text:
\begin{document}

\begin{frontmatter}

\title{ \textcolor{warmblack}{\bf An extension  of the Moore-Penrose Inverse of a Tensor via the Einstein product}}

\author{Krushnachandra Panigrahy$^\dag$$^a$, Debasisha Mishra$^\dag$$^b$}

\address{               $^{\dag}$Department of Mathematics,\\
                        National Institute of Technology Raipur,\\
                        Raipur, Chhattisgarh, India.\\
                        \textit{E-mail$^a$}: \texttt{kcp.224\symbol{'100}gmail.com }\\
                        \textit{E-mail$^b$}: \texttt{dmishra\symbol{'100}nitrr.ac.in. }
        }

\begin{abstract}
\textcolor{warmblack}{ In this paper, we first give an expression
for the Moore-Penrose inverse of the  product of two tensors via the
Einstein product. We then introduce a new generalized inverse of a
tensor called \textit{product Moore-Penrose inverse}. A necessary
and sufficient condition for the coincidence of the Moore-Penrose
inverse and the product Moore-Penrose inverse is also proposed.
Finally,  the \textit{triple reverse order law} of tensors  is
introduced.}
\end{abstract}

\begin{keyword}
Moore-Penrose inverse \sep Tensor \sep Einstein product \sep Reverse
order law.
\end{keyword}

\end{frontmatter}

\section{Introduction}\label{sec1}
Research on tensors has been very active recently \cite{bu1, bu2,
ding, klda, li, sh1, sh2, yw, yng} as tensors have many applications
in different fields like graph analysis, computer vision, signal
processing, data mining, and chemo-metrics, etc. (see  \cite{burmc,
cmn, cpr, eldn, hu, smilde, vla} and the references cited there in).
Let $\mathbb{C}^{I_{1}\times\cdots\times I_{N}}$  be the set of
order $N$ and dimension $I_1 \times \cdots \times I_N$ tensors over
the complex field $\mathbb{C}$.  $\mc{A} \in
\mathbb{C}^{I_{1}\times\cdots\times I_{N}}$ is a multiway array with
$N$-th order tensor, and $I_{1}, I_{2}, \cdots, I_{N}$ are
dimensions of the first, second,$\cdots$, $N$th way, respectively.
Each entry of $\mc{A}$ is denoted by $a_{i_{1}...i_{N}}$. Throughout
the paper, tensors are represented in calligraphic letters like
$\mc{A}$, and the notation $(\mc{A})_{i_{1}...i_N}=
a_{i_{1}...i_{N}}$ represents the scalars. Let $\mc{A}\in
{\C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times \cdots \times
J_{N}}$, then its {\it conjugate transpose}, denoted by
$\mc{A}^{H}$, is defined as $(\mc{A}^{H})_{j_{1}\hdots
j_{N}i_{1}\hdots i_{M}}=\overline{\mc{A}}_{i_{1}\hdots
i_{M}j_{1}\hdots j_{N}},$ where the over-line stands for the
conjugate of $\mc{A}_{i_{1}\hdots i_{M}j_{1}\hdots j_{N}}$. If the
tensor $\mc{A}$ is real, then its {\it transpose} is denoted by
$\mc{A}^{T}$, and is defined as $(\mc{A}^{T})_{j_{1}\hdots
j_{N}i_{1}\hdots i_{M}}=\mc{A}_{i_{1}\hdots i_{M}j_{1}\hdots
j_{N}}$. The Einstein product \cite{ein} $ \mc{A}\n\mc{B} \in
\mathbb{C}^{I_1\times\cdots\times I_N \times J_1 \times\cdots\times
J_M }$ of tensors $\mc{A} \in \mathbb{C}^{I_{1}\times\cdots\times
I_{N} \times K_{1} \times\cdots\times K_{N} }$ and $\mc{B} \in
\mathbb{C}^{K_{1}\times\cdots\times K_{N} \times J_{1}
\times\cdots\times J_{M} }$   is defined by the operation $\n$ via
\begin{equation*}\label{Eins}
(\mc{A}\n\mc{B})_{i_1...i_Nj_1...j_M}
=\displaystyle\sum_{k_1...k_N}a_{{i_1...i_N}{k_1...k_N}}b_{{k_1...k_N}{j_1...j_M}}.
\end{equation*}
The associative law of this tensor product holds. In the above
formula, if $\mc{B} \in \mathbb{C}^{K_1\times\cdots\times K_N}$,
then $\mc{A}\n\mc{B} \in \mathbb{C}^{I_1\times\cdots\times I_N}$ and
\begin{equation*}
(\mc{A}\n\mc{B})_{i_1...i_N} = \displaystyle\sum_{k_1...k_N}
a_{{i_1...i_N}{k_1...k_N}}b_{{k_1...k_N}}.
\end{equation*}
This product is  used in the study of the theory of relativity
\cite{ein} and in the area of continuum mechanics \cite{lai}. The
Einstein product $\1$  reduces to the standard matrix multiplication
as
$$(A\1B)_{ij}= \displaystyle\sum_{k=1}^{n} a_{ik}b_{kj},$$
for $A\in {\R}^{m \times n}$ and $B\in {\R}^{n \times l}$.

Brazell {\it et al.} \cite{BraliNT13} introduced the notion of the
ordinary tensor inverse,  as follows. A tensor $\mc{X} \in
\mathbb{C}^{I_1\times\cdots\times I_N \times I_1 \times\cdots\times
I_N}$ is called  the  {\it inverse} of $\mc{A}\in
\mathbb{C}^{I_1\times\cdots\times I_N \times I_1 \times\cdots\times
I_N}$  if it satisfies $\mc{A}\n\mc{X}=\mc{X}\n\mc{A}=\mc{I}$. It is
denoted  by $\mc{A}^{-1}$. Sun {\it et al.} \cite{sun} introduced
the Moore-Penrose  generalized inverse of a tensor and its
definition is recalled next.

\begin{definition}(Definition 2.2, \cite{sun})\label{defmpi}\\
Let $\mc{A} \in \mathbb{C}^{I_{1}\times\cdots\times I_{N} \times
J_{1} \times \cdots \times J_{M}}$. The tensor $\mc{X} \in
\mathbb{C}^{J_{1}\times\cdots\times J_{M} \times I_1
\times\cdots\times I_{N}}$ satisfying the following four tensor
equations:

\vspace{-1cm}

\begin{eqnarray}
\mc{A}\m\mc{X}\n\mc{A} &=& \mc{A};\\
\mc{X}\n\mc{A}\m\mc{X} &=& \mc{X};\\
(\mc{A}\m\mc{X})^{H} &=& \mc{A}\m\mc{X};\\
(\mc{X}\n\mc{A})^{H} &=& \mc{X}\n\mc{A},
\end{eqnarray}
is called  the \textbf{Moore-Penrose inverse} of $\mc{A}$, and is
denoted by $\mc{A}^{\dg}$.
\end{definition}
The authors of \cite{bm,  jiw1} further studied different
generalized inverses of tensors via the Einstein product.  Jin {\it
et al.} \cite{bai} again  introduced the Moore-Penrose inverse of  a
tensor using $t-$product. They showed the existence and uniqueness
of the Moore-Penrose inverse of an arbitrary tensor by using the
technique of fast Fourier transform, and discussed an application to
linear models.   Ji and Wei \cite{jiw1} introduced the weighted
Moore-Penrose inverse of an even-order tensor, and again  the
Drazin inverse of an even-order tensor \cite{jiw2}. They \cite{jiw2}
obtained an expression of  the Drazin inverse through the
core-nilpotent decomposition. Applications to find the Drazin
inverse solution of the singular linear tensor equation  $\mc{A} \n
\mc{X} = \mc{B}$ is also presented.  Many results on the generalized
inverses, the $\mc{X}$ which only satisfies some of the four
equations of the Definition \ref{defmpi}, can be found in \cite{bm,
kbm}. The vast work on  generalized inverses of a tensor and its
several multivariety extensions in different areas of mathematics in
the literature, motivate us to study further on theory of
generalized inverses of a tensor. In this paper, we introduce a new
type generalized inverse of tensor via the Einstein product called
\textit{product Moore-Penrose inverse} of tensor. This  is obtained
by extending the defining equations for the Moore-Penrose inverse of
a tensor. In addition, we present various expressions for the
Moore-Penrose inverse of products of tensors.

The paper is outlined as follows. In the next section, we discuss
some notations and definitions which are helpful in proving the main
results.  Section 3 contains the main results.

\section{Preliminaries}
For convenience, we first briefly explain some of the terminologies.
We  refer to ${\C}^{m \times n}$ as the  set of all complex ${m
\times n}$ matrices, where $\C$ denotes the set of complex scalars.
We denote ${\C}^{I_1\times\cdots\times I_N}$  as the set of order
$N$  complex tensors. Indeed, a matrix is a second order tensor, and
a vector is a first order tensor. A tensor $\mc{O}$ denotes the {\it
zero tensor} if  all the entries are zero. A tensor
$\mc{A}\in\mathbb{C}^{I_1\times\cdots\times I_N \times I_1
\times\cdots\times I_N}$ is {\it Hermitian}  if  $\mc{A}=\mc{A}^{H}$
and {\it skew-Hermitian} if $\mc{A}= - \mc{A}^{H}$. A tensor
$\mc{A}\in \mathbb{C}^{I_1\times\cdots\times I_N \times
I_1\times\cdots\times I_N}$  is  {\it idempotent}  if $\mc{A}\n
\mc{A}= \mc{A}.$ Next, we present a result for a Hermitian tensor
which is useful while proving our main results.

\begin{lemma}\label{ir}

If  $\mc{P}=\mc{P}^{H}$ for $\mc{P}\in {\C}^{I_{1}\times\cdots\times
I_{N}\times I_{1}\times \cdots \times I_{N}}$, then for any
$\mc{Q}\in {\C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times
\cdots \times I_{N}}$,
\begin{eqnarray}
    \mc{P}\n\mc{Q}=\mc{Q}&\text{ implies }& \mc{Q}^{\dg}\n\mc{P}=\mc{Q}^{\dg}.\label{ir1}\\
    \mc{Q}\n\mc{P}=\mc{Q}&\text{ implies }& \mc{P}\n\mc{Q}^{\dg}=\mc{Q}^{\dg}.\label{ir2}
\end{eqnarray}
\end{lemma}
\begin{proof}
Suppose $\mc{P}=\mc{P}^{H}$, and for any $\mc{Q}$,
$\mc{P}\n\mc{Q}=\mc{Q}$. Then $\mc{P}\n\mc{Q}=\mc{Q}$ leads to
$\mc{Q}\n\mc{Q}^{\dg}\n\mc{P}\n\mc{Q}=\mc{Q}$,
$\mc{Q}^{\dg}\n\mc{P}\n\mc{Q}\n\mc{Q}^{\dg}\n\mc{P}=\mc{Q}^{\dg}\n\mc{P}$,
and
$(\mc{Q}^{\dg}\n\mc{P}\n\mc{Q})^{H}=\mc{Q}^{\dg}\n\mc{P}\n\mc{Q}$.
$\mc{P}=\mc{P}^{H}$ and $\mc{P}\n\mc{Q}=\mc{Q}$ leads
$(\mc{Q}\n\mc{Q}^{\dg}\n\mc{P})^{H}=\mc{Q}\n\mc{Q}^{\dg}\n\mc{P}$.
Thus, by Definition \ref{defmpi} first part of the lemma follows.

