_k$ and
$$
\lambda_k=\mu_k=0
$$
in the case $i_k\times j_k\ne u_k\times v_k$.


Note that
$$
\sum_{k\in I(i)}\lambda_k=\sum_{k\in U(i)}\mu_k
$$
for $0\le i\le p$ and
$$
\sum_{k\in J(j)}\lambda_k=\sum_{k\in V(j)}\mu_k
$$
for $0\le j\le q$.

We see that
$$
\begin{array}{l}
 i_0=u_0=0\medskip\\
 j_0=v_0=0
\end{array}
$$
and
$$
\begin{array}{l}
 i_n=u_n=n\medskip\\
 j_n=v_n=n
\end{array}
$$
Let $k_1\ge 1$ be a minimal number such that
$$
\begin{array}{l}
 i_{k_1}=u_{k_1}=p_{1},\medskip\\
 j_{k_1}=v_{k_1}=q_{1}.
\end{array}
$$
If $k_1=1$, then
$$
I(0)=U(0)=0
$$
or
$$
J(0)=V(0)=0.
$$
Hence
$$
\lambda_0=\mu_0.
$$

Let $k_{1}\ge 2$. We may assume that for $1\le k\le k_{1}-1$ the first path is over the
second path. This means than
$$
\begin{array}{l}
 i_k<u_k\medskip\\
 j_k>v_k
\end{array}
$$
for all $1\le k\le k_{1}-1$. In particular, we have
$$
\begin{array}{ll}
 I(0)=\{0,1,\ldots,b\}  & U(0)=0, \medskip\\
 J(0)=0                 & V(0)=\{0,1,\ldots,a\},
\end{array}
$$
where $1\le a<p_{1}$ and $1\le b<q_{1}$.
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Thus
$$
\begin{array}{lll}
 \lambda_0+\lambda_1+\cdots+\lambda_{b}&=&\mu_0\medskip\\
 \lambda_0&=&\mu_0+\mu_1+\cdots+\mu_{a}
\end{array}
$$
Subtracting the first from the second, we get
$$
-(\lambda_1+\cdots+\lambda_{b})=\mu_1+\cdots+\mu_{a}.
$$
Therefore
$$
\lambda_1=\ldots=\lambda_{b}=\mu_1=\ldots=\mu_{a}=0
$$
and
$$
\lambda_0=\mu_{0}.
$$
Let us check that
$$
ø\lambda_1=\ldots=\lambda_{k_{1}-1}=\mu_1=\ldots=\mu_{k_{1}-1}=0.
$$
Assume the converse. Let $1\le m\ge k_{1}-1$ be a minimal number such that
$\lambda_{m}\ne 0$ or $\mu_{m}\ne 0$. For example, let $\mu_{m}\ne 0$. Since
$$
\mu_1=\ldots=\mu_{a}=0,
$$
we have $m\ge a+1$. We see that
$$
j=j_{m}\ge 1
$$
and
$$
ø\max J(j)<\min V(j).
$$
Hence
$$
ø1\le \min J(j)\le \max J(j)\le m-1.
$$
By definition of $m$, we have
$$
\sum_{k\in V(j)}\mu_k=\sum_{k\in J(j)}\lambda_k=0.
$$
Therefore $\mu_{m}=0$. The contradiction proves the statement.

Let $k_2\ge k_1+1$ be a minimal number such that
$$
\begin{array}{l}
 i_{k_2}=u_{k_2},\medskip\\
 j_{k_2}=v_{k_2}.
\end{array}
$$
Then analogously we obtain
$$
\lambda_{k_1}=\mu_{k_1}
$$
and
$$
\lambda_k=\mu_k=0
$$
for $k_1+1\le k\le k_2-1$.

We continue the process and obtain $m$ such that $k_{m}=n$. The last identity
$$
\lambda_n=\mu_n
$$
follows from the identity
$$
\lambda_0+\cdots+\lambda_n=\mu_0+\cdots+\mu_n=1.
$$

{\bf Part 4.} Let
$$
P=\{i_0\times j_0,\;i_1\times j_1,\;\ldots,\;i_n\times j_n\}
$$
be a maximal path, and let
$$
S(P)={\rm conv}\{A_{i_0}\times B_{j_0},A_{i_1}\times B_{j_1},\ldots,A_{i_n}\times
B_{j_n}\}
$$
be the maximal simplex corresponding to the maximal path. Denote by $T$ the set
consisting of all simplices $S(P)$ with their faces. Then
$$
X_{p}\times Y_{q}=\bigcup_{\Delta\in T}\Delta.
$$
We claim that $T$ is the triangulation of $X_{p}\times Y_{q}$.

$1^{0}$. Let $\Gamma$ be a face of $\Delta$ and $\Delta\in T$. Obviously $\Gamma\in T$.

$2^{0}$. Let $\Gamma,\Delta\in T$ and $\Gamma\subset \Delta$. Let $A_{i}\times B_{j}$ be
a vertex of $\Gamma$, and let $A_{i_{0}}\times B_{j_{0}},\ldots,A_{i_{m}}\times
B_{j_{m}}$ be all vertices of $\Delta$. Then
$$
øA_{i}\times B_{j}=\sum_{k=0}^{m}\lambda_{k}A_{i_{k}}\times B_{j_{k}},
$$
where $\lambda_{0}\ge 0,\ldots,\lambda_{m}\ge0$ and $\lambda_{0}+\cdots+\lambda_{m}=1$.
From
$$
øA_{i}=\sum_{k=0}^{m}\lambda_{k}A_{i_{k}}
$$
it follows that $\lambda_{k_{0}}=1$ for some $0\le k_{0}\le m$. This means that
$i=i_{k_{0}}$ and $j=j_{k_{0}}$. Hence $A_{i}\times B_{j}$ is a vertex of $\Delta$.
Therefore $\Gamma$ is a face of $\Delta$.

