text:
\section{Introduction}\label{IN}

Let $\mathbb{R}^n$ be the $n$-dimensional Euclidean space; thus,
a point $x\in \mathbb{R}^n$ is $n$-tuple $x=(x_1,\ldots,x_n)$ of real numbers.
Let $d : \mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}$ be the
$n$-dimensional Euclidean distance,
\[d(x,y)=\sqrt{\sum_{i=1}^n(x_i-y_i)^2},\]
where $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$.
Let $p$ be a given point in $\mathbb{R}^n$.
The mapping
$d_p : \mathbb{R}^n\rightarrow \mathbb{R}$, defined by $d_p(x)=d(p,x)$, is called a {\it distance function}.
\begin{defi}
{\rm Let $p_1,\ldots,p_\ell$ $(\ell\geq 1)$ be $\ell$ given points in $\mathbb{R}^n$.
The mapping
$d_{(p_1,\ldots,p_\ell)} : \mathbb{R}^n \rightarrow \mathbb{R}^\ell$ defined by
\[
d_{(p_1,\ldots,p_\ell)}(x)=(d(p_1,x),\ldots,d(p_\ell,x)),
\]
is called a {\it distance mapping}.}
\end{defi}
A distance mapping is one in which each component is a distance function.

Let $S^n$ be the $n$-dimensional unit sphere in $\mathbb{R}^{n+1}$.
\begin{pro} \label{d}
Let $i : S^1\rightarrow i(S^1)\subset \mathbb{R}^2$ be a homeomorphism.
Then, there exist two points $p_1$,$p_2$ $\in i(S^1)$ such that $d_{(p_1,p_2)}\circ i $ is a homeomorphism to
its image $\left(d_{(p_1,p_2)}\circ i\right)(S^1)$.
\end{pro}
\begin{figure}
\begin{center}
\includegraphics[width=6cm,clip]{Rhomeo.eps}
\caption{Figure of a mapping $i: S^1\to \mathbb{R}^2$ for which Proposition \ref{d} can be applied.}
\end{center}
\end{figure}\label{figure 1}
Proposition \ref{d} is proved in Section \ref{dP}.
Although Proposition \ref{d} is applicable even if a mapping $i$ is not differentiable
(see Figure \ref{figure 1}), it appears difficult to derive higher-dimensional extensions of the proposition.

On the other hand, the differentiable version of higher-dimensional extensions can be obtained as follows.
All manifolds and mappings in this paper belong to class $C^\infty $ except in Proposition \ref{d}
and the homeomorphism $\sqrt{\quad}: \mathbb{R}^\ell_+\to \mathbb{R}^\ell_+$ defined below.
\begin{defi}
{\rm Let $p_1,\ldots,p_\ell$ $(\ell\geq 1)$ be $\ell$ given points in $\mathbb{R}^n$.
The mapping $D_{(p_1,\ldots,p_\ell)} : \mathbb{R}^n \rightarrow \mathbb{R}^\ell$
defined by
\[
D_{(p_1,\ldots,p_\ell)}(x)=(d^2(p_1,x),\ldots,d^2(p_\ell,x)),
\]
is called a {\it distance-squared mapping}}.
\end{defi}
Note that $D_{(p_1,\ldots,p_\ell)}$ always has a singular point if $\ell \le n$.
Nevertheless, Theorems \ref{D1} and \ref{D2} below hold.
\begin{theo} \label{D1}
Let $M$ be an $m$-dimensional closed manifold $(m\geq 1)$, and let $i : M\to \mathbb{R}^\ell$
$(m+1\leq \ell)$ be an embedding. Then, there exist $p_1,\ldots,p_{m+1}\in i(M)$, $p_{m+2},\ldots,p_\ell\in \mathbb{R}^\ell $such that $D_{(p_1,\ldots,p_\ell)}\circ i : M\rightarrow \mathbb{R}^\ell$ is an embedding.
\end{theo}
\begin{coro}
Let $M$ be an $m$-dimensional closed manifold $(m\geq 1)$, and let $i : M\rightarrow \mathbb{R}^{m+1}$ be an embedding. Then, there exist $p_1,\ldots,p_{m+1}\in i(M)$
such that $D_{(p_1,\ldots,p_{m+1})}\circ i : M\rightarrow \mathbb{R}^\ell$ is an embedding.
\end{coro}
Define $\mathbb{R}_+^\ell=\{(x_1, \ldots, x_\ell)\; |\; x_i\ge 0\; (1\le i\le \ell)\}$ and let
$\sqrt{\quad} : \mathbb{R}_+^\ell\to \mathbb{R}_+^\ell$ be the homeomorphism defined by
$\sqrt{\quad}(x_1, \ldots, x_\ell)=(\sqrt{x_1}, \ldots, \sqrt{x_\ell})$.   It is clear that
$d_{(p_1,\ldots,p_\ell)}=\sqrt{\quad}\circ D_{(p_1,\ldots,p_\ell)}$.    Thus, the following corollary holds:
\begin{coro}
Let $M$ be an $m$-dimensional closed manifold $(m\geq 1)$, and let $i : M\to \mathbb{R}^\ell$
$(m+1\leq \ell)$ be an embedding.
Then, there exist $p_1,\ldots,p_{m+1}\in i(M)$, $p_{m+2},\ldots,p_\ell\in \mathbb{R}^\ell $such that $d_{(p_1,\ldots,p_\ell)}\circ i : M\rightarrow
\left(d_{(p_1,\ldots,p_\ell)}\circ i\right)(M)\subset \mathbb{R}^\ell$ is a homeomorphism.
\end{coro} \label{corollary 1.2}

Let $M$ and $N$ be manifolds. An immersion $f : M\rightarrow N$
is said to be
{\it with normal crossings at a point $y\in N$} if $f^{-1}(y)$ is a finite set $\{x_1,x_2,\ldots,x_k\}$ and for any subsets $\{\lambda _1,\lambda _2,\ldots,\lambda _s\}\subset
\{1,2,\ldots,k\}$ $(s\leq k)$,
\[{\rm codim}\,\Bigl(\bigcap _{j=1}^sdf_{x_{\lambda _{j}}}(T_{x_{\lambda _{j}}}M)\Bigr)=\sum_{j=1}^s{\rm codim}\,(df_{x_{\lambda _{j}}}(T_{x_{\lambda _{j}}}M)),\]
where ${\rm codim}\,H={\rm dim}\,T_yN-{\rm dim}\,H$ for any given linear subspace $H$ of $T_yN$.
An immersion $f : M\rightarrow N$ is said to be {\it with normal crossings} if $f$ is a mapping with normal crossings
at any point $y\in N$.
\begin{theo} \label{D2}
Let $M$ be an $m$-dimensional closed manifold $(m\geq 1)$, and let $i : M\rightarrow \mathbb{R}^\ell$ $(m+1\leq \ell)$ be an immersion with normal crossings. Then, there exist $p_1,\ldots,p_{m+1}\in i(M)$, $p_{m+2},\ldots,p_\ell\in \mathbb{R}^\ell$
such that $D_{(p_1,\ldots,p_\ell)}\circ i : M\rightarrow \mathbb{R}^\ell$ is an immersion with normal crossings.
\end{theo}
\begin{coro}
Let $M$ be an $m$-dimensional closed manifold $(m\geq 1)$, and let
$i : M\rightarrow \mathbb{R}^{m+1}$ be an immersion with normal crossings. Then, there exist $p_1,\ldots,p_{m+1}\in i(M)$ such that $D_{(p_1,\ldots,p_{m+1})}\circ i : M\rightarrow \mathbb{R}^{m+1}$ is an immersion with normal crossings.
\end{coro}

$L$-points $p_1,\ldots,p_\ell\in  \R^n$ $(1\leq \ell\leq n+1)$ are said to be {\it in general position}  if $\ell=1$ or
$\overrightarrow{p_1p_2},\ldots,\overrightarrow{p_1p_\ell}$ $(2\leq \ell\leq n+1)$ are linearly independent.
\par
A mapping $f : \R^n\rightarrow \R^\ell$ is said to be {\it $\mathcal{A}$-equivalent} to a mapping $g : \R^n\rightarrow \R^\ell$
if there exist diffeomorphisms $\varphi : \R^n\rightarrow \R^n$ and $\psi : \R^\ell\rightarrow \R^\ell$
such that $\psi \circ f\circ \varphi^{-1} =g$.
\par
A mapping $f : \R^n\rightarrow \R^\ell$ $(\ell\leq n)$ is called
{\it the normal form of definite fold mappings} if
$f(x_1,\ldots,x_n)=(x_1,\ldots,x_{\ell-1}, x^2_\ell+\cdots +x^2_n)$.
\par
The properties of distance-squared mappings are of significance in the proofs of Theorems
\ref{D1} and \ref{D2}.
It turns out that distance-squared mappings are $\mathcal{A}$-equivalent to the normal form of
definite fold mappings if $\ell$-points $p_1,\ldots,p_\ell \in \R^n$ $(2\leq \ell\leq n)$ are in general position
(see Proposition \ref{P}).
\medskip