The other part of the lemma follows by a similar proof.
\end{proof}

The next result states that the Moore-Penrose inverse coincides with
the tensor itself, for a Hermitian and idempotent tensor $\mc{P}$.

\begin{lemma}
  If $\mc{P}\in{\C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is a Hermitian idempotent tensor, then $\mc{P}^{\dg}=\mc{P}$.
\end{lemma}

Suppose, both $\mc{X}_{1}$ and $\mc{X}_{2}$ are solutions of
$\mc{X}\n\mc{A}\m\mc{X}=\mc{X}$, $\mc{A}\m\mc{X}=\mc{Z}_{1}$, and
$\mc{X}\n\mc{A}=\mc{Z}_{2}$, where $\mc{Z}_{1}$ and $\mc{Z}_{2}$ are
given tensors, independent of $\mc{X}$. Then
$\mc{X}_{1}=\mc{X}_{1}\n\mc{A}\m\mc{X}_{1}$ together with
$\mc{A}\m\mc{X}_{1}=\mc{Z}_{1}=\mc{A}\m\mc{X}_{2}$ and
$\mc{X}_{1}\n\mc{A}=\mc{Z}_{2}=\mc{X}_{2}\n\mc{A}$ results
$\mc{X}_{1}=\mc{X}_{2}$. The following lemma represents this result.

\begin{lemma}\label{ul}
For every $\mc{A}\in{\C}^{I_{1}\times\cdots\times I_{N}\times
J_{1}\times\cdots\times J_{M}}$ there is a unique
$\mc{X}\in{\C}^{J_{1}\times\cdots\times J_{M}\times
I_{1}\times\cdots\times I_{N}}$ satisfying
$\mc{X}\n\mc{A}\m\mc{X}=\mc{X}$, $\mc{A}\m\mc{X}=\mc{Z}_{1}$, and
$\mc{X}\n\mc{A}=\mc{Z}_{2}$, where
$\mc{Z}_{1}\in{\C}^{I_{1}\times\cdots\times I_{N}\times
I_{1}\times\cdots\times I_{N}}$ and
$\mc{Z}_{2}\in{\C}^{J_{1}\times\cdots\times J_{M}\times
J_{1}\times\cdots\times J_{M}}$ are given tensors, independent of
$\mc{X}$.
\end{lemma}

\section{Main Results}
In this section, we first obtain a result to find the Moore-Penrose
inverse of the product of two tensors. We next show the existence
and uniqueness of a new type generalized inverse, i.e., the
\textit{product Moore-Penrose inverse} of a tensor.  We then propose
a few necessary and sufficient conditions for coincidence this
product Moore-Penrose inverse with the Moore-Penrose inverses of
tensors.

Let $\mc{S}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
J_{1}\cdots \times J_{M}}$ and $\mc{T}\in {\C}^{J_{1}\times
\cdots\times J_{M}\times K_{1}\times\cdots \times K_{L}}$ be any two
tensors. Let $\mc{S}_{1}=\mc{S}\m\mc{T}_{1}\lp\mc{T}_{1}^{\dg}$,
where $\mc{T}_{1}=\mc{S}^{\dg}\n\mc{S}\m\mc{T}$. Then
$$\mc{S}_{1}\m\mc{T}_{1}=\mc{S}\m\mc{T}_{1}=\mc{S}\m\mc{T}.$$
Using this factorization, we obtain a representation for
$(\mc{S}\m\mc{T})^{\dg}$ in the following result.

\begin{theorem}
Let $\mc{S}\in{\C}^{I_{1}\times\cdots \times I_{N}\times
J_{1}\times\cdots \times J_{M}}$, $\mc{T}\in{\C}^{J_{1}\times\cdots
\times J_{M}\times K_{1}\times\cdots \times K_{L}}$ be any two
tensors. Let $\mc{S}_{1}=\mc{S}\m\mc{T}_{1}\lp\mc{T}_{1}^{\dg}$,
where $\mc{T}_{1}=\mc{S}^{\dg}\n\mc{S}\m\mc{T}$. Then
$$(\mc{S}\m\mc{T})^{\dg}=\mc{T}_{1}^{\dg}\m\mc{S}_{1}^{\dg}.$$
\end{theorem}

\begin{proof}
Let $\mc{A}=\mc{S}\m\mc{T}=\mc{S}_{1}\m\mc{T}_{1}$ and
$\mc{X}=\mc{T}_{1}^{\dg}\m\mc{S}_{1}^{\dg}$. It is sufficient to
show that $\mc{A}$ and $\mc{X}$ satisfy the equations in Definition
\ref{defmpi}. Using the definition of $\mc{S}_{1}$, we have
$\mc{S}_{1}\m\mc{T}_{1}\lp\mc{T}_{1}^{\dg}=\mc{S}_{1}$, which yields
$\mc{A}\lp\mc{X}=\mc{S}_{1}\m\mc{S}_{1}^{\dg}$. Thus,
$\mc{A}\lp\mc{X}\n\mc{A}=\mc{A}$, $\mc{X}\n\mc{A}\lp\mc{X}=\mc{X}$,
and also it follows that $\mc{A}\lp\mc{X}$ is Hermitian. Now, it
remains to show that $\mc{X}\n\mc{A}$ is Hermitian. We observe that
\begin{equation}\label{fteq1}
    \mc{S}^{\dg}\n\mc{S}_{1}=\mc{T}_{1}\lp\mc{T}_{1}^{\dg}.
\end{equation}
Also,
$\mc{S}_{1}^{\dg}\n\mc{S}_{1}\m\mc{T}_{1}\lp\mc{T}_{1}^{\dg}=\mc{S}_{1}^{\dg}\n\mc{S}_{1}$
and
$\mc{T}_{1}\lp\mc{T}_{1}^{\dg}\m\mc{S}_{1}^{\dg}\n\mc{S}_{1}=\mc{S}_{1}^{\dg}\n\mc{S}_{1}$
are equivalent due to the fact that both
$\mc{S}_{1}^{\dg}\n\mc{S}_{1}$ and  $\mc{T}_{1}\lp\mc{T}_{1}^{\dg}$
are Hermitian. Using equation \eqref{fteq1}, we now get
$\mc{S}_{1}^{\dg}\n\mc{S}_{1}=\mc{T}_{1}\lp\mc{T}_{1}^{\dg}.$

From this it follows $\mc{X}\n\mc{A}=\mc{T}_{1}^{\dg}\m\mc{T}_{1}$
is Hermitian. Hence,
$(\mc{S}\m\mc{T})^{\dg}=\mc{T}_{1}^{\dg}\m\mc{S}_{1}^{\dg}.$

\end{proof}

 Existence and uniqueness of a new type of generalized inverse of tensor is proposed in next theorem.

\begin{theorem}\label{Tpgi}
For any tensor $\mc{A}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
J_{1}\times\cdots \times J_{M}}$, let
$\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}$, where $\mc{R}\in
{\C}^{I_{1}\times \cdots\times I_{N}\times H_{1}\times\cdots \times
H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times \cdots\times H_{K}\times
G_{1}\times\cdots \times G_{L}}$, and $\mc{T}\in
{\C}^{G_{1}\times\cdots \times G_{L}\times J_{1}\times\cdots \times
J_{M}}$. Then, there is a unique matrix $\mc{X}\in
{\C}^{J_{1}\times\cdots \times J_{M}\times I_{1}\times \cdots\times
I_{N}}$ such that
\begin{eqnarray}
\mc{A}\m\mc{X}\n\mc{A}&=&\mc{A,}\label{pgi1}\\
\mc{X}\n\mc{A}\m\mc{X}&=&\mc{X},\label{pgi2}\\
(\mc{R}^{\dg}\n\mc{A}\m\mc{X}\n\mc{R})^{H}&=& \mc{R}^{\dg}\n\mc{A}\m\mc{X}\n\mc{R},\label{pgi3}\\
(\mc{T}\m\mc{X}\n\mc{A}\m\mc{T}^{\dg})^{H}&=& \mc{T}\m\mc{X}\n\mc{A}\m\mc{T}^{\dg},\label{pgi4}\\
\mc{X}\n\mc{R}\kp\mc{R}^{\dg}&=&\mc{X},\label{pgi5}\\
\mc{T}^{\dg}\lp\mc{T}\n\mc{X}&=&\mc{X}\label{pgi6}.
\end{eqnarray}
\end{theorem}
\begin{proof}
$\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}$ gives
\begin{equation}\label{pgipe1}
    \mc{R}\kp\mc{R}^{\dg}\n\mc{A}=\mc{A}=\mc{A}\m\mc{T}^{\dg}\lp\mc{T}.
\end{equation}
Equations \eqref{pgi1}, \eqref{pgi2}, \eqref{pgi3} and \eqref{pgi4}
together with equation \eqref{pgipe1} yield,
$(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp(\mc{T}\m\mc{X}$
$\n\mc{R})\kp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})=\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg}$,
$(\mc{T}\m\mc{X}\n\mc{R})\kp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp(\mc{T}\m\mc{X}\n\mc{R})=(\mc{T}$
$\n\mc{X}\n\mc{R})$,
$[(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp(\mc{T}\m\mc{X}\n\mc{R})]^{H}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp(\mc{T}\m\mc{X}\n\mc{R})$,
and $[(\mc{T}$
$\m\mc{X}\n\mc{R})\kp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})]^{H}=(\mc{T}\m\mc{X}\n\mc{R})\kp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})$,
respectively. We thus have
\begin{equation}\label{pgi}
    (\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}=\mc{T}\m\mc{X}\n\mc{R}.
\end{equation}
Conversely, suppose equation \eqref{pgi} holds. Using equation
\eqref{pgipe1},
$\mc{A}\m\mc{X}\n{A}=\mc{R}\kp\mc{R}^{\dg}\n\mc{A}\m$
$\mc{T}^{\dg}\lp\mc{T}=\mc{A}$,
$(\mc{R}^{\dg}\n\mc{A}\m\mc{X}\n\mc{R})^{H}=\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg}\lp\mc{T}\m\mc{X}\n\mc{R}=\mc{R}^{\dg}\n\mc{A}\m\mc{X}\n\mc{R}$,
$(\mc{T}\m$
$\mc{X}\n\mc{A}\m\mc{T}^{\dg})^{H}=\mc{T}\m\mc{X}\n\mc{R}\kp\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg}=\mc{T}\m\mc{X}\n\mc{A}\m\mc{T}^{\dg}$.
But, if equation \eqref{pgi5} and \eqref{pgi6} hold, then using
equation \eqref{pgipe1}
\begin{align*}
    \mc{X}\n\mc{A}\m\mc{X}&=\mc{T}^{\dg}\lp\mc{T}\m\mc{X}\n\mc{A}\m\mc{X}\n\mc{R}\kp\mc{R}^{\dg}\\
    &=\mc{T}^{\dg}\lp\mc{T}\m\mc{X}\n\mc{R}\kp\mc{R}^{\dg}\\
    &=\mc{X}.
\end{align*}
Thus, equations \eqref{pgi1}-\eqref{pgi6} is equivalent to equations
\eqref{pgi5}, \eqref{pgi6} and \eqref{pgi}. And from the latter
three equations
\begin{align}
    \mc{X}&=\mc{T}^{\dg}\lp\mc{T}\m\mc{X}\n\mc{R}\kp\mc{R}^{\dg}\nonumber\\
    &=\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}\label{epgi}.
\end{align}
Suppose there exist $\mc{X}$ and $\mc{Y}$ satisfying \eqref{pgi5},
\eqref{pgi6} and \eqref{pgi}, then
\begin{align*}
    \mc{X}&=\mc{T}^{\dg}\lp\mc{T}\m\mc{X}\n\mc{R}\kp\mc{R}^{\dg}\\
    &=\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}\\
    &=\mc{T}^{\dg}\lp\mc{T}\m\mc{Y}\n\mc{R}\kp\mc{R}^{\dg}\\
    &=\mc{Y}.
\end{align*}
Hence the uniqueness is established.
\end{proof}