$3^{0}$. Let $\Delta,\Gamma\in T$. Let $\Delta$ be the face of the simplex $S(P)$ with
$$
P=\{i_0\times j_0,\;i_1\times j_1,\;\ldots,\;i_n\times j_n\},
$$
and let $\Gamma$ be the face of the simplex $S(Q)$ with
$$
Q=\{u_0\times v_0,\;u_1\times v_1,\;\ldots,\;u_n\times v_n\}.
$$
Consider a point $x\times y\in \Delta\cap \Gamma$. Then
$$
x\times y=\sum_{k=0}^n\lambda_kA_{i_k}\times B_{j_k}= \sum_{k=0}^n\mu_kA_{u_k}\times
B_{v_k},
$$
where
$$
\lambda_{0}\ge 0,\ldots,\lambda_{n}\ge 0,\; \mu_{0}\ge 0,\ldots,\mu_{n}\ge 0
$$
and
$$
ø\lambda_{0}+\cdots+\lambda_{n}=\mu_{0}+\cdots+\mu_{n}=1.
$$
By Part 3, we have $\lambda_{k}=\mu_{k}$ for all $0\le k\le n$. Moreover, if
$\lambda_{k}\ne0$, then $i_{k}\times j_{k}=u_{k}\times v_{k}$. Hence
$$
ø\Delta\cap\Gamma={\rm conv}\{A_{i}\times B_{j}:\;A_{i}\times B_{j}\in\Delta\cap \Gamma\}
$$
This means that $\Delta\cap\Gamma\in T$. $\bullet$


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. Regular intersections}

Let $A_0,A_1,\ldots,A_n\in R^n$ be a nonsingular set. Consider a simplex
$$
øV={\rm conv}\{A_0,A_1,\ldots,A_n\}.
$$
Let the simplex $V$ be intersected by a plane
$$
\Gamma=\{x\in R^n:\;(w,x)=t\},
$$
where $w\ne0$. Put
$$
\begin{array}{lll}
 V^-&=&\{x\in V:\;(w,x)\le t\},\\
 V^0&=&\{x\in V:\;(w,x)=t\},\\
 V^+&=&\{x\in V:\;(w,x)\ge t\}.
\end{array}
$$
We are going to construct the compatible triangulations
$$
\begin{array}{ccc}
T^0&\subset&T^+\medskip\\ \cap&&\cap\medskip\\ T^-&\subset& T\medskip\\
\end{array}
$$
of the polytopes
$$
\begin{array}{ccc}
V^0&\subset&V^+\medskip\\ \cap&&\cap\medskip\\ V^-&\subset& V\medskip\\
\end{array}
$$

A level $t$ we call regular if
$$
ø(w,A_{i})\ne t
$$
for all $0\le i\le n$ and singular if
$$
ø(w,A_{i})=t
$$
for some $0\le i\le n$. In the section we consider regular intersections.

Let $t$ be regular. We assume that
$$
(w,A_{i})\le (w,A_{j})
$$
for all $0\le i\le j\le n$. Then
$$
ø(w,A_{i-1})<t
$$
for $i=1,\ldots,p$ and
$$
ø(w,A_{n-j+1})<t
$$
for $j=1,\ldots,q$, where $p+q=n+1$. Define
$$
ø\begin{array}{lllll}
  øA_{i0} &=&A_{i-1}   & ø{\rm for} & ø1\le i\le p \\
  øA_{0j} &=&A_{n-j-1} & ø{\rm for} & ø1\le j\le q \\
\end{array}
$$

Let $G_{pq}$ be the graph
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In the nontrivial case $p\ge1$ and $q\ge1$ the graph $G_{pq}$ has two initial vertices
$1\times0$ and $0\times1$, and one final vertex $p\times q$. Hence the graph has
$C^p_{p+q-1}+C^q_{p+q-1}$ maximal paths.

Let $G^-_{pq}\subset G_{pq}$ be the graph with vertices
$$
i\times j:\;1\le i\le p\;\; {\rm and}\;\; 0\le j\le q,
$$
let $G^0_{pq}\subset G_{pq}$ be the graph with vertices
$$
i\times j:\;1\le i\le p\;\; {\rm and}\;\; 1\le j\le q,
$$
and let $G^+_{pq}\subset G_{pq}$ be the graph with vertices
$$
i\times j:\;0\le i\le p\;\; {\rm and}\;\; 1\le j\le q.
$$


Denote by $A_{ij}$, where $1\le i\le p$ and $1\le j\le q$, the point of intersection of
the interval $[A_{i0},A_{0j}]$ with the plane $\Gamma$. We see that
$$
A_{ij}=(1-\lambda)A_{i0}+\lambda A_{0j},
$$
where
$$
\lambda=\frac{t-(w,A_{i0})}{(w,A_{0j})-(w,A_{i0})}.
$$
Let the vertex $i\times j\in G_{pq}$ be associated with the point $A_{ij}\in V$.

We claim that the triangulations $T,\,T^{-},\,T^0,\,T^{+}$ associated with the graphs
$G_{pq},\, G^-_{pq},\,G^0_{pq},\,G^+_{pq}$ are the compatible triangulations of the
polytopes $V,\,V^-,\,V^0,\,V^+$.
\medskip

\noindent {\bf Examples.} For $p=2$ and $q=1$ we have
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 \put(230,0){\scriptsize $A_{10}$}
 \put(313,0){\scriptsize $A_{20}$}
 \put(275,95){\scriptsize $A_{01}$}
 \put(242,50){\scriptsize $A_{11}$}
 \put(305,50){\scriptsize $A_{21}$}

%\input p15aug2.tex
\end{picture}
\end{center}
Note the simplicial homeomorphisms
$$
\begin{array}{lll}
 V^-&\approx& \Delta^{1}\times\Delta^1\\
 V^0&\approx& \Delta^{1}\times\Delta^{0}\\
 V^+&\approx& \Delta^2\times\Delta^{0}\\
\end{array}
$$
For $p=2$ and $q=2$ we have
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 \put(213,27){\footnotesize $A_{10}$}
 \put(325,0){\footnotesize $A_{20}$}
 \put(384,55){\footnotesize $A_{01}$}
 \put(295,125){\footnotesize $A_{02}$}


 \put(252,84){\footnotesize $A_{12}$}
 \put(313,84){\footnotesize $A_{22}$}
 \put(355,28){\footnotesize $A_{21}$}
 \put(272,45){\footnotesize $A_{11}$}

%\input p16aug1.tex
\end{picture}
\end{center}
Note the simplicial homeomorphisms
$$
\begin{array}{lll}
 V^-&\approx& \Delta^{1}\times\Delta^2\\
 V^0&\approx& \Delta^{1}\times\Delta^{1}\\
 V^+&\approx& \Delta^2\times\Delta^{1}\\
\end{array}
$$
\medskip