In Section \ref{SP}, several preliminaries for the proofs of
Theorems \ref{D1}, \ref{D2} and Proposition \ref{d} are given.
The proofs of Theorems \ref{D1} and \ref{D2} are given in Section \ref{SD}.
Finally, in Section \ref{dP}, the proof of Proposition \ref{d} is given for the sake of readers' convenience.
\section{Preliminaries}\label{SP}
\subsection{Preliminaries for the proofs of Theorems \ref{D1} and \ref{D2}}
\begin{pro}\label{P}
\ \\
$(I)$\ Let $\ell$,$n$ be integers such that $2\leq \ell\leq n$, and let $p_1,\ldots,p_\ell\in \mathbb{R}^n$ be in general position. Then, $D_{(p_1,\ldots,p_\ell)} : \R^n\rightarrow \R^\ell$ is $\mathcal{A}$-equivalent
to the normal form of definite fold mappings.\\
$(II)$\ Let $\ell$,$n$ be integers such that $1\leq n<\ell $, and let $p_1,\ldots,p_{n+1}\in \mathbb{R}^n$ be in general position. Then, $D_{(p_1,\ldots,p_\ell)} : \R^n\rightarrow \R^\ell$ is $\mathcal{A}$-equivalent to the inclusion $(x_1,\ldots,x_n)\mapsto (x_1,\ldots,x_n,0,\ldots,0)$.
\end{pro}
{\it Proof in the case $\ell=n=2$.}\qquad
Since the proof requires various elementary coordinate transformations,
we first prove Proposition \ref{P} in the case $\ell=n=2$ for the sake of clearness.
By definition, the mapping $D_{(p_1, p_2)}$ has the following form:
\begin{eqnarray*}
{} & { } & D_{(p_1, p_2)}(x_1, x_2) \\
{ } & = & \left((x_1-p_{11})^2+(x_2-p_{12})^2, (x_1-p_{21})^2+(x_2-p_{22})^2)\right) \\
{ } & = & \left(x_1^2+x_2^2-2\left( p_{11}x_1+p_{12}x_2\right) +p_{11}^2+p_{12}^2\right. ,  \\
{ } & { } & \qquad \qquad
x_1^2+x_2^2-2\left.\left( p_{21}x_1+p_{22}x_2\right) +p_{21}^2+p_{22}^2\right).
\end{eqnarray*}
\par
As Step 1, we remove the quadratic terms in one component.
This can be done by composing an affine diffeomorphism $h_1$ of the target space.
We may assume that the composition $h_1\circ D_{(p_1, p_2)}$ has the following form:
 \begin{eqnarray*}
&{}&(h_1\circ D_{(p_1,p_2)})(x_1,x_2)\\
&=&
\biggl(\sum_{j=1}^2 (p_{2j}-p_{1j})(x_j-p_{1j}), \sum_{j=1}^2(x_j-p_{1j})^2\biggl).
\end{eqnarray*}
\par
As Step 2, we reduce the first component to a linear function.
This can be done by composing  the affine diffeomorphism
$h_2(x_1, x_2)=(x_1+p_{11}, x_2+p_{12})$ of the source space.
The composition has the following form:
\begin{eqnarray*}
&{}&(h_1\circ D_{(p_1,p_2)}\circ h_2 )(x_1,x_2)\\
&=&
\biggl((p_{21}-p_{11})x_1+(p_{22}-p_{12})x_2, x_1^2+x_2^2 \biggl).
\end{eqnarray*}
\par
As Step 3, we reduce the linear function $(p_{21}-p_{11})x_1+(p_{22}-p_{12})x_2$ to $x_1$.
Since $p_1, p_2$ are in general position,
this can be done easily by composing  an affine diffeomorphism
$h_3$ of the target space and a linear diffeomorphism $h_4$ of the source space.
We may assume that the composition has the following form:
\[
(h_3\circ h_1\circ D_{(p_1,p_2)}\circ h_2 \circ h_4)(x_1,x_2)
=
\biggl(x_1, \left( x_1+\alpha x_2\right)^2+x_2^2 \biggl).
\]
\par
As Step 4, we complete the square of the second component with respect to the variable $x_2$ as follows:
\begin{eqnarray*}
\left(x_1+\alpha x_2\right)^2+x_2^2 & = &
x_1^2+2\alpha x_1 x_2+ \alpha^2x_2^2+x_2^2 \\
{ } & = &
\left(1+\alpha^2\right)\left(x_2+\frac{\alpha}{\left(1+\alpha^2\right)}x_1\right)^2+
\left(1-\frac{\alpha^2}{\left(1+\alpha^2\right)}\right)x_1^2.
\end{eqnarray*}
\par
As Step 5, eliminating the term
$\left(1-\frac{\alpha^2}{\left(1+\alpha^2\right)}\right)x_1^2$ is done
by composing a nonlinear diffeomorphism
$h_5$ of the target space.
\par
As the final step, the quadratic form
\[
\left(1+\alpha^2\right)\left(x_2+\frac{\alpha}{\left(1+\alpha^2\right)}x_1\right)^2
\]
is reduced to $x_2^2$ by composing an affine diffeomorphism $h_6$ of the source space.
\qed
\par
\medskip
{\it Proof of $(I)$ in general case.}\qquad
Let $H_1 : \mathbb{R}^\ell\rightarrow \mathbb{R}^\ell$ be the diffeomorphism given by
\begin{eqnarray*}
{}&{}&H_1(X_1,X_2,\ldots, X_\ell)\\
{}&=&\biggl(\frac{1}{2}\Bigl(X_1-X_2+\sum_{j=1}^n (p_{1j}-p_{2j})^2\Bigl),\ldots,
\frac{1}{2}\Bigl(X_1-X_\ell+\sum_{j=1}^n (p_{1j}-p_{\ell j})^2\Bigl),X_1\biggl).
\end{eqnarray*}
The composition of $D_{(p_1,\ldots,p_\ell)}$ and $H_1$ is given by
\begin{eqnarray*}
&{}&(H_1\circ D_{(p_1,\ldots,p_\ell)})(x_1,x_2,\ldots,x_n)\\
&=&
\biggl(\sum_{j=1}^n (p_{2j}-p_{1j})(x_j-p_{1j}),\ldots,
\sum_{j=1}^n (p_{\ell j}-p_{1j})(x_j-p_{1j}),\sum_{j=1}^n(x_j-p_{1j})^2\biggl).
\end{eqnarray*}

Let $H_2 : \mathbb{R}^n\rightarrow \mathbb{R}^n$ be the diffeomorphism defined by
\[H_2(x_1,x_2,\ldots,x_n)=(x_1+p_{11},x_2+p_{12},\ldots,x_n+p_{1n}).\]
The composition of $H_1\circ D_{(p_1,\ldots,p_\ell)}$ and $H_2$ is given by
\begin{eqnarray*}
{}&{}&(H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2)(x_1,x_2,\ldots,x_n)\\
{}&=&\biggl(\sum_{j=1}^n (p_{2j}-p_{1j})x_j,\ldots,\sum_{j=1}^n (p_{\ell j}-p_{1j})x_j,\,\sum_{j=1}^nx_j^2\biggl)\\
{}&=&
\begin{pmatrix}
x_1&x_2&\cdots &x_n
\end{pmatrix}
\begin{pmatrix}
p_{21}-p_{11}&\cdots &p_{\ell 1}-p_{11}&x_1\\
p_{22}-p_{12}&\cdots &p_{\ell 2}-p_{12}&x_2\\
\vdots & & \vdots &\vdots \\
\vdots &   & \vdots &\vdots \\
p_{2n}-p_{1n}&\cdots &p_{\ell n}-p_{1n}&x_n
\end{pmatrix}.
\end{eqnarray*}
Set
\begin{eqnarray*}
A=\begin{pmatrix}
p_{21}-p_{11}&\cdots &p_{\ell 1}-p_{11}\\
p_{22}-p_{12}&\cdots &p_{\ell 2}-p_{12}\\
\vdots & & \vdots \\
\vdots &   & \vdots \\
p_{2n}-p_{1n}&\cdots &p_{\ell n}-p_{1n}
\end{pmatrix}.
\end{eqnarray*}
Since $\ell$-points $p_1,\ldots,p_\ell$ are in general position, it is clear that the rank of $A$ is $\ell -1$.
Hence, by composiong linear coordinate transformations if necessary, we may assume
that there  exists an $(\ell-1)\times (\ell-1)$ regular matrix $B$ such that
\begin{eqnarray*}
AB=\begin{pmatrix}
  1            &     &      \bigzerol \\
                   & \ddots &       \\
   \bigzerou &     &        1     \\
   \alpha _{11}  & \cdots &\alpha _{1,\ell -1}\\
   \vdots & &\vdots \\
   \alpha _{n-\ell +1,1}  & \cdots &\alpha _{n-\ell +1,\ell -1}
\end{pmatrix}.
\end{eqnarray*}