 We termed the tensor $\mc{X}$ in the equation \eqref{epgi} as the \textit{product Moore-Penrose inverse} of $\mc{A}$, and denote it as $\mc{A}_{\pi \dg}$.
An alternative representation of the  product Moore-Penrose inverse
is given in the following theorem.

\begin{theorem}\label{pgif}

Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$, and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then
$$(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}=(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}.$$

\end{theorem}
\begin{proof}
In view of equation \eqref{epgi}, equation \eqref{pgi1} results
\begin{equation}\label{apgie1}
\mc{A}=\mc{A}\m\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}\n\mc{A}.
\end{equation}
Pre-multiplying $(\mc{A}\m\mc{T}^{\dg})^{\dg}$ and post-multiplying
$(\mc{R}^{\dg}\n\mc{A})^{\dg}$ to equation \eqref{apgie1}, we obtain
\begin{equation}
(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}.
\end{equation}
\end{proof}

One more expression of $\mc{A}_{\pi\dg}$ is provided below,

\begin{theorem}\label{aepgi}
$\mc{A}_{\pi\dg}=\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}$
is the unique tensor $\mc{X}$, such that
$\mc{X}\n\mc{A}\m\mc{X}=\mc{X}$,
$\mc{A}\m\mc{X}=\mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}$
and
$\mc{X}\n\mc{A}=\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}$.
\end{theorem}
\begin{proof}
By using Theorem \ref{pgif}z and equation \eqref{epgi}, we have
$$\mc{X}=\mc{A}_{\pi\dg}=\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}.$$
A simple calculation leads to $\mc{X}\n\mc{A}\m\mc{X}=\mc{X}$. Since
$\mc{R}\kp\mc{R}^{\dg}\n\mc{A}=\mc{A}=\mc{A}\m\mc{T}^{\dg}\lp\mc{T}$,
so
$\mc{A}\m\mc{X}=\mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}$
and
$\mc{X}\n\mc{A}=\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}$.
The uniqueness follows from Lemma \ref{ul}.

\end{proof}

Next, we present a result to find the Moore-Penrose inverse of a
tensor $\mc{A}$ with the factorization
$\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}$.

\begin{theorem}\label{aegi}Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times \cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where $\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times \cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$, and $\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times J_{1}\times\cdots \times J_{M}}$. Then
\begin{align*}\mc{A}^{\dg}=(\mc{R}\kp\mc{S}\lp\mc{T})^{\dg}&=(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\\
&=(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{S}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}.
\end{align*}

\end{theorem}
\begin{proof}
Using equation \eqref{pgipe1},
\begin{align}
    \mc{A}\m(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp(\mc{R}^{\dg}\n\mc{A})&=\mc{A},\label{ea1}\\
    (\mc{A}\m\mc{T}^{\dg})\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}&=\mc{A}.\label{ea2}
\end{align}
Pre-multiplying equation \eqref{ea1} and post-multiplying equation
\eqref{ea2} by $\mc{A}^{\dg}$, and using Lemma \ref{ir}, we have
\begin{align}
    (\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}\n\mc{A}=\mc{A}^{\dg}\n\mc{A},\label{Ea+a}\\
    \mc{A}\m\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}=\mc{A}\m\mc{A}^{\dg},\label{Eaa+}
\end{align}
respectively, since $\mc{A}^{\dg}\n\mc{A}$ and
$\mc{A}\m\mc{A}^{\dg}$ are Hermitian.
$\mc{A}^{\dg}=\mc{A}^{\dg}\n\mc{A}\m\mc{A}^{\dg}$ with the last two
expressions yields
$$\mc{A}^{\dg}=(\mc{R}\kp\mc{S}\lp\mc{T})^{\dg}=(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}.$$
And
$\mc{A}^{\dg}=(\mc{R}\kp\mc{S}\lp\mc{T})^{\dg}=(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{S}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg},$
is due to Lemma \ref{ir}.
\end{proof}

 The following corollary immediately follows from the above result when $\mc{S}=\mc{I}\in{\C}^{J_{1}\times\cdots\times J_{M}\times J_{1}\times \cdots\times J_{M}}$.

\begin{corollary}\label{caegi}
For any tensor $\mc{M}\in{\C}^{I_{1}\times\cdots\times I_{N}\times
J_{1}\times \cdots\times J_{M}}$ and
$\mc{N}\in{\C}^{J_{1}\times\cdots\times J_{M}\times H_{1}\times
\cdots\times H_{K}}$,
$(\mc{M}\m\mc{N})^{\dg}=(\mc{M}^{\dg}\n\mc{M}\m\mc{N})^{\dg}\n(\mc{M}\m\mc{N}\kp\mc{N}^{\dg})^{\dg}$.
\end{corollary}
\begin{proof} This follows from Theorem \ref{aegi}, by setting $\mc{R}=\mc{M}$, $\mc{S}=\mc{I}\in{\C}^{J_{1}\times\cdots\times J_{M}\times J_{1}\times \cdots\times J_{M}}$, $\mc{T}=\mc{N}$ and $\mc{A}=\mc{M}\n\mc{N}$.
\end{proof}

We observed that $\mc{P}^{\dg}=\mc{P}$ for a Hermitian idempotent
tensor $\mc{P}$. Combining this fact with the expression for
$(\mc{M}\n\mc{N})^{\dg}$ in Corollary \ref{caegi} gives the
following result.

\begin{corollary}
Let $\mc{M}\in{\C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times
\cdots\times I_{N}}$ and $\mc{N}\in{\C}^{I_{1}\times\cdots\times
I_{N}\times I_{1}\times \cdots\times I_{N}}$ be any two Hermitian
idempotent tensors, then $(\mc{M}\n\mc{N})^{\dg}$ is idempotent.

\end{corollary}

\subsection{Relation between $\mc{A}_{\pi\dg}$ and $\mc{A}^{\dg}$}

Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Let us also denote
$\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}$ and
$(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}$ as $\mc{B}$ and
$\mc{C}$, respectively. With these notations, Theorem \ref{aepgi}
and Theorem \ref{aegi} result
\begin{equation}\label{Eepgi}
    \mc{A}_{\pi \dg}=\mc{B}\n\mc{A}\m\mc{C},
\end{equation}
 and \begin{equation}\label{Egi}
  ~~ \mc{A}^{\dg}=\mc{C}\n\mc{A}\m\mc{B}.
\end{equation}
Also,
\begin{equation}\label{eu}
    \mc{A}_{\pi\dg}\n\mc{A}\m\mc{A}^{\dg}=\mc{B},
\end{equation}
and
\begin{equation}\label{ev}
\mc{A}^{\dg}\n\mc{A}\m\mc{A}_{\pi\dg}=\mc{C},
\end{equation}
follow from Theorem \ref{aepgi} and Lemma \ref{ir}. In the next
result, we represent $\mc{B}$ and $\mc{C}$ as product Moore-Penrose
inverse.

\begin{theorem}\label{Tapgi}

Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$.
\begin{itemize}
\item[(i)] If $\mc{A}=\mc{A}\m(\mc{A}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp\mc{T}$, then $\mc{A}_{\pi\dg}=\mc{B}$.
\item[(ii)] If $\mc{A}=\mc{R}\kp(\mc{R}^{\dg}\n\mc{A}\m\mc{A}^{\dg})\n\mc{A}$, then $\mc{A}_{\pi\dg}=\mc{C}$.
\end{itemize}
\end{theorem}
\begin{proof} $(i)$ Applying Theorem \ref{aepgi} to $\mc{A}\m(\mc{A}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp\mc{T}=\mc{A}\m\mc{T}^{\dg}\lp\mc{T}$ results
\begin{multline*}
[\mc{A}\m(\mc{A}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp\mc{T}]_{\pi\dg}=[\mc{A}\m\mc{T}^{\dg}\lp\mc{T}]_{\pi\dg}\\
=\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg}\lp\mc{T}\m\mc{T}^{\dg})^{\dg}\n\mc{A}\\
\m\mc{T}^{\dg}\lp\mc{T}\m(\mc{A}^{\dg}\n\mc{A}\m\mc{T}^{\dg}\lp\mc{T})^{\dg}\m\mc{A}^{\dg},
\end{multline*}
which reduces to
\begin{eqnarray*}
[\mc{A}\m(\mc{A}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp\mc{T}]_{\pi\dg}&=&\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}\n\mc{A}\m(\mc{A}^{\dg}\n\mc{A})^{\dg}\m\mc{A}^{\dg},
\end{eqnarray*}
due to equation \eqref{pgipe1}. Since $\mc{A}^{\dg}\n\mc{A}$ is
Hermitian, we have
$$[\mc{A}\m(\mc{A}^{\dg}\n\mc{A}\m\mc{T}^{\dg})\lp\mc{T}]_{\pi\dg}=\mc{T}^{\dg}\lp(\mc{A}\m\mc{T}^{\dg})^{\dg}=\mc{B},$$
by Lemma \ref{ir}.\\
$(ii)$ Similarly, we have
$\mc{C}=[\mc{R}\kp(\mc{R}^{\dg}\n\mc{A}\m\mc{A}^{\dg})\n\mc{A}]_{\pi\dg}.$
\end{proof}

Then using  Theorem \ref{Tpgi} and Theorem \ref{Tapgi}, it is easy
to show that $\mc{B}\n\mc{A}\m\mc{B}=\mc{B}$ and
$\mc{C}\n\mc{A}\m\mc{C}=\mc{C}$.