Let
$$
ø\begin{array}{lll}
  øE^{n}_{1} & ø= & ø(1,0,0,\ldots,0) \\
  øE^{n}_{2} & ø= & ø(0,1,0,\ldots,0) \\
  &\cdots&\\
  øE^{n}_{n} & ø= & ø(0,0,0,\ldots,1) \\
\end{array}
$$
be unit vectors of $\mathbb{R}^{n}$.


\begin{theorem}
\label{simper} Let $t$ be regular. The triangulations $T,\,T^{-},\,T^0,\,T^{+}$
associated with the graphs $G_{pq},\, G^-_{pq},\,G^0_{pq},\,G^+_{pq}$ are the compatible
triangulations of the polytopes $V,\,V^-,\,V^0,\,V^+$. The mappings
$$
\begin{array}{llllll}
F_-:&E^p_i\times E^{q+1}_{j+1}&\longrightarrow&A_{ij} & for&1\le i\le p,\;0\le j\le
q\medskip\\ F_0:&E^p_i\times E^q_j&\longrightarrow&A_{ij} & for&1\le i\le p,\;1\le j\le
q\medskip\\ F_+:&E^{p+1}_{i+1}\times E^q_j&\longrightarrow&A_{ij} & for&0\le i\le
p,\;1\le j\le q\medskip\\
\end{array}
$$
are extended to the simplicial homeomorphisms
$$
\begin{array}{llll}
F_-:&\Delta^{p-1}\times\Delta^q&\longrightarrow&V^-\medskip\\
F_0:&\Delta^{p-1}\times\Delta^{q-1}&\longrightarrow&V^0\medskip\\
F_+:&\Delta^p\times\Delta^{q-1}&\longrightarrow&V^+\medskip\\
\end{array}
$$
\end{theorem}
{\bf Proof.} Let us prove the statement for the polytope $V^0$. Without loss of
generality it can be assumed that $t=0$. Let
$$
\begin{array}{llr}
ø   t_i&=&-(w,A_{i0}), \\
 \tau_j&=& (w,A_{0j}).
\end{array}
$$
Then $t_{i}>0$ and $\tau_{j}>0$. Obviously
$$
A_{ij}=\frac{\tau_j}{t_i+\tau_j}A_{i0}+\frac{t_i}{t_i+\tau_j}A_{0j}.
$$

Consider a maximal path
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
on the graph $G^0_{pq}$. We note that
$$
1=i_1\le\cdots\le i_n=p
$$
and
$$
1=j_1\le\cdots\le j_n=q,
$$
where
$
i_k+j_k=k+1
$
for all $1\le k\le n$. Let
$$
\begin{array}{lll}
 I(i)&=&\{k:\; i_k=i\},\\
 J(j)&=&\{k:\; j_k=j\},\\
\end{array}
$$
where $1\le i\le p$ and $1\le j\le q$.

We claim that the set
$
A_{i_1j_1},\ldots,A_{i_nj_n}
$
is nonsingular. Assume that
$$
\lambda_1A_{i_1j_1}+\cdots+\lambda_nA_{i_nj_n}=0
$$
and
$$
\lambda_1+\cdots+\lambda_n=0.
$$
Since the set $A_{0},A_{1},\ldots,A_{n}$ is nonsingular, we have
$$
\sum_{k\in I(i)}\frac{\tau_{j_k}}{t_{i_k}+\tau_{j_k}} \lambda_k=0
$$
for $1\le i\le p$ and
$$
\sum_{k\in J(j)} \frac{t_{i_k}}{t_{i_k}+\tau_{j_k}} \lambda_k=0
$$
for $1\le j\le q$. From the system of linear equations we obtain
$$
\lambda_1=\ldots=\lambda_n=0.
$$
This means that the set
$
A_{i_1j_1},\ldots,A_{i_nj_n}
$
is nonsingular.

Let $x\in V^0$. There exists a maximal path
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
such that
$$
x\in {\rm conv}\{A_{i_1j_1},\ldots,A_{i_nj_n}\}.
$$
Indeed, let
$$
x=\zeta_1A_{10}+\cdots+\zeta_pA_{p0}+\eta_1A_{01}+\cdots+\eta_qA_{0q}.
$$
Since $(w,x)=0$, we have
$$
\zeta_1t_1+\cdots+\zeta_pt_p=\eta_1\tau_1+\cdots+\eta_q\tau_q=r
$$
for some $r>0$. Let
$$
u_i=\frac{1}{r}t_i\zeta_i
$$
for $1\le i\le p$ and
$$
v_j=\frac{1}{r}\tau_j\eta_j
$$
for $1\le j\le q$. Then
$$
u\times v\in\Delta^{p-1}\times\Delta^{q-1}.
$$
By Theorem 1, there exists a maximal path
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
such that
$$
u\times v\in{\rm conv}(E^p_{i_1}\times E^q_{j_1},\ldots,E^p_{i_n}\times E^q_{j_n}).
$$
Then
$$
øu\times v=\sum_{k=1}^{n}\gamma_{k}E^p_{i_k}\times E^q_{j_k}
$$
for $\gamma_{1}\ge 0,\ldots,\gamma_{n}\ge 0$ and $\gamma_{1}+ \cdots+\gamma_{n}=1$.
Therefore,
$$
øx=\sum_{k=1}^{n}\lambda_{k}A_{i_kj_k},
$$
where
$$
ø\lambda_{k}=r\frac{t_{i_{k}}+\tau_{j_{k}}}{t_{i_{k}}\tau_{j_{k}}}\gamma_{k}.
$$
Indeed,
\begin{eqnarray*}
ø\sum_{k=1}^{n}\lambda_{k}A_{i_kj_k}
  &=&\sum_{k=1}^{n}r\frac{t_{i_{k}}+\tau_{j_{k}}}{t_{i_{k}}\tau_{j_{k}}}\gamma_{k}
    \left(\frac{\tau_{j_{k}}}{t_{i_{k}}+\tau_{j_{k}}}A_{i_{k}0}+
          \frac{t_{i_{k}}}{t_{i_{k}}+\tau_{j_{k}}}A_{0j_{k}}\right)\medskip\\
  &=&\sum_{k=1}^{n}\frac{r\gamma_{k}}{t_{i_{k}}}A_{i_{k}0}+
     \sum_{k=1}^{n}\frac{r\gamma_{k}}{\tau_{j_{k}}}A_{0j_{k}}\medskip\\
  &=&\sum_{i=1}^{p}\frac{r}{t_{i}}\left(\sum_{k\in I(i)}\gamma_{k}\right)A_{i0}+
     \sum_{j=1}^{q}\frac{r}{\tau_{i}}\left(\sum_{k\in J(j)}\gamma_{k}\right)A_{0j}\medskip\\
  &=&\sum_{i=1}^{p}\frac{r}{t_{i}}u_{i}A_{i0}+
     \sum_{j=1}^{q}\frac{r}{\tau_{i}}v_{j}A_{0j}\medskip\\
  &=&\sum_{i=1}^{p}\zeta_{i}A_{i0}+\sum_{j=1}^{q}\eta_{i}A_{0j}\;=\;x
\end{eqnarray*}
Analogously
\begin{eqnarray*}
ø\sum_{k=1}^{n}\lambda_{k}
  &=&\sum_{k=1}^{n}r\frac{t_{i_{k}}+\tau_{j_{k}}}{t_{i_{k}}\tau_{j_{k}}}\gamma_{k}\medskip\\
  &=&\sum_{k=1}^{n}\frac{r\gamma_{k}}{t_{i_{k}}}+
     \sum_{k=1}^{n}\frac{r\gamma_{k}}{\tau_{j_{k}}}\medskip\\
  &=&\sum_{i=1}^{p}\frac{r}{t_{i}}\left(\sum_{k\in I(i)}\gamma_{k}\right)+
     \sum_{j=1}^{q}\frac{r}{\tau_{i}}\left(\sum_{k\in J(j)}\gamma_{k}\right)\medskip\\
  &=&\sum_{i=1}^{p}\frac{r}{t_{i}}u_{i}+
     \sum_{j=1}^{q}\frac{r}{\tau_{i}}v_{j}\medskip\\
  &=&\sum_{i=1}^{p}\zeta_{i}+\sum_{j=1}^{q}\eta_{i}\;=\;1
\end{eqnarray*}