Let $H_3 : \mathbb{R}^\ell\rightarrow \mathbb{R}^\ell$ be the diffeomorphism defined by
\begin{eqnarray*}
{}&{}&H_3(X_1,X_2,\ldots,X_\ell )=
\begin{pmatrix}
X_1&X_2&\cdots &X_\ell
\end{pmatrix}
\left(
\begin{array}{ccc|c}
&&&  0\\
&\mbox{\smash{\Large $B$}}&&\vdots\\
&&&0\\ \hline
0&\cdots &0&1
\end{array}
\right).
\end{eqnarray*}
The composition of $H_3$ and $H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2$ is given by
\begin{eqnarray*}
{}&{}&(H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2)(x_1,x_2,\ldots,x_n)\\
{}&=&\begin{pmatrix}
x_1&x_2&\cdots &x_n
\end{pmatrix}
\left(
\begin{array}{ccc|c}
&&&  x_1\\
&&&x_2\\
&\mbox{\smash{\Large $A$}}&&\vdots \\
&&&\vdots\\
&&&x_n
\end{array}
\right)
\left(
\begin{array}{ccc|c}
&&&  0\\
&\mbox{\smash{\Large $B$}}&&\vdots\\
&&&0\\ \hline
0&\cdots &0&1
\end{array}
\right)\\
{}&=&\begin{pmatrix}
x_1&x_2&\cdots &x_n
\end{pmatrix}
\begin{pmatrix}
  1            &     &      \bigzerol & x_1\\
                   & \ddots &     & \vdots  \\
   \bigzerou &     &        1  & x_{\ell -1}   \\
   \alpha _{11}  & \cdots &\alpha _{1,\ell -1} & x_\ell \\
   \vdots & &\vdots & \vdots \\
   \alpha _{n-\ell +1,1}  & \cdots &\alpha _{n-\ell +1,\ell -1} & x_n
\end{pmatrix}\\
{}&=&\biggl(x_1+\sum_{j=1}^{n-\ell +1}\alpha _{j1}x_{\ell +j-1},\ldots,
x_{\ell -1}+\sum_{j=1}^{n-\ell +1}\alpha _{j,\ell -1}x_{\ell +j-1},\sum_{j=1}^nx_j^2\biggl).
\end{eqnarray*}
Then, we have
\begin{eqnarray*}
\biggl(x_1+\sum_{j=1}^{n-\ell +1}\alpha _{j1}x_{\ell +j-1},\ldots,
x_{\ell -1}+\sum_{j=1}^{n-\ell +1}\alpha _{j,\ell -1}x_{\ell +j-1},x_\ell ,\ldots,x_n\biggl)\\
=\begin{pmatrix}
x_1&x_2&\cdots &x_n
\end{pmatrix}
\begin{pmatrix}
1     &    &      &     &  & &\\
     &  \ddots  &      &     &  &  &\\
 \bigzerou   && & &\text{\huge{0}}&&\\
 &  & 1 & &&&\\
\alpha _{11}      & \cdots  & \alpha _{1,\ell -1}   &1&&\\
\vdots &       &  \vdots    & 0 & \ddots & &\\
\vdots &       &    \vdots  & \vdots &  & \ddots &\\
\alpha _{n-\ell +1,1}      & \cdots & \alpha _{n-\ell +1,\ell -1}   &0&\cdots &0&1\\
\end{pmatrix}.
\end{eqnarray*}
Set
\begin{eqnarray*}
C=
\begin{pmatrix}
1     &    &      &     &  & &\\
     &  \ddots  &      &     &  &  &\\
 \bigzerou   && & &\text{\huge{0}}&&\\
 &  & 1 & &&&\\
\alpha _{11}      & \cdots  & \alpha _{1,\ell -1}   &1&&\\
\vdots &       &  \vdots    & 0 & \ddots & &\\
\vdots &       &    \vdots  & \vdots &  & \ddots &\\
\alpha _{n-\ell +1,1}      & \cdots & \alpha _{n-\ell +1,\ell -1}   &0&\cdots &0&1\\
\end{pmatrix}.
\end{eqnarray*}
The inverse matrix of $C$ is the following:
\begin{eqnarray*}
C^{-1}=
\begin{pmatrix}
1     &    &      &     &  & &\\
     &  \ddots  &      &     &  &  &\\
 \bigzerou   && & &\text{\huge{0}}&&\\
 &  & 1 & &&&\\
-\alpha _{11}      & \cdots  & -\alpha _{1,\ell -1}   &1&&\\
\vdots &       &  \vdots    & 0 & \ddots & &\\
\vdots &       &    \vdots  & \vdots &  & \ddots &\\
-\alpha _{n-\ell +1,1}      & \cdots & -\alpha _{n-\ell +1,\ell -1}   &0&\cdots &0&1\\
\end{pmatrix}.
\end{eqnarray*}

Let $H_4 : \R^n\rightarrow \R^n$ be the diffeomorphism defined by
\[
H_4(x)=xC^{-1}.
\]
The composition of $H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2$ and $H_4$ is as follows:
\begin{eqnarray*}
{}&{}&(H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2\circ H_4)(x_1,x_2,\ldots,x_n)\\
{}&=&\biggl(x_1,x_2,\ldots,x_{\ell -1},\sum_{k=1}^{\ell -1}\Bigl(x_k-\sum_{j=1}^{n-\ell +1}\alpha _{jk}x_{\ell+ j-1}\Bigl)^2+
\sum_{k=\ell}^nx_k^2\biggl).
\end{eqnarray*}
By completing the square of the last component function
with respect to variables $x_\ell, \ldots, x_n$, we have the following:
\begin{eqnarray*}
{ } & { } &
\sum_{k=1}^{\ell -1}\Bigl(x_k-\sum_{j=1}^{n-\ell +1}\alpha _{jk}x_{\ell+ j-1}\Bigl)^2+\sum_{k=\ell}^nx_k^2 \\
{ } & = &
\sum_{j=\ell}^n\left(\sum_{k=1}^{\ell -1}\alpha_{jk}^2+1\right)
\left(x_j-\frac{1}{\left(\sum_{k=1}^{\ell -1}\alpha_{jk}^2+1\right)}\sum_{k=1}^{\ell-1}\alpha_{j-\ell+1, k}x_k\right)^2 \\
{} & {} & \qquad \qquad
+\sum_{k=1}^{\ell-1}\left(\frac{\sum_{j=\ell}^n\alpha_{j-\ell+1, k}^2}{\left(\sum_{k=1}^{\ell-1}\alpha_{jk}^2+1\right)}
+1\right)x_k^2
\end{eqnarray*}
\par
Let $H_5 : \R^\ell\rightarrow \R^\ell$ be the diffeomorphism defined by
\begin{eqnarray*}
{}&{}&H_5(X_1,X_2,\ldots,X_\ell)\\
{}&=&\left(X_1,X_2,\ldots,X_{\ell -1},X_\ell-
\sum_{k=1}^{\ell-1}\left(\frac{\sum_{j=\ell}^n\alpha_{j-\ell+1, k}^2}{\left(\sum_{k=1}^{\ell-1}\alpha_{jk}^2+1\right)}
+1\right)X_k^2
\right).
\end{eqnarray*}
The composition of $H_5$ and $H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2\circ H_4$ is as follows:
\begin{eqnarray*}
{}&{}&(H_5\circ H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2\circ H_4)(x_1,x_2,\ldots,x_n)\\
{}&=&\left(x_1,x_2,\ldots,x_{\ell -1},
\sum_{j=\ell}^n\left(\sum_{k=1}^{\ell -1}\alpha_{jk}^2+1\right)
\left(x_j-\frac{1}{\left(\sum_{k=1}^{\ell -1}\alpha_{jk}^2+1\right)}\sum_{k=1}^{\ell-1}\alpha_{j-\ell+1, k}x_k\right)^2
\right).
\end{eqnarray*}
\par
Let $H_6 : \R^n\rightarrow \R^n$ be the diffeomorphism defined by
\begin{eqnarray*}
{}&{}&H_6(x_1,x_2,\ldots,x_n)\\
{}&=&\biggl(x_1,\ldots,x_{\ell -1},
\frac{x_\ell}{\sqrt{r_\ell}}+\varphi_\ell(x_1,\ldots,x_{\ell-1}),\ldots,\frac{x_n}{\sqrt{r_n}}+\varphi _n(x_1,\ldots,x_{\ell-1})
\biggl),
\end{eqnarray*}
where $r_j=\sum_{k=1}^{\ell -1}\alpha_{jk}^2+1$ and
$\varphi_j(x_1,\ldots,x_{\ell-1})=\frac{1}{\left(\sum_{k=1}^{\ell -1}\alpha_{jk}^2+1\right)}\sum_{k=1}^{\ell-1}\alpha_{j-\ell+1, k}x_k$ $(\ell \le j\le n)$.
It follows that
\begin{eqnarray*}
{}&{}&(H_5\circ H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2\circ H_4\circ H_6)(x_1,x_2,\ldots,x_n)\\
{}&=&(x_1,x_2,\ldots,x_{\ell -1},x_\ell^2+\cdots +x_n^2).
\end{eqnarray*}
\hfill\qed
\par
\smallskip
{\it Proof of $(II)$.}\qquad
Since $n<\ell$ and $p_1, \ldots, p_{n+1}$ are in general position,
there exists an $(\ell-1)\times (\ell-1)$ regular matrix $\widetilde{B}$ such that
\begin{eqnarray*}
A\widetilde{B}=
\begin{pmatrix}
1&&\mbox{\smash{\Large $0$}}&                                             0 & \cdots & 0   \\
&\ddots &&        \vdots &   & \vdots \\
\mbox{\smash{\Large $0$}}&&1&            0  &  \cdots  &  0\\
\end{pmatrix},
\end{eqnarray*}
where the matrix $A$ is the same as in the proof of $(I)$.
Similarly as the proof of $(I)$ of Proposition \ref{P}, the composition $H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2$
is as follows:
\begin{eqnarray*}
{}&{}&(H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2)(x_1,x_2,\ldots,x_n)\\
{}&=&\begin{pmatrix}
x_1&x_2&\cdots &x_n
\end{pmatrix}
\left(
\begin{array}{ccc|c}
&&&  x_1\\
&&&x_2\\
&\mbox{\smash{\Large $A$}}&&\vdots \\
&&&\vdots\\
&&&x_n
\end{array}
\right)
\left(
\begin{array}{ccc|c}
&&&  0\\
&\mbox{\smash{\Large $\widetilde{B}$}}&&\vdots\\
&&&0\\ \hline
0&\cdots &0&1
\end{array}
\right)\\
{}&=&\begin{pmatrix}
x_1&x_2&\cdots &x_n
\end{pmatrix}
\begin{pmatrix}
1&&\mbox{\smash{\Large $0$}}&                                             0 & \cdots & 0  &x_1 \\
&\ddots &&        \vdots &   & \vdots & \vdots  \\
\mbox{\smash{\Large $0$}}&&1&            0  &  \cdots  &  0 & x_n\\
\end{pmatrix}\\
{}&=&\biggl(x_1,\ldots,x_n,0,\ldots,0,\sum_{j=1}^nx_j^2\biggl).
\end{eqnarray*}