We next state two results on $\mc{B}$ and $\mc{C}$ with the help of
the expressions for
$\mc{A}\m\mc{B},~\mc{B}\n\mc{A},~\mc{A}\m\mc{C},$ and
$\mc{C}\n\mc{A}$ obtained from equation \eqref{eu} and equation
\eqref{ev}.

\begin{remark}\label{CaepgiU}
The system of tensor equations $\mc{X}\n\mc{A}\m\mc{X}=\mc{X}$,
$\mc{A}\m\mc{X}=\mc{A}\m\mc{A}^{\dg}$ and
$\mc{X}\n\mc{A}=\mc{A}_{\pi \dg}\n\mc{A}$
($\mc{A}\m\mc{X}=\mc{A}\m\mc{A}_{\pi \dg}$ and
$\mc{X}\n\mc{A}=\mc{A}^{\dg}\n\mc{A}$) has unique solution
$\mc{X}=\mc{B}$ $(\mc{X}=\mc{C})$.
\end{remark}

 In terms of $\mc{B}$ and $\mc{C}$, Theorem \ref{aepgi}, the equations \eqref{Ea+a} and \eqref{Eaa+}  can be rewritten as following.

\begin{remark}
$\mc{X}=\mc{A}^{\dg}(\mc{X}=\mc{A}_{\pi\dg})$ is the unique matrix
such that $\mc{X}\n\mc{A}\m\mc{X}=\mc{X}$,
$\mc{A}\m\mc{X}=\mc{A}\m\mc{B}$ and $\mc{X}\n\mc{A}=\mc{C}\n\mc{A}$
($\mc{A}\m\mc{X}=\mc{A}\m\mc{C}$ and
$\mc{X}\n\mc{A}=\mc{B}\n\mc{A}$).
\end{remark}

The next example confirms that  any two of
$\mc{A}^{\dg},~\mc{A}_{\pi\dg},~\mc{B}$ and $\mc{C}$ need not
coincide for a given factorization
$\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times \cdots\times
I_{N}\times J_{1}\times\cdots \times J_{M}}$, where $\mc{R}\in
{\C}^{I_{1}\times \cdots\times I_{N}\times H_{1}\times\cdots \times
H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times \cdots\times H_{K}\times
G_{1}\times\cdots \times G_{L}}$ and $\mc{T}\in
{\C}^{G_{1}\times\cdots \times G_{L}\times J_{1}\times\cdots \times
J_{M}}$.
\begin{example}
Consider the tensor $\mc{A}=(a_{ijkl})_{1\leq
i,j,k,l\leq2}\in{\C}^{2\times2\times2\times 2}$, such that

\vspace{0.5cm} \noindent
$$
a_{ij11}=
        \begin{pmatrix}
              1 & 0\\
              0 & -1
        \end{pmatrix},~
a_{ij12}=
        \begin{pmatrix}
              0 & -1\\
              0 & 0
        \end{pmatrix},~
a_{ij21}=
        \begin{pmatrix}
             0 & 1\\
             0 & -1
        \end{pmatrix},~
a_{ij22}=
        \begin{pmatrix}
            0 & 1\\
            0 & 0
        \end{pmatrix}.
$$

Suppose that $\mc{A}$ is factorized as  $\mc{A}=\mc{R}\2\mc{S}\2
\mc{T}$, where $\mc{R}=(r_{ijkl})\in{\C}^{2\times2\times2\times
2},~\mc{S}=(s_{ijkl})\in{\C}^{2\times2\times2\times
2},~\mc{T}=(t_{ijkl})\in{\C}^{2\times2\times2\times 2},~1\leq
i,j,k,l\leq2$ such that

\vspace{0.5cm} \noindent
$$
r_{ij11}=
         \begin{pmatrix}
                  0 & 1\\
                  1 & 0
         \end{pmatrix},~
r_{ij12}=
         \begin{pmatrix}
                      -1 & 0\\
                       1 & 0
          \end{pmatrix},~
r_{ij21}=
          \begin{pmatrix}
                       0 & 1\\
                       0 & 1
          \end{pmatrix},~
r_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     1 & 1
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
s_{ij11}=
        \begin{pmatrix}
                     1 & 0\\
                     1 & 0
        \end{pmatrix},~
s_{ij12}=
        \begin{pmatrix}
                     0 & -1\\
                     0 & 1
        \end{pmatrix},~
s_{ij21}=
        \begin{pmatrix}
                     1 & 0\\
                     0 & 0
        \end{pmatrix},~
s_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     1 & 0
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent and
$$
t_{ij11}=
        \begin{pmatrix}
                     1 & 0\\
                     0 & 0
        \end{pmatrix},~
t_{ij12}=
        \begin{pmatrix}
                     0 & 0\\
                     1 & 0
        \end{pmatrix},~
t_{ij21}=
        \begin{pmatrix}
                     0 & 1\\
                     0 & 0
        \end{pmatrix},~
t_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     1 & -1
        \end{pmatrix}.
$$

\vspace{0.5cm} \noindent Now, $\mc{A}^{\dg}=(a'_{ijkl})_{1\leq
i,j,k,l\leq2}\in{\C}^{2\times2\times2\times 2}$,
$\mc{R}^{\dg}=(r'_{ijkl})_{1\leq
i,j,k,l\leq2}\in{\C}^{2\times2\times2\times 2}$,
$\mc{S}^{\dg}=(s'_{ijkl})_{1\leq
i,j,k,l\leq2}\in{\C}^{2\times2\times2\times 2}$, and
$\mc{T}^{\dg}=(t'_{ijkl})_{1\leq
i,j,k,l\leq2}\in{\C}^{2\times2\times2\times 2}$ are given by

\vspace{0.5cm} \noindent
$$
a'_{ij11}=
         \begin{pmatrix}
                      1 & -1\slash 2\\
                      -1 & 1\slash2
        \end{pmatrix},~
a'_{ij12}=
         \begin{pmatrix}
                      0&-1\slash 2\\
                      0 & 1\slash2
        \end{pmatrix},~
a'_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 0
        \end{pmatrix},~
a'_{ij22}=
         \begin{pmatrix}
                      0 & -1\slash 2\\
                      -1 & 1\slash2
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
r'_{ij11}=
         \begin{pmatrix}
                      1 & -1\\
                      -1 & 1
         \end{pmatrix},~
r'_{ij12}=
         \begin{pmatrix}
                      0 & 0\\
                      1 & -1
        \end{pmatrix},~
r'_{ij21}=
         \begin{pmatrix}
                      1 & 0\\
                      -1 & 1
         \end{pmatrix},~
r'_{ij22}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 1
         \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
s'_{ij11}=
         \begin{pmatrix}
                      1\slash3 & 0\\
                      2\slash3 & -1\slash3
         \end{pmatrix},~
s'_{ij12}=
         \begin{pmatrix}
                      0 & -1\slash2\\
                      0 & 0
        \end{pmatrix},~
s'_{ij21}=
         \begin{pmatrix}
                      1\slash3 & 0\\
                      -1\slash3 & 2\slash3
         \end{pmatrix},~
s'_{ij22}=
         \begin{pmatrix}
                      0 & 1\slash2\\
                      0 & 0
         \end{pmatrix};
$$

\vspace{0.5cm} \noindent and
$$
t'_{ij11}=
         \begin{pmatrix}
                      1 & 0\\
                      0 & 0
         \end{pmatrix},~
t'_{ij12}=
         \begin{pmatrix}
                      0 & 0\\
                      1 & 0
         \end{pmatrix},~
t'_{ij21}=
         \begin{pmatrix}
                      0 & 1\\
                      0 & 1
         \end{pmatrix},~
t'_{ij22}=
         \begin{pmatrix}
                      0 & 1\\
                      0 & -1
         \end{pmatrix}.
$$

\vspace{0.2cm} \noindent
 Here, $\mc{R}^{\dg}\2\mc{A}\2\mc{T}^{\dg}=\mc{S}$, then  $\mc{A}_{\pi\dg}=\mc{T}^{\dg}\2(\mc{R}^{\dg}\2\mc{A}\2\mc{T}^{\dg})^{\dg}\2\mc{R}^{\dg}=\mc{T}^{\dg}\2\mc{S}^{\dg}\2\mc{R}^{\dg}=(x_{ijkl})$,  $\mc{B}=\mc{A}_{\pi\dg}\2\mc{A}\2\mc{A}^{\dg}=(u_{ijkl})$, and $\mc{C}=\mc{A}^{\dg}\2\mc{A}\2\mc{A}_{\pi\dg}=(v_{ijkl})$, where

\vspace{0.5cm} \noindent
 $$
 x_{ij11}=
         \begin{pmatrix}
                      1 & -1\slash3\\
                      -1 & 2\slash3
         \end{pmatrix},~
x_{ij12}=
         \begin{pmatrix}
                      0 & -1\slash3\\
                      0 & 2\slash3
         \end{pmatrix},~
x_{ij21}=
        \begin{pmatrix}
                     0 & 0\\
                     -1\slash2 & 1\slash2
        \end{pmatrix},~
x_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     -1 & 1
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
 u_{ij11}=
         \begin{pmatrix}
                      1 & -1\slash2\\
                      -1 & 1\slash2
         \end{pmatrix},~
u_{ij12}=
        \begin{pmatrix}
                     0 & -1\slash2\\
                     0 & 1\slash2
        \end{pmatrix},
u_{ij21}=
        \begin{pmatrix}
                     0 & -1\slash4\\
                     -1\slash2 & 1\slash4
        \end{pmatrix},~
u_{ij22}=
        \begin{pmatrix}
                     0 & -1\slash2\\
                     -1 & 1\slash2
        \end{pmatrix};
 $$

\vspace{0.5cm} \noindent and
$$
v_{ij11}=
        \begin{pmatrix}
                     1 & -1\slash3\\
                     -1 & 2\slash3
        \end{pmatrix},~
v_{ij12}=
        \begin{pmatrix}
                     0 & -1\slash3\\
                     0 & 2\slash3
        \end{pmatrix},~
v_{ij21}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 0
        \end{pmatrix},~~and~~
v_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     -1 & 1
        \end{pmatrix}.
$$
\end{example}
At this point one may be interested to know when does the
\textit{product Moore-Penrose inverse} coincide with the
\textit{Moore-Penrose inverse}? The answer to this question is
explained in the following theorem.