Consider two maximal paths
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
and
$$
u_1\times v_1,\;\ldots,\;u_n\times v_n.
$$
Let
$$
\sum_{k=1}^n\lambda_kA_{i_kj_k}= \sum_{k=1}^n\mu_kA_{u_kv_k}
$$
where
$$
\lambda_{1}\ge 0,\ldots,\lambda_{n}\ge 0,\; \mu_{1}\ge 0,\ldots,\mu_{n}\ge 0
$$
and
$$
ø\lambda_{1}+\cdots+\lambda_{n}=\mu_{1}+\cdots+\mu_{n}=1.
$$
We claim that
$$
\lambda_k=\mu_k
$$
in the case $i_k\times j_k=u_k\times v_k$ and
$$
\lambda_k=\mu_k=0
$$
in the case $i_k\times j_k\ne u_k\times v_k$.

Indeed, we see that
$$
\sum_{k\in I(i)} \frac{\tau_{j_k}}{t_{i_k}+\tau_{j_k}} \lambda_k= \sum_{k\in U(i)}
\frac{\tau_{v_k}}{t_{u_k}+\tau_{v_k}} \mu_k
$$
for $1\le i\le p$ and
$$
\sum_{k\in J(j)} \frac{t_{i_k}}{t_{i_k}+\tau_{j_k}} \lambda_k= \sum_{k\in V(j)}
\frac{t_{u_k}}{t_{u_k}+\tau_{v_k}} \mu_k
$$
¯à¨ $1\le j\le q$. Then we may continue as in Part 3 of the proof of Theorem 1.

Let
$$
P=\{i_1\times j_1,\;\ldots,\;i_n\times j_n\}
$$
be a maximal path, and let
$$
S(P)={\rm conv}\{A_{i_1j_1},\ldots,A_{i_nj_n}\}
$$
be the simplex corresponding to the path. Denote by $T^{0}$ the set consisting of all
simplices $S(P)$ with their faces. Then
$$
V^{0}=\bigcup_{\Delta\in T}\Delta.
$$
Obviously $T^{0}$ is the triangulation of $V^{0}$. Besides, the mapping
$$
F_0:\;\Delta^{p-1}\times\Delta^{q-1}\longrightarrow V^0
$$
is the simplicial homeomorphism.


Let us prove the statement about the compatible triangulations by induction on $n$. For
$n=1$ the statement is obvious. Assume that for $n-1$ the is statement true. By the
induction hypothesis, the statement is true for the simplex
$$
U={\rm conv}\{A_1,\ldots,A_n\}.
$$
Therefore the triangulation $\tilde T$ associated with the graph
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 \put(190,130){\circle*{3}}
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 \put(220,10){\circle*{3}}
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 \put(280,70){\circle*{3}}
 \put(280,100){\circle*{3}}
 \put(280,130){\circle*{3}}
 \put(280,160){\circle*{3}}



 \put(122,0){\scriptsize $2\times0$}
 \put(152,0){\scriptsize $3\times0$}
 \put(272,0){\scriptsize $p\times0$}

 \put(77,37){\scriptsize $1\times1$}
 \put(77,67){\scriptsize $1\times2$}
 \put(77,158){\scriptsize $1\times q$}

 \put(285,164){\scriptsize $p\times q$}

%\input p19aug.tex
\end{picture}
\end{center}
is a triangulation of the polytope $V_{0}\cup U_{-}$. Hence the triangulation $T^{-}$
associated with the graph $G_{pq}^{-}$ of the form
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 \put(90,0){\scriptsize $1\times0$}
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 \put(272,0){\scriptsize $p\times0$}

 \put(77,37){\scriptsize $1\times1$}
 \put(77,67){\scriptsize $1\times2$}
 \put(77,158){\scriptsize $1\times q$}

 \put(285,164){\scriptsize $p\times q$}

%\input p19aug1.tex
\end{picture}
\end{center}
is a tiangulation of the polytope $V^{-}$. The triangulation $T^{-}$ consists of all
simplices
$$
ø\Delta={\rm conv}\{A_{10},\Gamma\},
$$
where $\Gamma\in \tilde T$, and their faces. The construction of the triangulation
$T^{-}$ is shown in the picture
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 \put(130,0){\scriptsize $A_{10}$}
 \put(213,0){\scriptsize $A_{20}$}
 \put(175,95){\scriptsize $A_{01}$}
 \put(144,50){\scriptsize $A_{11}$}
 \put(205,50){\scriptsize $A_{21}$}
 \put(175,55){\scriptsize $V_0$}
 \put(214,31){\scriptsize $U_-$}