Let $\widetilde{H}_4 : \R^\ell\rightarrow \R^\ell $ be the diffeomorphism defined by
\begin{eqnarray*}
\widetilde{H}_4(X_1,X_2,\ldots,X_\ell)=\left(X_1,\ldots,X_n,\ldots,X_{\ell-1},X_\ell-\sum_{j=1}^nX_j^2\right).
\end{eqnarray*}
It follows that
\begin{eqnarray*}
{}&{}&(\widetilde{H}_4\circ H_3\circ H_1\circ D_{(p_1,\ldots,p_\ell)}\circ H_2)(x_1,x_2,\ldots,x_n)\\
{}&=&(x_1,x_2,\ldots,x_n,0,\ldots,0).
\end{eqnarray*}
\hfill\qed
\par
\smallskip
We have the following as a corollary of Proposition \ref{P} $(I)$.
\begin{coro}\label{corollary 2.1}
In the case that $n=\ell$ $(\geq 2)$, let $p_1,\ldots,p_\ell\in \mathbb{R}^\ell$ $(p_{1\ell}=\cdots =p_{\ell\ell}=a)$ be in general position. Then, there exists a diffeomorphism $H:\mathbb{R}^\ell \to \mathbb{R}^\ell$ such that $H\circ D_{(p_1,\ldots , p_\ell)}$
is a definite fold mapping whose singular set is the hyperplane $\mathbb{R}^{\ell-1}\times\{ a\}$.
\end{coro}
{\it Proof.}\qquad
The composition $H_1\circ D_{(p_1,\ldots,p_\ell)}$ can be expressed as follows:
\begin{eqnarray*}
{}&{}&(H_1\circ D_{(p_1,\ldots,p_\ell)})(x_1,x_2,\ldots,x_\ell)\\
&=&\tr{
\begin{pmatrix}
x_1-p_{11}\\
\vdots \\
\vdots \\
x_\ell-a \\
\end{pmatrix}}
\begin{pmatrix}
p_{21}-p_{11}&\cdots &p_{\ell 1}-p_{11}&x_1-p_{11}\\
p_{22}-p_{12}&\cdots &p_{\ell 2}-p_{12}&x_2-p_{12}\\
\vdots & & \vdots &\vdots \\
p_{2,\ell-1}-p_{1,\ell-1}&\cdots &p_{\ell,\ell-1}-p_{1,\ell-1}&x_{\ell-1}-p_{1,\ell-1}\\
0 & \cdots & 0 &x_\ell-a
\end{pmatrix},
\end{eqnarray*}
where $\tr{\widetilde{A}}$ means the transposed matrix of $\widetilde{A}$.
The composition of $H_3$ and $H_1\circ D_{(p_1,\ldots,p_\ell)}$ is as follows:
\begin{eqnarray*}
{}&{}&(H_3\circ H_1\circ D_{(p_1,..., p_\ell)})(x_1,\,x_2,\,.\,.\,.\,,\,x_\ell)\\
{}&=&\biggl(x_1-p_{11},\ldots,x_{\ell-1}-p_{1,\ell-1},\sum_{j=1}^{\ell-1}(x_j-p_{1j})^2+(x_\ell-a)^2\biggl).
\end{eqnarray*}
Let $H_4' : \R^\ell\rightarrow \R^\ell$ be the diffeomorphism defined by
\[H_4'(X_1,X_2,\ldots,X_\ell)=\left(X_1+p_{11},\ldots,X_{\ell-1}+p_{1,\ell-1},X_\ell-\sum_{j=1}^{\ell-1}X_j^2+a\right).\]
Set $H=H_4'\circ H_3\circ H_1$, and we have the following:
\begin{eqnarray*}
{}&{}&(H\circ D_{(p_1,\ldots,p_\ell)})(x_1,x_2,\ldots,x_\ell)\\
{}&=&(x_1,x_2,\ldots,x_{\ell-1},(x_\ell-a)^2+a).
\end{eqnarray*}
\hfill\qed
\par
\smallskip
\begin{lemm}[\cite{mather, bruce}]\label{M}
Let $M$ be a closed manifold and let $i:M\rightarrow \R^\ell$ be an immersion with normal crossings.
Then, there exists a subset $\Sigma\subset \mathcal{L}(\mathbb{R}^\ell,\mathbb{R})$ of Lebesgue measure $0$ such that for any $\pi \in \mathcal{L}(\mathbb{R}^\ell,\mathbb{R})-\Sigma$, the composition $\pi\circ i : M\rightarrow \mathbb{R}$ is a Morse function, where $\mathcal{L}(\mathbb{R}^\ell,\mathbb{R})$ stands for
the space of linear functions $\R^\ell\rightarrow \R$.
\end{lemm}
Let $M$ be a closed manifold and let $i:M\rightarrow \R^\ell$ be an immersion with normal crossings.
Then, by Lemma \ref{M} there exists a linear function
$\pi : \mathbb{R}^\ell\rightarrow \mathbb{R}$:
\[\pi (x)=\sum_{j=1}^{\ell}u_jx_j\hspace{5mm}(u=(u_1,\ldots,u_\ell),x=(x_1,\ldots,x_\ell))\]
such that $\pi\circ i : M\rightarrow \mathbb{R}$ is a Morse function.
By a change of basis of the linear space $\R^\ell$ if necessary, we may assume
that  $u=(0,\ldots,0,1)$.
Note that $\pi (x)=x_\ell$, $\pi \circ i=i_\ell$ ($i=(i_1,\ldots,i_\ell)$).
Since $M$ is compact and $\pi\circ i$ is continuous, $\pi\circ i$ has the minimal value $k_0$.
It is clear that $k_0$ is a critical value of $\pi\circ i$.
Since $\pi\circ i$ is a Morse function, the set $(\pi\circ i)^{-1}(k_0)$ consists of only one point $q$.
Let $(U,\varphi ,(t_1,\ldots,t_m))$ be a coordinate neighborhood of $q$.
By shrinking $U$ sufficiently small if necessary, we may assume that $i|_{(U)}:U\rightarrow \mathbb{R}^\ell$ is an embedding.

We show first that near $q$ a coordinate system can be chosen so that $i(M)$ can be expressed
locally as a graph of a map $\mathbb{R}^m\to \mathbb{R}^{\ell-m}$.
Let $(Ji)_q$ be the Jacobian matrix of the mapping $i$ at $q$. Note that
\[\frac{\partial (i_\ell\circ \varphi ^{-1})}{\partial t_j}(\varphi (q))=0\]
for any $j$ ($1\leq j\leq m$).
Thus, we have
\[(0,\ldots,0,1)(Ji)_q=(0,\ldots,0).\]
By a change of basis of the linear space $\R^\ell$ if necessary,
we may assume
that the matrix
\[
(E_m\mid O)(Ji)_q
\]
is invertible,
where $E_m$ is the $m\times m$ unit matrix and $O$ is the zero $m\times (\ell-m)$ matrix.

Consider the mapping $\psi : \varphi (U)\rightarrow \R^m\times \{0\}\subset \R^\ell$ defined by
\[
\psi =(i_1\circ \varphi ^{-1},\ldots,i_m\circ \varphi ^{-1}).\]
It follows that $(J\psi )_{\varphi (q)}$ is invertible.
By the inverse function theorem, there exist an open neighborhood $V$ of $\varphi (q)$ and
an open neighborhood $\widetilde {V}$ of $\psi$ ($\varphi (q)$) such that $\psi : V\rightarrow \widetilde {V}$ is
a diffeomorphism. We may suppose that $V$ and $\widetilde {V}$ are connected and $V\subset \varphi (U)$.
Then, $i(\varphi ^{-1}(V))$ can be expressed as follows:
\[i(\varphi ^{-1}(V))=\{(x,\xi _{m+1}(x),\ldots,\xi _\ell(x))\in \mathbb{R}^\ell
\mid x\in \widetilde {V}\},\]
where $x=(x_1,\ldots,x_m)$ and $\xi _j=i_j\circ \varphi ^{-1}\circ \psi ^{-1}$ for any $j$ $(m+1\leq j\leq \ell)$.
\par
Set
\[
k_1=\displaystyle\sup_{q'\in \varphi ^{-1}(V)}i_\ell(q').
\]
Then, $k_1>k_0$, and $i_\ell^{-1}(r)$ is a non-empty set for any $r\in [k_0,k_1)$.
Since $M-\varphi ^{-1}(V)$ is compact and $i_\ell$ is continuous,
there exists a minimal value
$k_2$ of $i_\ell$.
This value satisfies $k_2>k_0$.

Set
\[
k_3=\min\Bigl\{k_0+1,k_0+\displaystyle\frac{k_2-k_0}{3},k_1\Bigr\}.
\]
\begin{lemm}\label{lemma 2.1}
There exists a real number $a\in (k_0,k_3)$ such that
there exist $(m+1)$-points in general position in
$i(M)\cap (\mathbb{R}^{\ell-1}\times \{a\})$.
\end{lemm}
{\it Proof.}\qquad
Suppose that $m\geq 2$. By Sard's theorem, there exists a real number $a\in (k_0,k_3)$ such that $a$ is a regular value of the mapping $i_\ell : M\rightarrow \R$.
Since a real number $a$ is also an element of $(k_0,k_1)$, we get $i_\ell^{-1}(a)\not=\emptyset$.
Hence, $i_\ell^{-1}(a)$ is an $(m-1)$-dimensional closed manifold.
Note that $i(i_\ell^{-1}(a))=i(M)\cap (\R^{\ell-1}\times \{a\})$. Let $n_0$ be the maximal value of the number
of points which are in general position in $i(i_\ell^{-1}(a))$.
It is clear that $n_0\geq 2$.
Let $p_1,\ldots,p_{n_0}$ be $n_0$-points in general position in $i(i_\ell^{-1}(a))$, and set
\[
W=\Bigl\{\displaystyle\sum_{j=2}^{n_0}\alpha _j\overrightarrow{p_1p_j}\mid \alpha _j\in \R\Bigr\}.
\]
Then, $i(i_\ell^{-1}(a))\subset W$ and ${\rm dim}\;W=n_0-1$.
Since $i\mid _{i_\ell^{-1}(a)} : i_\ell^{-1}(a)\rightarrow W$ is an immersion and
${i_\ell^{-1}(a)}$ is closed, it follows that $m-1<n_0-1$.