\begin{theorem}\label{Tnscpgi}
Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then
$\mc{A}_{\pi\dg}=\mc{A}^{\dg}$ if and only if
$(\mc{R}^{\dg}\n\mc{A}\m
\mc{T}^{\dg})^{\dg}=\mc{T}\m\mc{A}^{\dg}\n\mc{R}$. In that case
$\mc{A}_{\pi\dg}=\mc{A}^{\dg}=\mc{B}=\mc{C}$.
\end{theorem}
\begin{proof}
Suppose
$(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}=\mc{T}\m\mc{A}^{\dg}\n\mc{R}$,
then by equation \eqref{epgi} and Lemma \ref{ir}, we have
\begin{eqnarray*}
\mc{A}_{\pi\dg}&=&\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}\\
&=&\mc{T}^{\dg}\lp\mc{T}\m\mc{A}^{\dg}\n\mc{R}\kp\mc{R}^{\dg}\\
&=&\mc{A}^{\dg}.
\end{eqnarray*}
And using equation \eqref{eu} and \eqref{ev}, we have
$\mc{B}=\mc{A}_{\pi\dg}\n\mc{A}\m\mc{A}^{\dg}=\mc{A}^{\dg}\n\mc{A}\m\mc{A}^{\dg}=\mc{A}^{\dg}=\mc{A}^{\dg}\n\mc{A}\m\mc{A}^{\dg}=\mc{A}^{\dg}\n\mc{A}\m\mc{A}_{\pi\dg}=\mc{C}.$

Conversely, suppose that $\mc{A}_{\pi\dg}=\mc{A}^{\dg}$. By using
equation \eqref{pgipe1} and Lemma \ref{ir}, we then have
\begin{align}
    \mc{T}\m\mc{A}^{\dg}\n\mc{R}&=\mc{T}\m\mc{A}_{\pi\dg}\n\mc{R}\nonumber\\
    &=\mc{T}\m\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}\n\mc{R}\label{Enscpgi}\\
    &=\mc{T}\m\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\nonumber\\
    &=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\nonumber.
\end{align}

\end{proof}

From the proof of Theorem \ref{Tnscpgi}, it is clear that
$\mc{A}_{\pi\dg}=\mc{A}^{\dg}$ implies $\mc{B}=\mc{C}$. However, the
converse is also true. Because if $\mc{B}=\mc{C}$, then the relation
$\mc{B}\n\mc{A}\m\mc{B}=\mc{B}$ with equation \eqref{Eepgi} and
equation \eqref{Egi} results $\mc{B}=\mc{A}_{\pi\dg}=\mc{A}^{\dg}$.
Thus, Theorem \ref{Tnscpgi} also can be restated as follows.

\begin{remark}
Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then,
$\mc{A}_{\pi\dg}=\mc{A}^{\dg}$ if and only if $\mc{B}=\mc{C}$, in
which case $\mc{A}_{\pi\dg}=\mc{A}^{\dg}=\mc{B}=\mc{C}$.
\end{remark}

One may be interested on the case that either of $\mc{B}$ or
$\mc{C}$ coincides with $\mc{A}^{\dg}$ but $\mc{B}\neq\mc{C}$. The
answer is shown below.

\begin{theorem}
Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then,
$\mc{B}=\mc{A}^{\dg}(\mc{B}=\mc{A}_{\pi\dg})$ if and only if
$\mc{C}=\mc{A}_{\pi\dg}(\mc{C}=\mc{A}^{\dg})$.
\end{theorem}
\begin{proof}
Suppose that $\mc{B}=\mc{A}^{\dg}$. By equation \eqref{Eepgi}, we
have  $\mc{A}_{\pi\dg}=\mc{A}^{\dg}\n\mc{A}\m\mc{C}$. But
$\mc{C}=(\mc{R}^{\dg}\n\mc{A})^{\dg}\kp\mc{R}^{\dg}$ results
$\mc{A}_{\pi\dg}=\mc{C}$.

Conversely, suppose that $\mc{C}=\mc{A}_{\pi\dg}$. By equation
\eqref{ev} we get
$\mc{A}_{\pi\dg}=\mc{A}^{\dg}\n\mc{A}\m\mc{A}_{\pi\dg}$.
Pre-multiplying $\mc{A}^{\dg}\n\mc{A}$ to equation \eqref{eu}
results
$\mc{A}^{\dg}\n\mc{A}\m\mc{B}=\mc{A}_{\pi\dg}\n\mc{A}\m\mc{A}^{\dg}=\mc{B}.$
By Remark \ref{CaepgiU}, we have
$\mc{A}\m\mc{B}=\mc{A}\m\mc{A}^{\dg}$. Thus $\mc{A}^{\dg}=\mc{B}$.
\end{proof}

The next result gives a necessary and sufficient condition for
$\mc{A}_{\pi\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$.

\begin{theorem}
Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then,
$\mc{A}_{\pi\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$ if and
only if
$\mc{S}^{\dg}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}+\mc{Y}$,
where $\mc{Y}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
H_{1}\times\cdots \times H_{K}} $ satisfies
$\mc{T}^{\dg}\lp\mc{Y}\kp\mc{R}^{\dg}=\mc{O}$.
\end{theorem}
\begin{proof}
Suppose that
$\mc{S}^{\dg}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}+\mc{Y}$. By
using equation \eqref{epgi}, we then have
$\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}=\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}+\mc{T}^{\dg}\lp\mc{Y}\kp\mc{R}^{\dg}=\mc{A}_{\pi\dg}$.

Conversely, suppose
that$\mc{A}_{\pi\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$,
then by equation \eqref{epgi} and the last equality of equation
\eqref{Enscpgi}
\begin{equation}
    \mc{T}\m\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}\n\mc{R}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}.
\end{equation}
Hereafter, by applying equation \eqref{ir2} to
$\mc{T}\m\mc{T}^{\dg}\lp\mc{X}\kp\mc{R}^{\dg}\n\mc{R}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}$,
it follows that there exists a $\mc{Z}$ such that
\begin{equation}
    \mc{S}^{\dg}=(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}+\mc{Z}-\mc{T}\m\mc{T}^{\dg}\lp\mc{Z}\kp\mc{R}^{\dg}\n\mc{R},
\end{equation}
due to the equation \eqref{Enscpgi} and the fact
$(\mc{T}\m\mc{T}^{\dg})^{\dg}=\mc{T}\m\mc{T}^{\dg}$ and
$(\mc{R}^{\dg}\n\mc{R})^{\dg}=\mc{R}^{\dg}\n\mc{R}$. Thus, if
$\mc{Y}=\mc{Z}-\mc{T}\m\mc{T}^{\dg}\lp\mc{Z}\kp\mc{R}^{\dg}\n\mc{R}$,
then
\begin{eqnarray*}
    \mc{T}^{\dg}\lp\mc{Y}\kp\mc{R}^{\dg}&=&\mc{T}^{\dg}\lp\mc{Z}\kp\mc{R}^{\dg}-\mc{T}^{\dg}\lp\mc{T}\m\mc{T}^{\dg}\lp\mc{Y}\kp\mc{R}^{\dg}\n\mc{R}\kp\mc{R}^{\dg}\\
    &=&\mc{O}.
\end{eqnarray*}
\end{proof}

Next, we provide an example to show that it may be possible that
$\mc{A}^{\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$ while
$\mc{A}_{\pi\dg}\neq\mc{A}^{\dg}$.

\begin{example}\label{exmppgi}
Let $\mc{A}=(a_{ijkl})\in{\C}^{2\times2\times2\times 2},~1\leq
i,j,k,l\leq2$. Suppose that $\mc{A}$ is factorized as
$\mc{A}=\mc{R}\2\mc{S}\2 \mc{T}$, where
$\mc{R}=(r_{ijkl})\in{\C}^{2\times2\times2\times
2},~\mc{S}=(s_{ijkl})\in{\C}^{2\times2\times2\times
2},~\mc{T}=(t_{ijkl})\in{\C}^{2\times2\times2\times 2},~1\leq
i,j,k,l\leq2$ such that

\vspace{0.5cm} \noindent
$$
a_{ij11}=
        \begin{pmatrix}
              1 & 1\\
              1 & 1
        \end{pmatrix},~
a_{ij12}=
        \begin{pmatrix}
              0 & 0\\
              0 & 0
        \end{pmatrix},~
a_{ij21}=
        \begin{pmatrix}
             0 & 1\\
             1 & 1
        \end{pmatrix},~
a_{ij22}=
        \begin{pmatrix}
            0 & 0\\
            0 & 0
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent

$$
r_{ij11}=
         \begin{pmatrix}
                  1 & 0\\
                  0 & 0
         \end{pmatrix},~
r_{ij12}=
         \begin{pmatrix}
                      0 & 1\\
                      1 & 0
          \end{pmatrix},~
r_{ij21}=
          \begin{pmatrix}
                       0 & 0\\
                       1 & 0
          \end{pmatrix},~
r_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 0
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
s_{ij11}=
        \begin{pmatrix}
                     1 & 0\\
                     0 & 0
        \end{pmatrix},~
s_{ij12}=
        \begin{pmatrix}
                     0 & 1\slash2\\
                     -3\slash2 & 0
        \end{pmatrix},~
s_{ij21}=
        \begin{pmatrix}
                     0 & 1\slash2\\
                     1\slash2 & 0
        \end{pmatrix},~
s_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 0
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent and
$$
t_{ij11}=
        \begin{pmatrix}
                     1 & 1\\
                     1 & 1
        \end{pmatrix},~
t_{ij12}=
        \begin{pmatrix}
                     0 & 1\\
                     1 & 1
        \end{pmatrix},~
t_{ij21}=
        \begin{pmatrix}
                     0 & 1\\
                     1 & 1
        \end{pmatrix},~
t_{ij22}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 0
        \end{pmatrix}.
$$

\noindent Now, $\mc{A}^{\dg}=(a'_{ijkl})$,
$\mc{R}^{\dg}=(r'_{ijkl})$, $\mc{S}^{\dg}=(s'_{ijkl})$, and
$\mc{T}^{\dg}=(t'_{ijkl})$ are given by
$$
a'_{ij11}=
         \begin{pmatrix}
                      1 & 0\\
                      -1 & 0
        \end{pmatrix},~
a'_{ij12}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash3 & 0
        \end{pmatrix},~
a'_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash3 & 0
        \end{pmatrix},~
a'_{ij22}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash3 & 0
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
r'_{ij11}=
         \begin{pmatrix}
                      1 & 0\\
                      0 & 0
         \end{pmatrix},~
r'_{ij12}=
         \begin{pmatrix}
                      0 & 1\\
                      -1 & 0
        \end{pmatrix},~
r'_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      1 & 0
         \end{pmatrix},~
r'_{ij22}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 0
         \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
s'_{ij11}=
         \begin{pmatrix}
                      1 & 0\\
                      0 & 0
         \end{pmatrix},~
s'_{ij12}=
         \begin{pmatrix}
                      0 & 1\slash2\\
                      3\slash2 & 0
        \end{pmatrix},~
s'_{ij21}=
         \begin{pmatrix}
                      0 & -1\slash2\\
                      1\slash2 & 0
         \end{pmatrix},~
s'_{ij22}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 0
         \end{pmatrix};
$$