%\input pis1zx.tex
\end{picture}
\end{center}
We analogously prove that the triangulation $T^{+}$ associated with the graph
$G_{pq}^{+}$ is a tiangulation of the polytope $V^{+}$. $\bullet$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{4. Singular intersections}

We consider the same simplex
$$
øV={\rm conv}\{A_{0},A_1,\ldots,A_n\}
$$
as in the previous section. In this section we denote the plane in the form
$$
\Gamma_{t}=\{x\in R^n:\;(w,x)=t\}
$$
and the polytopes in the form
$$
\begin{array}{lll}
 V^-_{t}&=&\{x\in V:\;(w,x)\le t\},\\
 V^0_{t}&=&\{x\in V:\;(w,x)=t\},\\
 V^+_{t}&=&\{x\in V:\;(w,x)\ge t\}.
\end{array}
$$

Let $t^{0}$ be a fixed singular level. We assume that
$$
(w,A_{i})\le (w,A_{j})
$$
for all $0\le i\le j\le n$. Then
$$
ø(w,A_{i-1})<t^{0}
$$
for $i=1,\ldots,p$,
$$
ø(w,A_{n-j+1})>t^{0}
$$
for $j=1,\ldots,q$, and
$$
ø(w,A_{n-q-k})=t^{0}
$$
for $k=1,\ldots,r$, where $p+q+r=n+1$. Define
$$
ø\begin{array}{lllll}
  øA_{i0} &=&A_{i-1}   & ø{\rm for} & ø1\le i\le p \\
  øA_{0j} &=&A_{n-j-1} & ø{\rm for} & ø1\le j\le q \\
  øA_{pk} &=&A_{n-q-k} & ø{\rm for} & ø1\le k\le r \\
\end{array}
$$

Let $G_{pqr}$ be the graph
\begin{center}
\begin{picture}(360,170)

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\put(122,0){\scriptsize $1\times0$} \put(152,0){\scriptsize $2\times0$}
\put(272,0){\scriptsize $p\times0$}

\put(77,37){\scriptsize $0\times1$} \put(77,98){\scriptsize $0\times q$}

\put(285,98){\scriptsize $p\times q$}

\put(285,164){\scriptsize $p\times (q+r)$}

%\input pis2z.tex
\end{picture}
\end{center}
To obtain the graph $G_{pqr}$ we add vertices
$$
p\times (q+1),p\times (q+2),\ldots,p\times (q+r)
$$
and corresponding vertical edges to the graph $G_{pq}$. Analogously we define the graphs
$G^-_{pqr},G^0_{pqr}$ and $G^+_{pqr}$.

Denote by $A_{ij}$, where $1\le i\le p$ and $1\le j\le q$, the point of intersection of
the interval $[A_{i0},A_{0j}]$ with the plane $\Gamma_{t^{0}}$. Let the vertex $i\times
j\in G_{pqr}$ be associated with the point $A_{ij}\in V$.

\begin{theorem}
\label{simsing} Let $t^{0}$ be singular. The triangulations associated with graphs
$G_{pqr},G^-_{pqr},G^0_{pqr},G^+_{pqr}$ are the compatible triangulations of the
polytopes $V,\,V^-_{t^{0}},\,V^0_{t^{0}},\,V^+_{t^{0}}$. The mappings
$$
\begin{array}{ll}
F_-:& \left\{\begin{array}{lllll} E^p_i\times E^{q+1}_{j+1}\times
0^r&\longrightarrow&A_{ij} &for&1\le i\le p,\;0\le j\le q\medskip\\ 0^p\times
0^{q+1}\times E^r_k&\longrightarrow&A_{p,q+k} &for&1\le k\le r\medskip\\
\end{array}\right.\medskip\\
F_0:& \left\{\begin{array}{lllll} E^p_i\times E^q_j\times 0^r&\longrightarrow&A_{ij}
&for&1\le i\le p,\;1\le j\le q\medskip\\ 0^p\times 0^q\times
E^r_k&\longrightarrow&A_{p,q+k} &for&1\le k\le r\medskip\\
\end{array}\right.\medskip\\
F_+:& \left\{\begin{array}{lllll} E^{p+1}_{i+1}\times E^q_j\times
0^r&\longrightarrow&A_{ij} &for&0\le i\le p,\;1\le j\le q\medskip\\ 0^{p+1}\times
0^q\times E^r_k&\longrightarrow&A_{p,q+k} &for&1\le k\le r\medskip\\
\end{array}\right.\medskip\\
\end{array}
$$
are extanded to the simplicial homeomorphisms
$$
\begin{array}{llll}
F_-:&(\Delta^{p-1}\times\Delta^q)*\Delta^{r-1}&\longrightarrow&V^-_{t^{0}}
\medskip\\
F_0:&(\Delta^{p-1}\times\Delta^{q-1})*\Delta^{r-1}&\longrightarrow&V^0_{t^{0}}
\medskip\\
F_+:&(\Delta^p\times\Delta^{q-1})*\Delta^{r-1}&\longrightarrow&V^+_{t^{0}}
\medskip\\
\end{array}
$$
\end{theorem}
{\bf Proof.} Let us prove the statement for the polytope $V^{0}_{t^{0}}$. Without loss of
generality it can be assumed that $t^{0}=0$. Let
$$
\begin{array}{llr}
ø   t_i&=&-(w,A_{i0}), \\
 \tau_j&=& (w,A_{0j}),
\end{array}
$$
where $1\le i\le p$ and $1\le j\le q$. Then $t_{i}>0$ and $\tau_{j}>0$. Obviously
$$
A_{ij}=\frac{\tau_j}{t_i+\tau_j}A_{i0}+\frac{t_i}{t_i+\tau_j}A_{0j}
$$
where $1\le i\le p$ and $1\le j\le q$.

Consider a maximal path
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
on the graph $G^0_{pqr}$. Let $m=p+q$. We note that
$$
1=i_1\le\cdots\le i_m=p
$$
and
$$
1=j_1\le\cdots\le j_m=q,
$$
where
$
i_k+j_k=k+1
$
for all $1\le k\le m$. Let
$$
\begin{array}{lll}
 I(i)&=&\{k:\; i_k=i\},\\
 J(j)&=&\{k:\; j_k=j\},\\
\end{array}
$$
where $1\le i\le p$ and $1\le j\le q$.