Suppose that $m=1$.
Since $\widetilde {V}$ is a bounded connected open set of $\R$, we can set $\widetilde {V}=(b_1,b_2)$.
Then, the set $i(\varphi ^{-1}(V))$ can be expressed as follows:
\[i(\varphi ^{-1}(V))=\{(x,\xi _2(x),\ldots,\xi _\ell(x))\in \R^\ell \mid x\in (b_1,b_2)\}.\]
Note that $b_1<i_1(q)<b_2$.    Define the following two closed intervals:
\begin{eqnarray*}
B_1&=&\Bigl[b_1+\frac{i_1(q)-b_1}{2},\,i_1(q)\Bigr],\\
B_2&=&\Bigl[i_1(q),\,i_1(q)+\frac{b_2-i_1(q)}{2}\Bigr].
\end{eqnarray*}
Then,
\begin{eqnarray*}
\xi _\ell\Bigl(b_1+\frac{i_1(q)-b_1}{2}\Bigr)&>&\xi _\ell(i_1(q))=k_0,\\
\xi _\ell\Bigl(i_1(q)+\frac{b_2-i_1(q)}{2}\Bigr)&>&\xi _\ell(i_1(q))=k_0.
\end{eqnarray*}
We set
\begin{eqnarray*}
a=\min\Bigl\{\xi _\ell\Bigl(b_1+\frac{i_1(q)-b_1}{2}\Bigr),\,\xi _\ell\Bigl(i_1(q)+\frac{b_2-i_1(q)}{2}\Bigr),\,\frac{k_0+k_3}{2}\Bigr\}.
\end{eqnarray*}
It follows that $a\in (k_0,k_3)$.
Since $B_1$ and $B_2$ are compact and $\xi _\ell$ is continuous, by the intermediate value theorem, there exists
a real number $c_1\in B_1$ such that $\xi _\ell(c_1)=a$ and there exists a real number $c_2\in B_2$ such that $\xi _\ell(c_2)=a$. Since $c_1\not=c_2$, there exist at least two points
$(c_1,\xi _2(c_1),\ldots,\xi _\ell(c_1))$, $(c_2,\xi _2(c_2),\ldots,\xi _\ell(c_2))$
in $i(M)\cap (\R^{\ell-1}\times \{a\})$.
\hfill\qed
\par
\smallskip
\subsection{Preliminaries for the proof of Proposition \ref{d}}
The proof of Proposition \ref{d} requires
the following four lemmas.
\begin{lemm}[\cite{Berger}, p.354]\label{AH}
Let $A\subset \R^2$ be a closed convex set, and let $x$ be a point in the boundary $\partial A$ of $A$.
Then, there exist the following sets
\begin{eqnarray*}
L=\{(x_1,x_2)\in \R^2\mid \alpha x_1+\beta x_2+\gamma =0\},\\
L_{\leq }=\{(x_1,x_2)\in \R^2\mid \alpha x_1+\beta x_2+\gamma \leq 0\}
\end{eqnarray*}
such that $x\in L$, $A\subset L_{\leq }$ $(\alpha \not=0 \mbox{ or } \beta \not=0)$.
\end{lemm}
\begin{lemm}[\cite{Berger}, p.339]\label{AK}
Let $A$ be a subset of $\R^2$. Then, for any point $x=(x_1,x_2)$ in the convex hull $\mathrm{conv}(A)$ of $A$,  there exist three points $q_1=(q_{11},q_{12})$, $q_2=(q_{21},q_{22})$, $q_3=(q_{31},q_{32})$ $\in A$
and real numbers $t_1$,$t_2$,$t_3$ such that
\begin{eqnarray*}
x_1&=&t_1q_{11}+t_2q_{21}+t_3q_{31}\\
x_2&=&t_1q_{12}+t_2q_{22}+t_3q_{32}\\
&{}&(t_1\geq 0,\,t_2\geq 0,\,t_3\geq 0,\,t_1+t_2+t_3=1).
\end{eqnarray*}
\end{lemm}
\begin{lemm}[\cite{Berger}, p.343]\label{AS}
Let $A$ be a subset of $\R^2$ homeomorphic to $S^1$. Then, $\partial \mathrm{conv}(A)$ is homeomorphic to $S ^1$.
\end{lemm}
Now, we define the following mapping which is important in the proof of Proposition \ref{d}.
\begin{defi}\label{definition 2.1}
{\rm Let $\pi _\theta : i(S^1)\rightarrow \R$ be the mapping defined by
\[\pi _\theta (x_1,x_2)=(\cos\theta )x_1+(\sin\theta )x_2.\]}
\end{defi}
Since $i(S^1)$ is compact and $\pi _\theta $ is continuous,
$\pi _\theta $ has the maximal value $k_\theta$.
Then, $(I)$ or $(II)$ holds.
\medskip \\
$(I)$ For any $\theta \in [0,2\pi )$, the set $\pi _\theta ^{-1}(k_\theta )$ has only one point.\\
$(II)$ There exists $\theta \in [0,2\pi )$ such that the set $\pi _\theta ^{-1}(k_\theta )$ has at least two points.
\begin{lemm}\label{I}
In the circumstance of $(I)$, $i(S^1)=\partial \mathrm{conv}(i(S^1))$.
\end{lemm}
{\it Proof.}\qquad
First, we prove that $\partial \mathrm{conv}(i(S^1))\subset i(S^1)$.
Suppose that there exists $a\in$ $\partial \mathrm{conv}(i(S^1))$ such that $a\not\in i(S^1)$ $(a=(a_1,a_2))$.
Then, by Lemma \ref{AH}, there exist the following sets:
\begin{eqnarray*}
L=\{(x_1,x_2)\in \R^2\mid \alpha x_1+\beta x_2+\gamma =0\},\\
L_{\leq }=\{(x_1,x_2)\in \R^2\mid \alpha x_1+\beta x_2+\gamma \leq 0\}
\end{eqnarray*}
such that $a\in L$, $\mathrm{conv}(i(S^1))\subset L_{\leq }$.
By changing the origin and basis of the linear space $\R^2$ if necessary, we may assume that
\begin{eqnarray*}
a&=&(0,0),\\
L&=&\R\times \{0\},\\
L_{\leq }&=&\R\times \{x\in \R\mid x\leq 0\}.
\end{eqnarray*}

Note that $(a_1,a_2)=(0,0)$, and by Lemma \ref{AK} there exist three points $q_1=(q_{11},q_{12})$,
$q_2=(q_{21},q_{22})$, $q_3=(q_{31},q_{32})$ $\in i(S^1)$ and
real numbers $t_1$,$t_2$,$t_3$ such that
\begin{eqnarray}
&&t_1q_{11}+t_2q_{21}+t_3q_{31}=0\\
&&t_1q_{12}+t_2q_{22}+t_3q_{32}=0\\
&&(t_1\geq 0,\,t_2\geq 0,\,t_3\geq 0,\,t_1+t_2+t_3=1).\nonumber
\end{eqnarray}
Note that $q_{12}\leq 0$, $q_{22}\leq 0$, $q_{32}\leq 0$, and
there exists $j$ $(1\leq j\leq 3)$ such that $q_{j2}=0$.
Then, we may assume that $q_{12}=0$.

Thus, the maximal value of the mapping $\pi _{\pi /2}$ is $0$.
Since the set $\pi _{\pi /2}^{-1}(0)$ has only one point, it follows that $q_{22}<0$, $q_{32}<0$.
Hence, by $(2)$, it is necessary that $t_2=t_3=0$.
Since we have $t_1=1$, we obtain $a=q_1$ by $(1)$ and $(2)$. This is inconsistent with $a\not\in i(S^1)$.

Next, we prove that $i(S^1)\subset \partial \mathrm{conv}(i(S^1))$. Suppose that there exists $a\in i(S^1)$ such that
$a\not\in \partial \mathrm{conv}(i(S^1))$. Since $\partial \mathrm{conv}(i(S^1))\subset i(S^1)$, we obtain
$\partial \mathrm{conv}(i(S^1))\subset i(S^1)-\{a\}$.
This means that $\partial \mathrm{conv}(i(S^1))$ is not homeomorphic to $S^1$.
This is inconsistent with Lemma \ref{AS}. \qed
\section{Proofs of Theorems \ref{D1} and \ref{D2}}\label{SD}
Except the last parts, the proof of Theorem \ref{D1} is completely
the same as the proof of Theorem \ref{D2} as follows.
\par
By Lemma \ref{M} there exists a linear function
$\pi : \mathbb{R}^\ell\rightarrow \mathbb{R}$:
\[\pi (x)=\sum_{j=1}^{\ell}u_jx_j\hspace{5mm}(u=(u_1,\ldots,u_\ell),x=(x_1,\ldots,x_\ell))\]
such that $\pi\circ i : M\rightarrow \mathbb{R}$ is a Morse function.
Let $k_0$ be the minimal value of $\pi\circ i$ and let $q\in M$ be the unique point satisfying
$(\pi\circ i)(q)=k_0$ as in Section 2.
Then, by Lemma \ref{lemma 2.1},
there exists a real number $a\in (k_0,k_3)$ such that there exist $(m+1)$-points $p_1,\ldots,p_{m+1}$
in general position in
$i(M)\cap (\R^{\ell-1}\times \{a\})$.
Then, we take $\ell-(m+1)$-points \;$p_{m+2},\ldots,p_\ell\in \mathbb{R}^{\ell-1}\times \{a\}$\;
such that $\ell$-points $p_1,\ldots,p_\ell$ are in general position.