\vspace{0.5cm} \noindent and
$$
t'_{ij11}=
         \begin{pmatrix}
                      1 & -1\slash2\\
                      -1\slash2 & 0
         \end{pmatrix},~
t'_{ij12}=
         \begin{pmatrix}
                      0 & 1\slash6\\
                      1\slash6 & 0
         \end{pmatrix},~
t'_{ij21}=
         \begin{pmatrix}
                      0 & 1\slash6\\
                      1\slash6 & 0
         \end{pmatrix},~
t'_{ij22}=
         \begin{pmatrix}
                     0 & 1\slash6\\
                      1\slash6 & 0
         \end{pmatrix}.
$$

\vspace{0.2cm} \noindent Here,
$\mc{T}^{\dg}\2\mc{S}^{\dg}\2\mc{R}^{\dg}=(y_{ijkl})$, where
 \begin{eqnarray*}
y_{ij11}=
         \begin{pmatrix}
                      1 & 0\\
                      -1 & 0
        \end{pmatrix},~
y_{ij12}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash3 & 0
        \end{pmatrix},~
y_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash3 & 0
        \end{pmatrix},~
y_{ij22}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash3 & 0
        \end{pmatrix}.
\end{eqnarray*}
But,
$\mc{A}_{\pi\dg}=\mc{T}^{\dg}\2(\mc{R}^{\dg}\2\mc{A}\2\mc{T}^{\dg})^{\dg}\2\mc{R}^{\dg}=(x_{ijkl})$,
where
 \begin{eqnarray*}
 x_{ij11}=
         \begin{pmatrix}
                      1 & 1\slash2\\
                      -1 & 0
         \end{pmatrix},~
x_{ij12}=
         \begin{pmatrix}
                      0 & -1\slash6\\
                      1\slash3 & 0
         \end{pmatrix},~
x_{ij21}=
        \begin{pmatrix}
                     0 & -1\slash6\\
                      1\slash3 & 0
        \end{pmatrix},~
x_{ij22}=
        \begin{pmatrix}
                     0 & -1\slash6\\
                      1\slash3 & 0
        \end{pmatrix}.
\end{eqnarray*}
Here, in this example $\mc{A}_{\pi\dg}\neq\mc{A}^{\dg}$, while
$\mc{A}^{\dg}=\mc{T}^{\dg}\2\mc{S}^{\dg}\2\mc{R}^{\dg}$.
\end{example}

In case of the Moore-Penrose inverse, we observed that the
Moore-Penrose inverse of $\mc{A}^{\dg}$ reduces to $\mc{A}$ for
every tensor $\mc{A}$ \cite{sun}. This fact leads to a natural
question ``Does the product Moore-Penrose inverse of
$\mc{A}_{\pi\dg}$ also reduces to $\mc{A}$?'' The answer is
\textit{affirmative}, if we use the same factorization employed in
forming $\mc{A}_{\pi\dg}$. The following theorem proves this fact.

\begin{theorem}
Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then,
$$(\mc{A}_{\pi\dg})_{\pi\dg}=[\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}]_{\pi\dg}=\mc{A}.$$
\end{theorem}
\begin{proof}
Applying equation \eqref{epgi} to
$\mc{A}_{\pi\dg}=\mc{T}^{\dg}\lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}$
gives
\begin{equation}\label{le}
    (\mc{A}_{\pi\dg})_{\pi\dg}=\mc{R}\kp[\mc{T}\m\mc{T}^{\dg} \lp(\mc{R}^{\dg}\n\mc{A}\m\mc{T}^{\dg})^{\dg}\kp\mc{R}^{\dg}\n\mc{R}]^{\dg}\lp\mc{T}.
\end{equation}
With the help of Lemma \ref{ir}, equation \eqref{le} gives
$(\mc{A}_{\pi\dg})_{\pi\dg}=\mc{A}$.

\end{proof}

\section{Conclusion}

Note that in Example \ref{exmppgi},
$\mc{A}^{\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$ holds.  A
natural question arises now  regarding the trueness of the same
equality which is called as the \textit{triple reverse order law} of
tensors via the Einstein product, i.e., $(\mc{R} \kp \mc{S} \lp
\mc{T})^{\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$.  However,
this is not true in general, and  is shown next through an example.

\begin{example}
Let $\mc{R}=(r_{ijkl})\in{\C}^{2\times2\times2\times
2},~\mc{S}=(s_{ijkl})\in{\C}^{2\times2\times2\times
2},~\mc{T}=(t_{ijkl})\in{\C}^{2\times2\times2\times 2},~1\leq
i,j,k,l\leq2$ such that

\vspace{0.5cm}

\noindent
$$
r_{ij11}=
         \begin{pmatrix}
                  1 & 0\\
                  -1 & 0
         \end{pmatrix},~
r_{ij12}=
         \begin{pmatrix}
                      1 & 1\\
                      0 & 1
          \end{pmatrix},~
r_{ij21}=
          \begin{pmatrix}
                       0 & 1\\
                       0 & 1
          \end{pmatrix},~
r_{ij22}=
        \begin{pmatrix}
                     -1 & 0\\
                     1 & 0
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
s_{ij11}=
        \begin{pmatrix}
                     0 & 1\\
                     0 & -1
        \end{pmatrix},~
s_{ij12}=
        \begin{pmatrix}
                     1 & 1\\
                     1 & 0
        \end{pmatrix},~
s_{ij21}=
        \begin{pmatrix}
                     1 & 0\\
                     1 & 0
        \end{pmatrix},~
s_{ij22}=
        \begin{pmatrix}
                     0 & -1\\
                     0 & 1
        \end{pmatrix};
$$

\vspace{0.5cm} \noindent and
$$
t_{ij11}=
        \begin{pmatrix}
                     1 & 0\\
                     0 & 0
        \end{pmatrix},~
t_{ij12}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 1
        \end{pmatrix},~
t_{ij21}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 0
        \end{pmatrix},~
t_{ij22}=
        \begin{pmatrix}
                     0 & 1\\
                     0 & 0
        \end{pmatrix}.
$$

\vspace{0.2cm} \noindent
And let $\mc{A}=\mc{R}\2\mc{S}\2 \mc{T}=(a_{ijkl})\in{\C}^{2\times2\times2\times 2},~1\leq i,j,k,l\leq2$, where \\

\vspace{0.2cm} \noindent
$$
a_{ij11}=
        \begin{pmatrix}
              -1 & -1\\
              0 & 1
        \end{pmatrix},~
a_{ij12}=
        \begin{pmatrix}
              1 & 0\\
              0 & 1
        \end{pmatrix},~
a_{ij21}=
        \begin{pmatrix}
             1 & 1\\
             0 & 0
        \end{pmatrix},~
a_{ij22}=
        \begin{pmatrix}
            1 & 1\\
            0 & -1
        \end{pmatrix}.
$$

\vspace{0.2cm} \noindent
Now, $\mc{A}^{\dg}=(a'_{ijkl})$, $\mc{R}^{\dg}=(r'_{ijkl})$, $\mc{S}^{\dg}=(s'_{ijkl})$, and $\mc{T}^{\dg}=(t'_{ijkl})$ are given by \\

\vspace{0.2cm} \noindent
$$
r'_{ij11}=
         \begin{pmatrix}
                      0 & 1\\
                      -1 & 0
         \end{pmatrix},~
r'_{ij12}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash 2 & 0
        \end{pmatrix},~
r'_{ij21}=
         \begin{pmatrix}
                      -1\slash2 & 1\\
                      -1 & 1\slash 2
         \end{pmatrix},~
r'_{ij22}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash 2 & 0
         \end{pmatrix};
$$

\vspace{0.5cm} \noindent
$$
s'_{ij11}=
         \begin{pmatrix}
                      10 & 0\\
                      1\slash 2 & 0
         \end{pmatrix},~
s'_{ij12}=
         \begin{pmatrix}
                      0 & 1\\
                      -1 & 0
        \end{pmatrix},~
s'_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      1\slash2 & 0
         \end{pmatrix},~
s'_{ij22}=
         \begin{pmatrix}
                      -1\slash 2 & 1\\
                      -1 & 1\slash 2
         \end{pmatrix};
$$

\vspace{0.5cm} \noindent and
$$
t'_{ij11}=
         \begin{pmatrix}
                      1 & 0\\
                      0 & 0
         \end{pmatrix},~
t'_{ij12}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 1
         \end{pmatrix},~
t'_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 0
         \end{pmatrix},~
t'_{ij22}=
         \begin{pmatrix}
                     0 & 1\\
                     0 & 0
         \end{pmatrix}.
$$

\vspace{0.2cm} \noindent
Here, $\mc{A}^{\dg}=(\mc{R}\2\mc{S}\2\mc{T})^{\dg}=(a'_{ijkl})$, where \\

\vspace{0.2cm} \noindent
$$
a'_{ij11}=
         \begin{pmatrix}
                      -1\slash 2 & 1\\
                      -1 & 1\slash 2
        \end{pmatrix},~
a'_{ij12}=
         \begin{pmatrix}
                      1\slash 2 & -1\\
                      2 & -1\slash 2
        \end{pmatrix},~
a'_{ij21}=
         \begin{pmatrix}
                      0 & 0\\
                      0 & 0
        \end{pmatrix},~
a'_{ij22}=
         \begin{pmatrix}
                      1\slash 2 & 0\\
                      1 & -1\slash 2
        \end{pmatrix}.
$$

\vspace{0.2cm} \noindent But,
$\mc{T}^{\dg}\2\mc{S}^{\dg}\2\mc{R}^{\dg}=(x_{ijkl})$, where
 \begin{eqnarray*}
x_{ij11}=
         \begin{pmatrix}
                      -1\slash 4 & 1\slash2\\
                      -1\slash 2 & 1\slash 4
         \end{pmatrix},~
x_{ij12}=
         \begin{pmatrix}
                      1\slash 2 & -3\slash2\\
                      9\slash4 & -1\slash2
         \end{pmatrix},~
x_{ij21}=
        \begin{pmatrix}
                     0 & 0\\
                     0 & 0
        \end{pmatrix},~
x_{ij22}=
        \begin{pmatrix}
                     1\slash2 & -1\\
                      3\slash2 & -1\slash 2
        \end{pmatrix}.
\end{eqnarray*}
Here, in this example
$(\mc{R}\2\mc{S}\2\mc{T})^{\dg}\neq\mc{T}^{\dg}\2\mc{S}^{\dg}\2\mc{R}^{\dg}$.
\end{example}

The following theorem gives a necessary and sufficient condition for
the triple reverse order law.