We claim that the set
$
A_{i_1j_1},\ldots,A_{i_nj_n}
$
is nonsingular. Assume that
$$
\lambda_1A_{i_1j_1}+\cdots+\lambda_nA_{i_nj_n}=0
$$
and
$$
\lambda_1+\cdots+\lambda_n=0.
$$
Since the set $A_{0},A_{1},\ldots,A_{n}$ is nonsingular, we have
$$
\sum_{k\in I(i)}\frac{\tau_{j_k}}{t_{i_k}+\tau_{j_k}} \lambda_k=0
$$
for $1\le i\le p$,
$$
\sum_{k\in J(j)} \frac{t_{i_k}}{t_{i_k}+\tau_{j_k}} \lambda_k=0
$$
for $1\le j\le q$, and
$$
ø\lambda_{k}=0
$$
for $m+1\le k\le n$. From the system of linear equations we obtain
$$
\lambda_1=\ldots=\lambda_n=0.
$$
This means that the set
$
A_{i_1j_1},\ldots,A_{i_nj_n}
$
is nonsingular.

Let $x\in V^0_{t^{0}}$. There exists a maximal path
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
such that
$$
x\in {\rm conv}\{A_{i_1j_1},\ldots,A_{i_nj_n}\}.
$$
Indeed, let
$$
x=\zeta_1A_{10}+\cdots+\zeta_pA_{p0}+\eta_1A_{01}+\cdots+\eta_qA_{0q}+
 \theta_1A_{p1}+\cdots+\theta_rA_{pr}.
$$
Since
$$
(w,x)=(w,A_{p1})=\ldots=(w,A_{pr})=0
$$
we have
$$
\zeta_1t_1+\cdots+\zeta_pt_p=\eta_1\tau_1+\cdots+\eta_q\tau_q=r
$$
for some $r>0$. Then we continue as in the proof of Theorem 2.

Consider two maximal paths
$$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$$
and
$$
u_1\times v_1,\;\ldots,\;u_n\times v_n.
$$
Let
$$
\sum_{k=1}^n\lambda_kA_{i_kj_k}= \sum_{k=1}^n\mu_kA_{u_kv_k}
$$
where
$$
\lambda_{1}\ge 0,\ldots,\lambda_{n}\ge 0,\; \mu_{1}\ge 0,\ldots,\mu_{n}\ge 0
$$
and
$$
ø\lambda_{1}+\cdots+\lambda_{n}=\mu_{1}+\cdots+\mu_{n}=1.
$$
We see that
$$
\lambda_k=\mu_k
$$
in the case $i_k\times j_k=u_k\times v_k$ and
$$
\lambda_k=\mu_k=0
$$
in the case $i_k\times j_k\ne u_k\times v_k$.

Let
$$
P=\{i_1\times j_1,\;\ldots,\;i_n\times j_n\}
$$
be a maximal path, and let
$$
S(P)={\rm conv}\{A_{i_1j_1},\ldots,A_{i_nj_n}\}
$$
be the simplex corresponding to the path. Denote by $T^{0}$ the set consisting of all
simplices $S(P)$ with their faces. Then
$$
V^{0}_{t^{0}}=\bigcup_{\Delta\in T}\Delta.
$$
Obviously $T^{0}$ is the triangulation of $V^{0}_{t^{0}}$. Besides, the mapping
$$
F_0:\;(\Delta^{p-1}\times\Delta^{q-1})*\Delta^{r-1}\longrightarrow V^0_{t^{0}}
$$
is the simplicial homeomorphism. $\bullet$
\medskip

\noindent{\bf Remark.} The last theorem may be proved by continuity. For example,
consider the polytope $V^{0}_{t_{0}}$. By continuity, there exists $\varepsilon>0$ such
that $t$ is regular whenever $t_0-\varepsilon<t<t_0$. By Theorem 2, the polytope
$V^{0}_{t}$ has a triangulation associated with the graph $G_{p,q+r}$.
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%\input p20aug.tex
\end{picture}
\end{center}
We note that $G_{p,q,r}^{0}$ is a subgraph of $G_{p,q+r}^{0}$. Obviously
$$
ø\lim_{t\to t_0}A_{ij}^{t}=A_{ij}
$$
for all $1\le i\le p$ and $1\le j\le q+r$. Consider a maximal path
$
i_1\times j_1,\;\ldots,\;i_n\times j_n
$
on the graph $G_{p,q+r}^{0}$. By Theorem 2, the set
$$
A_{i_{1},j_{1}}^{t},\;\ldots,A_{i_{n},j_{n}}^{t}
$$
is nonsingular. Obviously
$$
ø\lim_{t\to t_0}(A_{i_{1},j_{1}}^{t},\;\ldots,A_{i_{n},j_{n}}^{t})=
 (A_{i_{1},j_{1}},\;\ldots,A_{i_{n},j_{n}}).
$$
Let $m$ be a maximal number such that $j_{k}=q$. The set
$$
øA_{i_{1},j_{1}},\;\ldots,A_{i_{m},j_{m}},\;A_{p,q+1},\;\ldots,A_{p,q+r}
$$
is nonsingular and
$$
ø{\rm conv}\{A_{i_{1},j_{1}},\;\ldots,A_{i_{n},j_{n}}\}={\rm
conv}\{A_{i_{1},j_{1}},\;\ldots,A_{i_{m},j_{m}},\;A_{p,q+1},\;\ldots,A_{p,q+r}\}.
$$
We see that the right simplex belongs to the triangulation of the polytope
$V^{0}_{t^{0}}$.
\bigskip

\noindent{\bf Example.} Let
$$
V={\rm conv}\{A_{0},A_{1},A_{2},A_{3}\}
$$
be a three-dimensional simplex
\begin{center}
\begin{picture}(400,130)



 \put(116,27){\footnotesize $A_0$}
 \put(225,0){\footnotesize $A_1$}
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 \put(195,125){\footnotesize $A_3$}


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\end{center}
and let
$
t_{0}<t_{1}<t_{2}<t_{3}
$
be singular levels. When the level $t$ increases from $t_{0}$ to $t_{3}$, we see the
following graphs $G_{p,q}$ and $G_{p,q,r}$.
\begin{center}
\begin{picture}(410,110)
\input p22aug.tex
\end{picture}
\end{center}
Here $\circ$ is the origin.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{5. Cylinders of simplicial mappings}

Let $V$ be a simplex of dimension $n$, let $S$ be a simplex of dimension $k$, and let
$$
F:\;V\longrightarrow S
$$
be a simplicial surjection. Recall the notion of the cylinder ${\rm Cyl} F$ of the
mapping $F$. Let the equivalence relation $\sim$ in the cylinder ${\rm cyl} V=V\times
[0,1]$ have the following classes of equivalence $[x\times t]$. If $0<t\le 1$, then
$[x\times t]=x\times t$. If $t=0$, then $[x\times 0]=F^{-1}\{F(x)\}\times 0$. By
definition, the cylinder ${\rm Cyl} F$ of the mapping $F$ is the factor-space ${\rm cyl}
V/\sim$.