Note that $p_{1\ell}=\cdots=p_{\ell\ell}=a$.
Thus, by Corollary \ref{corollary 2.1}, the mapping $H\circ D_{(p_1,\ldots,p_\ell)}$ is a definite fold mapping whose singular set is the hyperplane
$\R^{\ell-1}\times \{a\}$. Hence, any singular point of $(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}$ must be
contained in $i(M)\cap (\R^{\ell-1}\times \{a\})$. Moreover, since the intersection $i(M)\cap (\R^{\ell-1}\times \{a\})$
is contained in $i(\varphi ^{-1}(V))$,
the following equalities hold for any point in $i(M)\cap (\R^{\ell-1}\times \{a\})$:
\begin{eqnarray*}
{}&{}&(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}(x,x_{m+1},\ldots,x_\ell)\\
{}&=&(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}(x,\xi _{m+1}(x),\ldots,\xi _\ell(x))\\
{}&=&(x,\xi _{m+1}(x),\ldots,\xi _{\ell-1}(x),(\xi _\ell(x)-a)^2+a),
\end{eqnarray*}
where $x=(x_1,\ldots,x_m)$.
Note that the rank of the Jacobian matrix of $(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}$ at any point of $i(M)\cap (\R^{\ell-1}\times \{a\})$ is always $m$, which implies that $H\circ D_{(p_1,\ldots,p_\ell)}\circ i$ is non-singular.
\par
Next, we show that $(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}$ is injective.
For a real number $r$, set
\begin{eqnarray*}
A_{<r}&=&i(M)\cap (\mathbb{R}^{\ell-1}\times \{x\in \mathbb{R}\mid x<r\}),\\
A_{\geq r}&=&i(M)\cap (\mathbb{R}^{\ell-1}\times \{x\in \mathbb{R}\mid x\geq r\}).
\end{eqnarray*}
Since $H\circ D_{(p_1,\ldots,p_\ell)}$ is a definite fold mapping whose singular set is the hyperplane $\R^{\ell-1}\times \{a\}$ by Corollary \ref{corollary 2.1}, the following holds.
\[
(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)} \mbox{ is injective }
\Leftrightarrow  (H\circ D_{(p_1,\ldots,p_\ell)})(A_{<a})\cap A_{\geq a}=\emptyset.
\]

In order to prove $(H\circ D_{(p_1,\ldots,p_\ell)})(A_{<a})\cap A_{\geq a}=\emptyset$, it is sufficient to show the following $(3)$ and $(4)$:
\begin{eqnarray}
(H\circ D_{(p_1,\ldots,p_\ell)})(A_{<a})\cap A_{\geq k_2}&=&\emptyset,  \\
(H\circ D_{(p_1,\ldots,p_\ell)})(A_{<a})\cap A_{<k_2}&=&\emptyset.
\end{eqnarray}

First, we prove $(3)$. Suppose that $(H\circ D_{(p_1,\ldots,p_\ell)})(A_{<a})\cap A_{\geq k_2}\not=\emptyset $.
Then, there exists a real number $\widetilde {a}\in [k_0,a)$ such that $(\widetilde {a}-a)^2+a\geq k_2$.
Then, we have $(k_0-a)^2+a\geq (\widetilde {a}-a)^2+a$.  It follows that $(k_0-a)^2+a\geq k_2$.
On the other hand, since $k_0<a<k_0+1$, we obtain $0<a-k_0<1$.
Thus, we have $(k_0-a)^2+a<2(a-k_0)+a$. Since $a<k_0+(k_2-k_0)/3$, it follows that $2(a-k_0)+a<k_2$.
Therefore, we obtain $(k_0-a)^2+a<k_2$. However, this is inconsistent with $(k_0-a)^2+a\geq k_2$.

For $(4)$,
suppose that $(H\circ D_{(p_1,\ldots,p_\ell)})(A_{<a})\cap A_{<k_2}\not=\emptyset$.
Note that $A_{<k_2}\subset i(\varphi ^{-1}(V))$, and
there exist two points \[(x,\xi _{m+1}(x),\ldots,\xi _\ell(x))\in A_{<a},\ (x',\xi _{m+1}(x'),\ldots,\xi _\ell(x'))\in A_{<k_2}\]
such that
\begin{eqnarray*}
(H\circ D_{(p_1,\ldots,p_\ell)})(x,\xi _{m+1}(x),\ldots,\xi _\ell(x))=(x',\xi _{m+1}(x'),\ldots,\xi _\ell(x')),
\end{eqnarray*}
where $x=(x_1,\ldots,x_m)$ and $x'=(x_1',\ldots,x_m')$.
Since $x=x'$ and $\xi _\ell(x')=(\xi_\ell(x)-a)^2+a$, we get $\xi _\ell(x)\geq a$. However, this contradicts to
$\xi _\ell(x)<a$.
Therefore, $(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}$ must be injective.
\subsection{Proof of Theorem \ref{D1}}\label{D1P}
We have already proved that $H\circ D_{(p_1,\ldots,p_\ell)}\circ i$ is an immersion and
$(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}$ is injective.
Since $i$ is an embedding and $H$ is a diffeomorphism,
the mapping $D_{(p_1,\ldots,p_\ell)}\circ i$ must be an embedding.
\hfill\qed
\par
\subsection{Proof of Theorem \ref{D2}}
It is sufficient to show that  $H\circ D_{(p_1,\ldots,p_\ell)}\circ i$ is with normal crossings.
Since $(H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)}$ is injective,
for any $y\in (H\circ D_{(p_1,\ldots,p_\ell)}\circ i)(M)$, the set $((H\circ D_{(p_1,\ldots,p_\ell)})\mid _{i(M)})^{-1}(y)$
has only one point.
Let $x$ be that point.

Suppose that $x\in i(M)\cap (\R^{\ell-1}\times \{a\})$. Since $i^{-1}(x)\subset \varphi ^{-1}(V)$ and
$i\mid _{\varphi ^{-1}(V)} :  \varphi ^{-1}(V) \rightarrow \R^\ell$ is an embedding, the set $i^{-1}(x)$ has
only one point.
It follows that $H\circ D_{(p_1,\ldots,p_\ell)}\circ i$ is a mapping with normal crossings at the
point.

Suppose that $x\in i(M)-(\R^{\ell-1}\times \{a\})$. Since $i$ is an immersion with normal crossings,
the set $(H\circ D_{(p_1,\ldots,p_\ell)}\circ i)^{-1}(y)$ can be expressed as follows:
\[(H\circ D_{(p_1,\ldots,p_\ell)}\circ i)^{-1}(y)=i^{-1}(x)=\{q_1,\ldots,q_n\}.\]
Since $d(H\circ D_{(p_1,\ldots,p_\ell)})_x : T_x\R^\ell\rightarrow T_y\R^\ell$ is a bijective linear mapping, for any subset
$\{\lambda _1,\lambda _2,\ldots,\lambda _s\}\subset \{1,2,\ldots,n\}$ $(s\leq n)$, the following equalities hold.
\begin{eqnarray*}
{}&{}&{\rm codim}\,\Bigl(\bigcap _{j=1}^sd(H\circ D_{(p_1,\ldots,p_\ell)}\circ i)_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M)\Bigr)\\
{}&=&\ell-{\rm dim}\,\bigcap _{j=1}^sd(H\circ D_{(p_1,\ldots,p_\ell)}\circ i)_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M)\\
{}&=&\ell-{\rm dim}\,d(H\circ D_{(p_1,\ldots,p_\ell)})_x\Bigl(\bigcap _{j=1}^sdi_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M)\Bigr)\\
&=&\ell-{\rm dim}\,\Bigl(\bigcap _{j=1}^sdi_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M)\Bigr)\\
&=&\sum_{j=1}^s(\ell-{\rm dim}\,di_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M))\\
&=&\sum_{j=1}^s(\ell-{\rm dim}\,d(H\circ D_{(p_1,\ldots,p_\ell)}\circ i)_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M))\\
&=&\sum_{j=1}^s{\rm codim}\,d(H\circ D_{(p_1,\ldots,p_\ell)}\circ i)_{q_{\lambda _{j}}}(T_{q_{\lambda _{j}}}M)
\end{eqnarray*}
Hence, $H\circ D_{(p_1,\ldots,p_\ell)}\circ i$ is an immersion with normal crossings.
\hfill\qed
\section{Proof of Proposition \ref{d}}\label{dP}
Since $i(S^1)$ is compact, $\R^2$ is a Hausdorff space, $d_{(p_1,p_2)}\mid _{i(S^1)}$ is continuous
and $i$ is homeomorphic to the image,
in order to prove that $d_{(p_1,p_2)}\circ i$ is homeomorphic to the image, it is sufficient to show that
$d_{(p_1,p_2)}\mid _{i(S^1)}$ is injective.
Recall that $k_\theta$ is the maximal value of the map $\pi_\theta$  defined in Definition \ref{definition 2.1} and
one of the following $(I)$ and $(II)$ holds for $k_\theta$.
\par
\smallskip
\noindent
$(I)$ For any $\theta \in [0,2\pi )$, the set $\pi _\theta ^{-1}(k_\theta )$ has only one point.\\
$(II)$ There exists $\theta \in [0,2\pi )$ such that the set $\pi _\theta ^{-1}(k_\theta )$ has at least two points.
\par
We consider first the case $(I)$.
Let $d : \R^2\times \R^2\rightarrow \R$ be the two-dimensional Euclidean distance defined in Section \ref{IN}.
Since the set $i(S^1)\times i(S^1)$ is compact and $d$ is continuous, $d\mid _{i(S^1)\times i(S^1)}$
has a maximal value $k$.
Let $(a,a')$ $\in i(S^1)\times i(S^1)$ be an element of the set $(d\mid _{i(S^1)\times i(S^1)})^{-1}(k)$.
By changing the origin and basis of the linear space $\R^2$ if necessary,
we may assume
that $a=(0,0)$, $a'=(1,0)$. Then, $k=1$.

Set $\theta_0 =\pi /2$, and define $\pi _{\theta _0} : i(S^1)\rightarrow \R$ ; thus
\begin{eqnarray*}
\pi _{\theta _0}(x_1,x_2)=x_2.
\end{eqnarray*}
Since $i(S^1)$ is compact and $\pi _{\theta _0}$ is continuous, $\pi _{\theta _0}$ has the maximal value
$k_{\theta _0}$ and the minimal value $\widetilde k_{\theta _0}$.
Since there exist two points $a=(0,0)$, $a'=(1,0)$ in $i(S^1)$, it is clear that
$\widetilde k_{\theta _0}\leq 0\leq k_{\theta _0}$.
Then, the set $\pi _{\theta _0}^{-1}(k_{\theta _0})$ has only one point. Since $-\widetilde k_{\theta _0}$ is
also the maximal value of $\pi_{3\pi/2}$ and $\pi _{\theta _0}^{-1}(\widetilde k_{\theta _0})=\pi_{3\pi/2}^{-1}(-\widetilde k_{\theta _0})$, the set $\pi _{\theta _0}^{-1}(\widetilde k_{\theta _0})$
also has only one point.
It follows that
$\widetilde k_{\theta _0}<0<k_{\theta _0}$.
\par
Again by changing of the basis of the linear space $\R^2$ if necessary, we may assume
that $\mid \widetilde k_{\theta _0}\mid \leq \mid k_{\theta _0}\mid $.