\begin{theorem}\label{trol1}
Let $\mc{A}=\mc{R}\kp\mc{S}\lp\mc{T}\in {\C}^{I_{1}\times
\cdots\times I_{N}\times J_{1}\times\cdots \times J_{M}}$, where
$\mc{R}\in {\C}^{I_{1}\times \cdots\times I_{N}\times
H_{1}\times\cdots \times H_{K}}$, $\mc{S}\in {\C}^{H_{1}\times
\cdots\times H_{K}\times G_{1}\times\cdots \times G_{L}}$ and
$\mc{T}\in {\C}^{G_{1}\times\cdots \times G_{L}\times
J_{1}\times\cdots \times J_{M}}$. Then,
$\mc{A}^{\dg}=\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg}$ if and
only if
\begin{itemize}
    \item[(i)] $\mc{R}^{\dg}\n(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})\m\mc{T}^{\dg}=\mc{R}^{\dg}\n(\mc{R}\kp\mc{S}\lp\mc{T})\m\mc{T}^{\dg},$
    \item[(ii)] $\mc{T}\m (\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n\mc{R}=\mc{T}\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n\mc{R},$
    \item[(iii)]$[\mc{R}^{H}\n(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n\mc{R}]^{H}=\mc{R}^{H}\n(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n\mc{R},$
    \item[(iv)]$[\mc{T} \m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})\m\mc{T}^{H}]^{H}=\mc{T}\m (\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})\m\mc{T}^{H}$.
\end{itemize}

\end{theorem}
\begin{proof}
It can be esily seen that $(i)$ and $(ii)$ are equivalent to
$(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\\\n(\mc{R}\kp\mc{S}\lp\mc{T})=\mc{R}\kp\mc{S}\lp\mc{T}$
and $
(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})=\mc{T}^{\dg}\lp\\\mc{S}^{\dg}\kp\mc{R}^{\dg}$,
respectively. $(iii)$ is equivalent to
$[(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})]^{H}=(\mc{R}\kp\mc{S}\lp\\\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})$.
Because
$[(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})]^{H}=(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\\\kp\mc{R}^{\dg})$
imply $(iii)$ and conversely, by pre-multiplying  $\mc{R}^{\dg *}$
and post-multiplying $\mc{R}^{\dg}$ to $(iii)$ gives
$[(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})]^{H}=(\mc{R}\kp\mc{S}\lp\mc{T})\m(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})$.
Similarly, $(iv)$ is equivalent to
$[(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})]^{H}=
(\mc{T}^{\dg}\lp\mc{S}^{\dg}\kp\mc{R}^{\dg})\n(\mc{R}\kp\mc{S}\lp\mc{T})$.
\end{proof}

 Note that the triple reverse order law  presented in the above theorem contains a number of conditions. We conclude the article with the note that  simpler characterization of triple reverse order law will be tried in near future.

{\small {\bf Acknowledgments.}\\

The authors acknowledge the support provided by Science and
Engineering Research Board, Department of Science and Technology,
New Delhi, India, under the grant number YSS/2015/000303. }

\section*{References}

\bibliographystyle{amsplain}


TITLE: Show that a point x€ Cl(A) iff there exists a sequence {xn} of points on A such that xn tends to X as n tends to infinity.
QUESTION [0 upvotes]: Suppose the sequence converges to X.Then |xn-x|k,now xn€A for all n€N.Therefore epsilon nbd of x contains  atleast one point of A other than X.So x is a limit point of A.x€A'.so x€ Cl(A).
Now let,x€Cl(A).Then x€A or x€A'.When x€A' then it is done that x is a limit of the sequence {xn}.But if we assume that x€A and X does not belong to A' then how to prove that the sequence {xn} converges to x?

REPLY [0 votes]: For such case, we can just set $x_n=x$ that is contained in $A$. Actually, such point $x \in A \backslash A'$ is called an isolated point.


TITLE: Solve $\int_{[0,\infty[}(\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)) d\lambda(y)$
QUESTION [0 upvotes]: Hint Required for :
$\int_{[0,\infty[}(\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)) d\lambda(y)$
My ideas:
Looking at the first inetgral $\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)$
I would say $\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)=\sin{y}\int_{[0,\infty[}e^{-xy}\sin{x}d\lambda(x)$
And we know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
So $\sin{y}\int_{[0,\infty[}e^{-xy}\sin{x}d\lambda(x)=\sin{y}\int_{[0,\infty[}e^{-xy}\frac{e^{ix}-e^{-ix}}{2i}d\lambda(x)$
and then I do not know what to do. Should I use this route or is partial integration better suited?
I've just started with multivariable integration, so any pointers would be of great assistance.

REPLY [1 votes]: HINT: note that the inner integral is equivalent to the imaginary part of the integral
$$\int_{[0,\infty)}e^{-xy}(\cos x+i\sin x)\, dx=\int_{[0,\infty)} e^{-x(y-i)}\, dx=-\frac{e^{-x(y-i)}}{y-i}\bigg|_0^\infty=\frac1{y-i}$$
where the result is taken assuming that $y> 0$ (the integral doesn't exists for $y=0$, however we can ignore this point because the set $\{0\}$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $y>0$, hence Lebesgue integrable.
Of course you can also use the identity $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ as you was doing, and you will find the same answer.

However the outer integral doesn't have a closed form solution.

REPLY [0 votes]: If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $x$ first, then $y$ second. Note that $\sin(x)$ is the imaginary part of $e^{ix}$, that should help you compute your integral better (you know the integral from $0$ to $\infty$ of $t \longmapsto e^{-pt}$ for reasonable complex numbers $p$, right?)


TITLE: Existence of complex branch for real exponents
QUESTION [4 upvotes]: I've recently encountered a problem regarding complex branches that made me feel like there is something fundamental about branches I do not understand:
The problem
Let $G$ be an open subset of $\mathbb{C}$ and $z_1,z_2,...z_m \in \mathbb C \setminus G$ be distinct points such that there are real numbers $a_1,a_2,...a_m$ satisfying:
$$\sum_{j=1}^ma_jInd_\gamma(z_j)=0$$
for every closed path $\gamma \subset G$.
Show that there is a holomorphic branch of $(z-z_1)^{a_1}(z-z_2)^{a_2}...(z-z_m)^{a_m}$ in $G$.
My confusion
The basic idea behind the statement of the problem seems intuitively obvious to me, but I've had great difficulty expressing a solution rigurously, which makes me believe there's something fundamental I'm missing here.
My approach was to try and utilize a theorem that states $logf$ has a holomorphic branch in $G$ if and only if $$\int_\gamma \frac{f'}{f}dz=0$$
for every closed path $\gamma \subset G$.
However, the only sensible way I can think of to define $f$ is by $f=\exp[\sum_{j=1}^{m}a_jlog(z-z_j)]$
Obviously if the exponent is holomorphic, so is $f$, but how can it be shown that it is indeed holomorphic? Induction does not work because the separate $log(z-z_j)$ may not exist on their own, even though their weighted sum does.
It is also quite simple to show that $f$ as defined in the problem satisfies the conditions of the theorem, and hence the sum of logarithms in the above exponent is indeed holomorphic.. but that presumes that $f$ exists (and is holomorphic) in the first place.
Can $f$ be defined in some other way that makes this all fall into place?

REPLY [3 votes]: I think it should be possible to prove an even stronger statement than the one stated in the question. Suppose that, instead of assuming that $\sum_j a_j {\rm \ Ind}_{\gamma} (z_j) = 0$ for all closed loops $\gamma \subset G$, we merely assume that
$$ \sum_j a_j {\rm \ Ind}_{\gamma} (z_j) \in \mathbb Z$$
for all closed loops $\gamma \subset G$. With this weaker assumption, I think it should still be possible to prove that $ \prod_{j = 1}^m (z - z_j)^{a_j}$ has a holomorphic branch in $G$.