We are going to prove that the cylinder ${\rm Cyl} F$ is homeomorphic to the ball
$D^{n+1}$. Namely, we prove that the cylinder ${\rm Cyl} F$ is homeomorphic to the join
${\rm cyl}\Delta^{k}\,*\,\Delta^{n-k-1}$. In particular, the cylinder ${\rm Cyl} F$ has a
natural triangulation associated with the graph
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%\input p23aug.tex
\end{picture}
\end{center}
The triangulation has $k+1$ simplices of dimension $n+1$.

In the cases $k=n$ and $k=0$ the statement is obvious. In the case $k=n$ the mapping $F$
is the simplicial homeomorphism. Hence
$$
ø{\rm Cyl}F\approx\Delta^{n}\times[0,1].
$$
Therefore ${\rm Cyl} F$ has a triangulation with $n+1$ simplicies of dimension $n+1$. In
the case $k=0$ the image of the mapping $F$ is a point. Hence
$$
ø{\rm Cyl}F\approx\Delta^{n+1}.
$$
Therefore ${\rm Cyl} F$ has a triangulation with $1$ simplex of dimension $n+1$.
\medskip

\noindent{\bf Example.} Let $n=2$. The cases $k=2,1,0$ are shown in the pictures
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 \put(315,0){\footnotesize ${\rm cyl}\,\Delta^{0}*\Delta^{1}$}

%\input pq13.tex
\end{picture}
\end{center}
In the first picture we see the cylinder ${\rm cyl}\Delta^{2}$. In the second picture we
see the join of the cylinder ${\rm cyl}\Delta^{1}$ and the simplex $\Delta^{0}$. In the
third picture we see the join of the cylinder ${\rm cyl}\Delta^{0}$ and the simplex
$\Delta^{1}$. The triangulations are associated with the graphs
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 \put(301,45){\scriptsize $0\times1$}
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%\input p23aug1.tex
\end{picture}
\end{center}


\begin{theorem} The cylinder ${\rm Cyl} F$ of the simplicial surjection
$$
F:\;V\longrightarrow S
$$
is homeomorphic to the join ${\rm cyl}\Delta^{k}\,*\,\Delta^{n-k-1}$.
\end{theorem}
{\bf Proof.} We construct a convex $(n+1)$-dimensional polytope $M$ and a continuous
surjection
$$
\Phi:\;{\rm cyl}V\longrightarrow M
$$
such that
$$
{\rm Cyl} F={\rm cyl}V/{\rm zer}\Phi.
$$
Here ${\rm zer}\Phi$ is the equivalence relation in the cylinder ${\rm cyl} V$ such that
$x\times t$ is equivalent to $x'\times t'$ if and only if $\Phi(x\times t)=\Phi(x'\times
t')$. Then the proof of the theorem follows from the homeomorphism
$$
{\rm cyl}V/{\rm zer}\Phi\approx M.
$$
The representation
$$
M={\rm cyl}\Delta^{k}\,*\,\Delta^{n-k-1}
$$
is obtained automatically.


We may assume that
$
S\subset V\subset \mathbb{R}^n.
$
Let $p=k+1$, and let $A_{10},\ldots,A_{p0}$ be the vertices of the simplex $S$. Let
$p+q=n+1$, and let $A_{10},\ldots,A_{p0},A_{01},\ldots,A_{0q}$ be the vertices of the
simplex $V$. We may assume that
$$
øF(A_{i0})=A_{i0}
$$
for $1\le i\le p$ and
$$
øF(A_{0j})=A_{\nu(j)0}
$$
for $1\le j\le q$, where
$$
\nu:\;\{1,2,\ldots,q\}\longrightarrow\{1,2,\ldots,p\}.
$$
Let us consider the polytope
$$
M={\rm conv}\{A_{10}\times 0,\ldots,A_{p0}\times 0;\;A_{10}\times 1,\ldots,A_{p0}\times
1;\;A_{01}\times1,\ldots,A_{0q}\times1\}.
$$
The polytope $M$ is shown in the pucture
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 \put(185,5){\footnotesize $S$}
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%\input p23aug2.tex
\end{picture}
\end{center}
The polytope $M$ has a triangulation with $k+1$ simplices of dimension $n+1$. The
simplices have the form
$$
{\rm conv}\{A_{10}\times 0,\ldots,A_{i0}\times 0,A_{i0}\times 1,\ldots, A_{p0}\times
1,A_{01}\times1,\ldots,A_{0q}\times1\},
$$
where $1\le i\le p$. This means that the triangulation is associated with the graph.

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%\input p23aug.tex
\end{picture}
\end{center}
In particular, the triangulation of the polytope $M$ is the same as the triangulation of
the join ${\rm cyl}\Delta^{k}\,*\,\Delta^{n-k-1}$.