We shall prove that the set $\pi _{\theta _0}^{-1}(2k_{\theta _0}/3)$ has at least two points.
Note that $\pi _{\theta _0}^{-1}(2k_{\theta _0}/3)=i(S^1)\cap (\R\times \{2k_{\theta _0}/3\})$. Suppose that $\pi _{\theta _0}^{-1}(2k_{\theta _0}/3)=\emptyset $.
Since $i(S^1)\cap (\R\times \{k_{\theta _0}\})\not=\emptyset$ and $i(S^1)\cap (\R\times \{0\})\not=\emptyset $,
this is inconsistent with the fact that $i(S^1)$ is connected. Suppose that the set $\pi _{\theta _0}^{-1}(2k_{\theta _0}/3)$ has only one point, and
put $\pi _{\theta _0}^{-1}(2k_{\theta _0}/3)=\{q\}$.
Since $i(S^1)-\{q\}$ is not connected in this case, an contradiction yields.
Hence, the set $\pi _{\theta _0}^{-1}(2k_{\theta _0}/3)$ has at least two points.
Let $p_1$,$p_2$ be the two points such that
\begin{eqnarray*}
p_1=\Bigl(p_{11},\frac{2}{3}k_{\theta _0}\Bigr),\ p_2=\Bigl(p_{21},\frac{2}{3}k_{\theta _0}\Bigr)\in \pi _{\theta _0}^{-1}(2k_{\theta _0}/3)\ (p_{11}<p_{21}).
\end{eqnarray*}

Suppose that there exist two points $b$, $b'\in i(S^1)$ $(b\not=b')$ such that
\[d_{(p_1,p_2)}\mid _{i(S^1)}(b)=d_{(p_1,p_2)}\mid _{i(S^1)}(b'),\leqno{(5)}\]
where $b=(b_1,b_2)$ and $b'=(b'_1,b'_2)$. Note that $b\not=b'$, and by $(5)$ we obtain
\begin{eqnarray*}
b'_1&=&b_1,\\
b'_2&=&-b_2+\frac{4}{3}k_{\theta _0}.
\end{eqnarray*}
Since $b\not=b'$ and $b_1=b'_1$, we may assume that $b_2<b'_2$. Now, we show that $0<b_1<1$.
If $b_1>1$ (resp., $b_1<0$), we have $d\mid _{i(S^1)\times i(S^1)}(a,b)>1$ (resp., $d\mid _{i(S^1)\times i(S^1)}(a',b)>1$).
These are contradictory with the fact that the maximal value of $d\mid _{i(S^1)\times i(S^1)}$
is $1$.
If $b_1=0$, there exist two points $b$, $b'$ in $\pi _\pi ^{-1}(0)$ and if $b_1=1$, there exist two points
$b$, $b'$ in $\pi _0^{-1}(1)$. Since the maximal value of $\pi _\pi$ is $0$ and  the maximal value of $\pi _0$ is $1$, these are inconsistent with $(I)$. Hence, we have $0<b_1<1$.
Since $b_2<4k_{\theta _0}/3$ (resp., $b'_2<4k_{\theta _0}/3$), by
$b'_2=-b_2+\frac{4}{3}k_{\theta _0}$,
we have $b'_2>0$ (resp., $b_2>0$).

Set
\begin{eqnarray*}
\bigtriangleup aa'b'=\{t_1a+t_2a'+t_3b'\in \R^2\mid t_1,t_2,t_3\geq 0,\ t_1+t_2+t_3=1\}.
\end{eqnarray*}

\begin{figure}
\begin{center}
\includegraphics[width=6cm,clip]{Rtriangle.eps}
\end{center}
\caption{$\bigtriangleup aa'b'$ and the point $b$}
\end{figure}
Since the set $i(S^1)$ has three points $a$,$a'$,$b'$ and by Lemma \ref{I} $i(S^1)=\partial \mathrm{conv}(i(S^1))$,
it is clearly seen that $\bigtriangleup aa'b'\subset \mathrm{conv}(i(S^1))$.
Then, $(\bigtriangleup aa'b')^\circ \subset (\mathrm{conv}(i(S^1)))^\circ $, where $A^\circ \subset \R^2$
is the set which consists of the interior points of $A$.
It follows that $b\in (\bigtriangleup aa'b')^\circ $ (see Figure 2).
Hence, we have $b\in (\mathrm{conv}(i(S^1)))^\circ $.
This is inconsistent with $b\in \partial \mathrm{conv}(i(S^1))$.
\par
\smallskip
We consider now the case $(II)$.
There exists $\theta _1\in [0,2\pi )$ such that the set $\pi _{\theta _1}^{-1}(k_{\theta _1})$ has at least
two points, where $k_{\theta _1}$ is the maximal value of $\pi_{\theta _1}$.
Let $p_1$,$p_2$ be two points in $\pi _{\theta _1}^{-1}(k_{\theta _1})$.
We may assume
that $p_1=(0,0)$, $p_2=(1,0)$, and $i(S^1)\subset \R\times \{x\in \R\mid x\leq 0\}$.
Then, note that $\theta _1=\pi /2$, $k_{\theta_1}=0$.
Now, suppose that there exist $c=(c_1,c_2)$, $c'=(c'_1,c'_2)\in i(S^1)$ $(c\not=c')$ such that
\[d_{(p_1,p_2)}\mid _{i(S^1)}(c)=d_{(p_1,p_2)}\mid _{i(S^1)}(c').\leqno{(6)}\]

Note that $c\not=c'$, and by $(6)$, we obtain
\begin{eqnarray*}
c_1&=&c'_1,\\
c_2&=&-c'_2.
\end{eqnarray*}
Since $c_2\leq 0$ and $c'_2\leq 0$, it follows that $c_2=c'_2=0$.    This contradicts to the assumption  $c\not=c'$.
\hfill\qed
\section*{Acknowledgement}
The authors are most grateful to the anonymous reviewer for his/her careful reading
of the first manuscript of this paper and invaluable suggestions.
They would like, also, to thank O. Saeki and M. Nishioka for their valuable comments.
T. N. was partially supported
by JSPS and CAPES under the Japan--Brazil research cooperative
program.

\bibliographystyle{model1-num-names}

redpajama_set_name: RedPajamaArXiv


Simone Bennett-Smith portrays Olivia Mitchell.
Carols by Candlelight to be held at Royal Pines Resort, and the release of Chris Nyst's new feature film Crooked Business.
She's been passionate about performing since age 4, and trained as a classical ballet dancer until she graduated high-school. In fact, Simone was invited to attend the National Theatre Ballet School when she was just 15-years-old. She completed her studies and dance training here, but rather than move to America to dance (as was an option for her at the time), she chose to stay in Australia and further study drama and performance.
Since the mid-1990s Simone has appeared in many advertising and corporate campaigns, worked as a television and film actor and presenter. Additionally she has performed in numerous high-profiled stage shows here inAustralia including Too Hot To Handle (1994), My Fair Lady and Aida (1992) with the Victorian State Opera, and Kevin Jacobson's A Chorus Line (1993) where she played the lead role of Judy Turner.


So you've decided that your fleet could be better managed. Perhaps you've entertained a few bids for location-based services. Or maybe you've taken the plunge and are in the process of implementing a new fleet management system. Along the way, have you asked, "Am I doing the right thing? Have I chosen the right company? What is the best way to maximize this new tool?" Well, three experts in the business have a few things for you to consider while you implement this useful, cost-saving technology.


Amazing tour, great value for money, well worth it!
Had a great time with my family, the tour is amazing worth the money. Our tour guide is kind, funny and knowledgeable. Even though some of the tourists were with, were always late coming back he waited for them, overall enjoyed this trip, highly recommended..
Excellent tour, lovely places and Jenny was a lovely guide!
This tour is great, as you visit amazing locations. Although most of the time is spent on the bus it is definitely worth it and you get amazing views while on the road. Our guide Jenny was absolutely amazing. She had a good balance between interesting facts, humour, and history so there wasn't a single boring moment! She was so kind and funny and she even had photo props according to the locations that we visited, so it was good fun taking pictures with them! Also she was very clear when speaking (no harsh accent) so even my mom who doesn't speak English that well could understand what was happening. I've done this same tour with a different tour company and this one was better by far! Highly recommend!
My husband and I enjoyed the trip very much.
Me and my husband enjoyed the sceneries. Marty, the tourist guide/driver was superb. He appeared like he was really enjoying evry minute of his job. He narrated the history of the places that we visited with a bit of humor so the tourist were loughing and didn"t feel bored while we were travelling. A great value for money.


Since 1997, Dr. Tracy R. Patterson has been providing exceptional orthodontic treatment to patients of all ages! She is an accredited orthodontist, meaning she is a dentist who has completed an advanced post-doctoral course, accredited by the American Dental Association, of at least two academic years in the specialty of orthodontics. She attended the University of Connecticut School of Dental Medicine to obtain her Doctor of Dental Medicine and Temple University where she earned her orthodontic specialty certificate and a Masters of Science in Dentistry.
After dental school, Dr. Patterson completed an Advanced Education in General Dentistry residency at the University of Connecticut Health Center. She was also a Fellow and Chief Resident at the University of Connecticut Health Center. She then completed a Roth-Williams Philosophy two-year continuing education program.
Dr. Patterson feels it is important to stay on the cutting-edge of dentistry in order to offer the most advanced treatments. Thus, she maintains memberships in many prestigious professional organizations, such as the American Dental Association, Massachusetts Dental Association, American Association of Orthodontists, Roth-Williams International Society of Orthodontists, Northeast Society of Orthodontists, and the Spear Dental Study Club.
When she's not at the office, Dr. Patterson has a busy family life with her husband, Pierre, and two children, Eva and Remy.