Let $w\in \mathbb{R}^n$ be any vector such that
$$
(w,A_{i0}-A_{10})=0
$$
for $1\le i\le p$,
$$
(w,A_{0j}-A_{01})=0
$$
for $1\le j\le q$, and
$$
(w,A_{10})<(w,A_{01}).
$$
Consider the plane
$$
\Gamma_{t}=\{x\in R^n:\;(w,x)=(1-t)(w,A_{10})+t(w,A_{01})\}
$$
and the polytopes
\begin{eqnarray*}
øV_{t}^{-}&=&\{x\in V:\;(w,x)\le(1-t)(w,A_{10})+t(w,A_{01})\},\medskip\\ ø
 V_{t}^{+}&=&\{x\in V:\;(w,x)\ge(1-t)(w,A_{10})+t(w,A_{01})\}.
\end{eqnarray*}
We see that $V_{0}^{-}=S$ and $V_{1}^{-}=V$

Denote by $A_{ij}^{t}$, where $1<t<1$, the point of intersection of the interval
$[A_{i0},A_{0j}]$ with the plane $\Gamma_{t}$. It is clear that
$$
A_{ij}^{t}=(1-t)A_{i0}+tA_{0j}.
$$
Let
$$
B_{ij}^{t}=(1-t^2)A_{i0}+t^2A_{0j}.
$$
We note that
$$
ø\frac{\|B_{ij}^{t}-A_{i0}\|}{\|A_{ij}^{t}-A_{i0}\|}=
 \frac{\|A_{ij}^{t}-A_{i0}\|}{\|A_{0j}-A_{i0}\|}.
$$
Using the compatible triangulations of the polytopes $V^{-}_{t},V^{+}_{t},V$, define the
simplicial mapping
$$
\varphi_{t}:\;V\longrightarrow V^{-}_{t}
$$
by the rule
$$
\begin{array}{lllll}
 A_{i0}&\longmapsto& A_{i0}&&\medskip\\
 A_{0j}&\longmapsto& A_{\nu(j)j}^{t}&&\medskip\\
 A_{ij}^{t}&\longmapsto&A_{ij}^{t}&{\rm for}&i\ne\nu(j)\medskip\\
 A_{ij}^{t}&\longmapsto& B_{ij}^{t}&{\rm for}&i=\nu(j)\\
\end{array}
$$
If $0<t<1$, then $\varphi_{t}$ is a homeomorphism. Let $\varphi_{1}={\rm id}$ and
$\varphi_{0}=F$, where ${\rm id}$ is the identical mapping.

Example. Let $n=2$ and $k=1$. Let the mapping
$$
F:\;V\longrightarrow S
$$
be such that
$$
ø\begin{array}{ccc}
  øA_{10} & ø\longmapsto & øA_{10}, \\
  øA_{20} & ø\longmapsto & øA_{20}, \\
  øA_{01} & ø\longmapsto & øA_{10}. \\
\end{array}
$$
Then
$$
\nu:\;1\longmapsto1
$$
and the homeomorphism $\varphi_{t}$ is such that
$$
ø\begin{array}{lll}
  øA_{10} & ø\longmapsto & øA_{10}, \\
  øA_{20} & ø\longmapsto & øA_{20}, \\
  øA_{01} & ø\longmapsto & øA_{11}^{t}, \\
  øA_{21}^{t} & ø\longmapsto & øA_{21}^{t}, \\
  øA_{11}^{t} & ø\longmapsto & øB_{11}^{t}. \\
\end{array}
$$
The homeomorphism $\varphi_{t}$ is shown in the picture
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 \put(217,30){\footnotesize $B_{11}^{t}$}


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%\input pq15.tex
\end{picture}
\end{center}

We see that $x\times t\in M$ if and only if $x\in V^{-}_{t}$ and $0\le t\le 1$. Define
the surjection
$$
\Phi:\;{\rm cyl} V\longrightarrow M
$$
by the rule
$$
\Phi:\;x\times t\longmapsto \varphi_{t}(x)\times t.
$$
We claim that $\Phi$ is continuous. To prove this we use the following statement. Let
$X,Y,Z$ be topologycal spaces and let
$$
\begin{array}{c}
 f:\;X\longrightarrow Y\\
 h:\;Y\longrightarrow Z\\
\end{array}
$$
be surjections. Let $X$ be compact and let $Y$ be Hausdorff. If $f$ and $g=h\circ f$ are
continuous, then $h$ is continuous.

Let
$$
P=\{i_0\times j_0,\;i_1\times j_1,\;\ldots,\;i_n\times j_n\}
$$
be a maximal path on the graph $G_{pq}$. Define the continuous mapping
$$
f_{_{P}}:\;{\rm cyl}\Delta^n\longrightarrow{\rm cyl} V
$$
by the rule
$$
f_{_{P}}:\;\lambda\times t\longmapsto\mu\times t,
$$
where
$$
\mu=\lambda_0A_{i_0j_0}^{t}+\cdots+\lambda_{n}A_{i_{n}j_{n}}^{t}.
$$
Let
$$
C_{ij}^{t}=\left\{ø\begin{array}{lll}
  øA_{ij}^{t} & ø{\rm for} & øi\ne\nu(j)\medskip \\
  øB_{ij}^{t} & ø{\rm for} & øi=\nu(j)
\end{array}\right.
$$
Define the continuous mapping
$$
g_{_{P}}:\;{\rm cyl}\Delta^n\longrightarrow M
$$
by the rule
$$
g_{_{P}}:\;\lambda\times t\longmapsto\mu\times t,
$$
where
$$
\mu=\lambda_0C_{i_0j_0}^{t}+\cdots+\lambda_{n}C_{i_{n}j_{n}}^{t}.
$$
We see that
$$
g_{_{P}}=\Phi\circ f_{_{P}}.
$$
By the above statement, the mapping
$$
\Phi|_{f_{_{P}}\{{\rm cyl}\Delta^n\}}:\;f_{_{P}}\{{\rm cyl}\Delta^n\}\longrightarrow M.
$$
is continuous.  Since
$$
ø{\rm cyl} V=\bigcup_{P}f_{_{P}}\{{\rm cyl}\Delta^n\},
$$
the mapping $\Phi$ is continuous.


Let
$$
{\rm fact}\Phi:\;{\rm cyl} V/{\rm zer}\Phi\longrightarrow M
$$
be the mapping such that
$$
ø{\rm fact}\Phi:\;[x\times t]\longmapsto \Phi(x\times t).
$$
Since ${\rm cyl} V$ is compact and since $M$ is Hausdorff, the mapping ${\rm fact}\Phi$
is a homeomorphism. We see that
$$
{\rm cyl} V/{\rm zer}\Phi={\rm Cyl} F.
$$
Hence ${\rm Cyl} F\approx M$. $\bullet$




\begin{thebibliography}{9}

\bibitem{Rokh1}
V.A.Rokhlin, D.B.Fuchs. An innitial course in topology. Moscow, "Nauka", 1977.

\end{thebibliography}


\end{document}