The production of a fountain pen is a high-precision operation. Take a look behind the scenes and watch the fabrication of Pelikan's finest fountain pens - from the nib to the sleeve.
This short film shows how the Japanese artist Mushu Yamazaki elaborately forms a Pelikan fountain pen into a piece of Maki-e art, by hand.
Find out the reasons why the purchase of a Pelikan piston fountain pen makes a lot of sense, and discover the hidden qualities of these highly valuable writing instruments.


text:
Over 2 lakh to appear for Delhi University entrance test
More than 2.22 lakh students, including including M.Phil and PhD aspirants, have applied for Delhi University Entrance Test (DUET), which has begun from Wednesday and will continue till July 8.
Notably, the National Testing Agency (NTA) has been given the responsibility of conducting the first-ever DUET for over 180 undergraduate and postgraduate courses.
DUET is being conducted as a Computer Based Test.
Here's more.
81,307 candidates to sit for UG courses, 1,28,699 for PG
The test will last for two hours, and will be conducted in three shifts per day: morning, afternoon and evening.
For the undergraduate entrance test, 81,307 candidates, out of 222361 applicants will be appearing, as per the press released issued by NTA.
The release also informed that 1,28,699 and 12,355 candidates will be appearing for admissions in postgraduate courses and M.Phil/PhD, respectively.
1,929 PwD/Special Category candidates have applied for the exam
NTA's data also said that 1,929 candidates, who have applied for the entrance examination this year, belong to the PwD/special category. Notably, the three shifts - morning shift, afternoon shift and evening shift - will commence at 8 am, 12 noon and 4 pm, respectively.
NTA has set-up examination centers in 18 cities across India
That's not all.
NTA is also conducting the examination in 18 cities across the country, and has appointed 15 city coordinators and 150 observers.
In order to defuse use of any electronic device, 3,600 jammers have been set-up.
Moreover, for students' practice, NTA has even set-up 4000 test practice centers and as per its reports, almost 10 lakh students have availed this facility.
Nine undergraduate courses' entrance exams to be conducted by NTA
While admissions to most undergraduate courses are subject to cut-off lists, there are nine subjects for which entrance exam is being conducted by NTA.
These courses include BA (Honors) Business Economics, Bachelor of Management Studies, Bachelor of Business Administration (Financial Investment Analysis), BTech, BA (Honors) Humanities and Social Sciences, among others.
To note, a merit list will be prepared based on the marks scored.
NTA aims to conduct tests to assess candidate's competency
In an information brochure released earlier, the Delhi University stated that the aim of the NTA is to conduct "efficient, transparent and international standards tests in order to assess the competency of candidates for admissions to premier higher education institutions."
Notably, NTA also conducted the Joint Entrance Examination (Main) and National Eligibility cum Entrance Test-UG [NEET (UG)] this year.
Candidates must report to the examination centre 1.5hrs before
Meanwhile, the candidates must note that they have to report at the center 1.5 hours before the commencement of the exam and registration desks will close 30-minutes prior. For any other query related to DUET, the candidates can reach out on du.pg.helpdesk2019@gmail.com.

redpajama_set_name: RedPajamaCommonCrawl


text:
Ahmed Abdelnabi
+971 (0)4 364 1641 a.abdelnabi@tamimi.com Dubai International Financial Centre, United Arab Emirates
Ahmed is an Associate in the Litigation team in Dubai. He has over 6 years of experience in Egypt and the UAE. His practice focuses on Banking, Real Estate and Construction related disputes before the UAE Courts and in arbitration.
Ahmed has advised on several contentious and non-contentious construction-related matters in both the litigation and arbitration practices. He has had exposure to construction claims that are mainly focused on the technical aspect as well as claims with commercial aspects only.
In particular, Ahmed has expertise in construction claims that had disputed variations, prolongation costs and extension of time applications. He had also been involved in claims that focused on the validity of termination of construction contracts and its consequences.
Ahmed is also well-familiarised with the FIDIC suite of contracts, and has acted in many disputes that are governed by the FIDIC conditions of contract before the UAE Courts and in arbitration, with a special focus on the validity of the arbitration agreements usually incorporated in such forms.
Advised and represented a major contractor in relation to its insurance/construction related litigation in connection with the eruption of a fire in one of Dubai's most recognizable hotels.
Advised and represented an unincorporated joint venture acting as main contractor of one Abu Dhabi's pivotal infrastructural projects in their dispute against the employer (a state-owned public joint-stock company that oversees the airports of the Emirate of Abu Dhabi) in relation to the validity of the contract termination as well as the validity of its claim against the main contractor's bank guarantees, with a total claim value of approximately AED 5B.
Advised and represented a main contractor in its claim against the employer in relation to the validity of the contract termination and its claim for payment of overdue entitlements.
Advised and represented an unincorporated joint venture acting as a main contractor in its circa AED 600M dispute with its MEP subcontractor in relation to the construction of one of Abu Dhabi's most recognisable landmarks.
Advised an represented one of the UAE's largest MEP contractors in its circa AED 230M contentious dispute against the main contractor in relation to the construction of one of the UAE's largest children's specialty hospital, which involved jurisdictional challenge, joinder of third parties (employer and engineer), pre-maturity of claims arguments, technical claims relating to re-measurable works.
Advised and represented an Abu Dhabi based contractor in its circa AED 500M dispute against the employer in relation to one of the Emirate's largest developments spanning over 22 million square meters area in Al Ain City, Abu Dhabi, in addition to defending the main contractors against claims brought by its subcontractors that valued over AED 350M.
LLB (Bachelor of Laws), English Section, Alexandria University
2015 – Bar Association Alexandria Egypt


text:
Allen Madison
School of Law 213
Email: Allen.Madison@usd.edu
Website: https://sites.google.com/a/usd.edu/allen-madison/home
Allen Madison cultivates understanding in the area of federal tax law and enforcement at the University of South Dakota School of Law. Prior to academia, Professor Madison represented or advised participants on all sides of federal tax disputes—litigating against the IRS, serving in the IRS Chief Counsel Office, and clerking for a United States Tax Court Judge. Through these experiences, Professor Madison has realized the great fortune of living in a country where a taxpayer with a legitimate tax dispute may challenge the government without fear of reprisal. Professor Madison's teaching and research examine the engagement between the government and its people in the context of taxation.
Federal tax law and enforcement.
Tax history, tax theory, tax policy, tax psychology, statutory interpretation, logic, and tax procedure.
LLM, Taxation and Securities Regulation, Georgetown University Law Center, 1998
JD, Law, Hofstra University School of Law, 1994
MBA, Business, Michigan State University Broad Graduate School of Management, 1992
BA, Psychology, Michigan State University College of Social Science, 1989
BA, Marketing, Michigan State University Collge of Business, 1989
Low Income Taxpayer Clinic, United States Internal Revenue Service. Grant Role: Principal. (2015 - 2015)
Citations listed below are presented in a standardized, modified format for display purposes only. They do not necessarily reflect the preferred style and conventions of the faculty member or discipline.
Lombardi, Paul, and Madison, Allen. Blurred Justice. In Entertainment, Publishing and the Arts Handbook, edited by Karen B. Tripp. Thomson Reuters (West), 2020.
Madison, Allen. Fundamentals of Federal Tax Procedure and Enforcement, Wolters Kluwer (2019)..
Madison, Allen. Teacher's Manual, Fundamentals of Federal Tax Procedure and Enforcement, Wolters Kluwer (2019)..
Lombardi, Paul, and Madison, Allen. Blurred Justice, 64. Vol. 38. Loyola of Los Angeles Entertainment Law Review , 2019.
Madison, Allen. Sharyn Fisk, The Tax Court's Home, 157 Tax Notes 1249 (2017)..
Madison, Allen. The IRS's Tax Determination Authority, 71 Tax Law. 143 (2017)..
Madison, Allen. The Legal Framework for Tax Compliance, 70 Tax Law. 497 (2017)..
Madison, Allen. The Legal Consequences of Noncompliance with Federal Tax Laws, 70 Tax Law. 367 (2016)..
Madison, Allen. The Denial of IRS Access to Its Adversary's Playbook, 60 S.D. L. Rev. 232 (2015)..
Madison, Allen. The Taxation of Gratuitous Services Gone Out of Control, 45 U. Mem. L. Rev. 115 (2014)..
Madison, Allen. The Futility of Tax Protester Arguments, 36 T. Jefferson L. Rev. 253 (2013)..
Madison, Allen. Rationalizing Tax Law by Breaking the Addiction to Economic Substance, 47 Idaho L. Rev. 441 (2011)..
Madison, Allen. In-House Basics, in Careers In Tax Law: Perspectives On the Tax Profession an What It Holds For You (John Gamino, Robb A. Longman & Matthew R. Sontag, eds. 2009)..
Madison, Allen. Revisiting Access to the Tax Court's Deliberative Process, 103 Tax Notes 749 (2004)..
Madison, Allen. Important Developments During the Year: Court Procedure and Practice, 56 Tax Law. 1013 (2003)..
Madison, Allen. The Tension Between Textualism and Substance-Over-Form Doctrines in Tax Law, 43 Santa Clara L. Rev. 699 (2003)..
Madison, Allen. Point & Counterpoint: Disclosure and Private Enforcement, ABA Tax Sec. News Q., Fall 2002, at 15 (with Theodore Sims; contribution titled Point: Don't Publicize Corporate Tax Returns or Privatize Enforcement)..
Madison, Allen. Access to the Tax Court's Deliberative Process, 91 Tax Notes 2247 (2001)..
Madison, Allen. An Analysis of the IRS's Voluntary Disclosure Policy, 54 Tax Law. 729 (2001)..
Madison, Allen. Derivatives Regulation in the Context of the Shingle Theory, 1999 Colum. Bus. L. Rev. 271 (1999)..
Madison, Allen. The Uncopyrightability of Jokes, 35 San Diego L. Rev. 111 (1998)..
Madison, Allen. The Context of Employment Discrimination in Japan, 74 U. Det. Mercy L. Rev. 187 (1997)..
Fellow, American College of Tax Counsel, 2014
Presentations:
Lombardi, Paul, Madison, Allen. Blurred Justice. Presented at the College Music Society Great Plains Chapter, Waverly, IA. (March 18, 2018)