Before we start addressing this question, we shall prove some useful lemmas.
Lemma 1: Suppose $f$ is a holomorphic function on an open set $U$ such that $\oint_\gamma f(z')dz' = 0$ for all closed contours $\gamma \subset U$. Then the function
$$F(z) = \int_{\gamma_{c \to z}} f(z') dz'$$
is well-defined and holomorphic in $U$, with derivative $F'(z) = f(z)$. (Here, $c$ is an arbitrary fixed point in $U$, and for every $z\in U$, $\gamma_{c \to z }$ denotes any path from $c$ to $z$ in $U$; the "well-defined" part of the statement is that the value of the integral in this definition does not depend on the choice of path $\gamma_{c \to z}$, as long as this path is contained within $U$.)
Proof: The well-definedness follows immediately from the fact that the difference between two paths $c \to z$ is a closed contour, on which the integral wanishes.
For holomorphicity, pick any $w \in U$, and pick any open ball $U_w \subset U$ of $w$. On this open ball, we can write
$$ F(z) = \int_{[w \to z]} f(z') dz' + F(w)$$
where $[w \to z]$ denotes the straight line segment from $w$ to $z$. Hence, for any $z \in U_w \setminus \{ w \}$,
$$  \frac{F(z) - F(w)}{z - w} - f(w) = \frac{1}{z - w} \int_{[w \to z]} \left( f(z') - f(w) \right) dz' $$
which tends to zero as $z$ tends to $w$, by the continuity of $f(z')$ at $w$.
Lemma 1a: Suppose $f$ is a holomorphic function on an open set $U$, such that $\oint_\gamma f(z') dz' \in 2\pi i \mathbb Z$ for all closed contours $\gamma \subset U$. Then the function
$$ g(z) = \exp\left( \int_{\gamma_{c \to z} }f(z') dz' \right)$$
is well-defined and holomorphic in $U$. (Again, for every $z \in U$, $\gamma_{c \to z}$ denotes any path from $c$ to $z$ that is contained within $U$.)
Proof: The well-definedness follows from the fact that if $\gamma_{c \to z}$ and $\gamma'_{c \to z}$ are two paths from $c$ to $z$, contained within $U$, then
$$ \exp \left( \oint_{\gamma'_{c \to z}} f(z') dz' \right) = \exp \left( \oint_{\gamma_{c \to z}} f(z') dz' + 2\pi i n\right) = \exp \left( \oint_{\gamma_{c \to z}} f(z') dz' \right). $$
To prove holomorphicity, pick any $w \in U$, and pick an open ball $U_w \subset U$ of $w$. Fix any path $\gamma_{c \to w}$, and consider the function $F : U_w \to \mathbb C$ defined by
$$ F(z) = \int_{\gamma_{w \to z}} f(z') dz' + k_c,$$
where $k_c = \int_{\gamma_{c \to w}} f(z') dz'$ with $\gamma_{c \to w}$ being our chosen fixed path, and where $\gamma_{w \to z}$ denotes any path from $w$ to $z$ within $U_w$. $F(z)$ is well-defined and holomorphic on the open ball $U_w$ by virtue of Lemma 1, since for any closed loop $\gamma \subset U_w$, we have $\oint_\gamma f(z') dz' = 0$ because the open ball $U_w$ is simply-connected.
Since $g(z) = \exp(F(z))$ on $U_w$, it follows that $g(z)$ is also holomorphic on $U_w$. As our choice of $w$ was arbitrary, this proves that $g(z)$ is holomorphic everywhere in $U$.
Lemma 2: For any $z_0 \in \mathbb C$, and for any path $\gamma_{a \to b} \subset \mathbb C \setminus \{ z_0 \}$ from $a$ to $b$, we have
$$ \exp\left(\int _{\gamma_{a \to b}} \frac{1}{z' - z_0} dz'\right) = \frac{b - z_0}{a - z_0}.$$
[Really, this lemma is saying that $\int _{\gamma_{a \to b}} \frac{1}{z' - z_0} dz' = \log(b - z_0) - \log(a - z_0)$.  But I have deliberately avoided writing it this way so that we don't have to worry about ambiguities in defining logarithms.]
Proof:
We first prove this statement for the special case where $\gamma_{a \to b}$ is contained inside some simply-connected subset $U$ of $\mathbb C \setminus \{ z_0 \}$. In this special case, Lemma 1 tells us that the function
$$ F(z) = \int_{\gamma_{a \to z}} \frac{1}{z' - z_0} dz'$$
is well-defined on $U$ (i.e. the integral in this definition does not depend on the choice of the path $\gamma_{a \to z}$ from $a$ to $z$, as long as this path $\gamma_{a \to z}$ is contained within our simply-connected open neighbourhood $U$). Lemma 1 also tells us that $$F'(z) = \frac{1}{z - z_0}.$$ By the product rule for differentiation, we have
$$ \frac{d}{dz} \left( \frac{\exp(F(z))}{z - z_0}\right) = 0$$
on $U$. Therefore, $ \exp(F(z))$ is a constant multiple of $z - z_0$. We can identify the value of the constant as $1/(a - z_0)$ from the fact that $\exp(F(a)) = 1$. Thus
$$ \exp(F(z)) = \frac{z - z_0}{a - z_0}$$
for all $z \in U$, and the result follows by taking $z = b$.
For the general case, we cover $\mathbb C \setminus \{ z_0 \}$ with the two simply-connected open sets $$U = \mathbb C \setminus [z_0 -\infty, z_0], \ \  \ \ \ V = \mathbb C \setminus [z_0, z_0 + \infty],$$ where $[z_0 - \infty, z_0]$ (resp. $[z_0, z_0 + \infty]$) denote the left-pointing (resp. right-pointing) straight horizontal half-lines originating at $z_0$.
Any path $\gamma_{a \to b} \subset \mathbb C \setminus \{ z_0 \} $, can be broken down into finitely many segments,
$$ \gamma_{a \to c_1}, \ \gamma_{c_1 \to c_2}, \ \dots, \ \gamma_{c_{n-1} \to b},$$
where each segment $\gamma_{c_{i} \to c_{i+1}}$ is either wholly contained in $U$ or is wholly contained in $V$. (The fact that we can find finitely many such segments is a consequence of  $\gamma_{a \to b}$ being compact.)
By the special case, we have
$$ \exp \left( \int_{\gamma_{c_i \to c_{i+1}}} \frac{1}{z' - z_0}dz'\right) = \frac{c_{i+1} - z_0}{c_i - z_0},$$
for each segment $\gamma_{c_i \to c_{i+1}}$, and hence
$$  \exp \left( \int_{\gamma_{a, b}} \frac{1}{z' - z_0}dz'\right) = \frac{c_1 - z_0}{a - z_0}. \frac{c_2 - z_0}{c_1 - z_0}. \dots .\frac{b - z_0}{c_{n-1} - z_0} = \frac{b - z_0}{a - z_0} $$

Proof of main result: By definition,
$$ \prod_{j=1}^m (z - z_j)^{a_j} = \exp \left( \sum_{j=1}^m a_j \log (z - z_j)  \right). $$
So to prove that $\prod_{j=1}^m (z - z_j)^{a_j}$ has a holomorphic branch on our open set $G \subset \mathbb C \setminus \{ z_1, \dots, z_n \}$, we must prove that there exists a
holomorphic function $g(z)$, defined on the whole of $G$, such that for any $w \in G$, there exists an open neighbourhood $U_w \subset  G$ and holomorphic functions $u_{w, j}(z)$ defined on $U_w$ (for $j \in \{ 1, \dots m \}$) such that
$$ g(z) = \exp \left( \sum_{j=1}^m a_j u_{w, j}(z) \right) $$
for all $z \in U_w$, and such that
$$ \exp(u_{w,j}(z)) = z - z_j$$
for all $z \in U_w$ and for all $j \in \{1, \dots m \}$.
[Effectively, the function $u_{w,j}(z)$ is a local representative of $\log(z - z_j)$ in the neighbourhood $U_w$ of $w$.]
To construct such a $g(z)$, we observe that for any closed contour $\gamma \subset G$, we have
$$ \oint_\gamma  \sum_{j=1}^m \frac{a_j}{z' - z_j} dz' = 2\pi i\sum_{j=1}^m a_j {\rm \ Ind}_{\gamma}(z_j) \in 2\pi i \mathbb N.$$
This observation enables us to define the function $g : G \to \mathbb C$ using the expression,
$$ g(z) = \exp \left( \int_{\gamma_{c \to z}} \sum_{j=1}^m \frac{a_j}{z' - z_j} dz' + \sum_{j = 1}^m a_j  k_{c,j} \right).$$
Here, $c$ is some arbitrary fixed point in $G$, and for each $j$, $k_{c,j}$ is a constant such that $$\exp(k_{c, j}) = c - z_j.$$ Meanwhile, $\gamma_{c \to z}$ denotes any path in $G$ from $c $ to $z$. By Lemma 1a, the integral in this definition only depends on the choice of $\gamma_{c \to z}$ up to additive shifts of $2\pi i$, ensuring that $g(z)$ is well-defined in $G$. Lemma 1a also tells us that the function $g(z)$ defined in this way is holomorphic in $G$.
Now pick any $w \in G$, and pick a path $\gamma_{c\to w} \subset G$ from $c$ to $w$. Let $U_w$ be an open ball around $w$. Because $U_w$ is simply-connected, we have $$\oint_\gamma \frac{1}{z' - z_j} dz'= 0$$ for all $j$ and for all closed contours $\gamma \in U_w$. So it makes sense to define functions $u_{w, j} : U_w \to \mathbb C$ using the expression,
$$ u_{w, j} (z) = \int_{\gamma_{w \to z}} \frac{1}{z' - z_j} dz' + k_{w, j}.$$
Here, the constants $k_{w, j}$ are defined by the formula
$$ k_{w, j} = k_{c, j} + \int_{\gamma_{c \to w}} \frac{1}{z' - z_j} dz', $$
where $\gamma_{c \to w}$ is the path in $G$ from $c$ to $w$ that we have fixed.  Meanwhile, in the definition of $u_{w,j}(z)$,  $\gamma_{w \to z}$ denotes any path in $U_w$ from $w$ to $z$; Lemma 1 ensures that the integral in the definition does not depend on the precise choice of path, and that the function $u_{w, j}(z)$ defined in this way is holomorphic in $U_w$.
It is clear from the definitions that
$$ g(z) = \exp \left( \sum_{j = 1}^m a_j u_{w, j}(z) \right)$$
for all $z \in U_w$. Applying  Lemma 2 to the definition of $k_{w, j}$ gives $$\exp(k_{w, j}) = w - z_j$$ for each $j$, and by applying Lemma 2 once more, this time to the definition of $u_{w,j}(z)$, we learn that
$$ \exp(u_{w, j}(z)) = z - z_j$$
for all $z \in U_w$ and for all $j$. So $g(z)$ and the $u_{w, j}(z)$'s obey all the desired properties on $U_w$. Since the choice of $w$ was arbitrary, the proof is complete.


TITLE: Complete valuation field has no other valuations
QUESTION [0 upvotes]: Let $K$ be a complete discrete valuation field. That is, suppose that $K$ is a field, and that there is a map $\nu : K^{\times} \rightarrow \mathbb{Z}$ such that $\nu(a+ b) \geq \nu(a) \vee \nu(b)$, $\nu(1) = 0$, and $\nu(ab) = \nu(a) + \nu(b)$.
How might I show that there are no other valuations on this field?
For instance, when we complete $\mathbb{Q}$ to $\mathbb{Q}_p$, the other valuations $\nu_q$ disappear for $q \neq p$.

REPLY [2 votes]: There are plenty of valuations on $\Bbb{Q}_p$ (for a cardinality reason $\overline{\Bbb{Q}}_p\cong \overline{\Bbb{Q}}_q$ when assuming the axiom of choice), but (up to a constant) there is only one discrete non-trivial valuation. This is because for $1+a p\in 1+p\Bbb{Z}_p$ then $(1+ap)^{1/n}\in \Bbb{Q}_p$ whenever $p\nmid n$, whence $v(1+ap)=0$ for $v$ to be discrete.
Then $$\Bbb{Q}_p^\times = \langle \zeta_{p-1}\rangle \times 1+p\Bbb{Z}_p \times p^\Bbb{Z}$$
Since $\zeta_{p-1}$ is a root of unity you'll have $v(\zeta_{p-1})=0$, for $v$ being non-trivial you need $v(p)\ne 0$, and $v(p)= v(1+\ldots+1)\ge v(1)=0$ gives $v(p)>0$ ie. up to a constant $v$ is the $p$-adic valuation.
It works the same way on any field complete for a discrete valuation $w$, the elements $a\in F$ with $w(a-1)>0$ still have $n$-th roots for $n$ coprime with the characteristic of the residue field so $v(a)=0$ for $v$ being discrete, then for $w(b)=0$ we must have $v(b)\ge 0$ since otherwise $v(1+\pi_F b^r)$ wouldn't be $0$ for $r$ large enough, and hence $v$ is determined by $v(\pi_F)$ which must be $\ge 0$ for the same reason.


TITLE: What does "removing a point" have to do with homeomorphisms?
QUESTION [28 upvotes]: I am self-studying topology from Munkres.  One exercise asks, in part, to show that the spaces $(0,1)$ and $(0,1]$ are not homeomorphic.  An apparent solution is as follows: If you remove a point, $x$, from $(0,1)$, you get a disconnected space; however, you can remove the point $\{1\}$ from the space $(0,1]$ and the space will still be connected.  This apparently means the spaces aren't homeomorphic. I don't quite see the connection to the definition of homeomorphic spaces (that two spaces are homeomorphic if there is a bicontinuous function between them).   So, what is the connection between the "bicontinuous" definition of homeomorphisms and the "removing a point" procedure?  Sorry for missing something obvious.

REPLY [0 votes]: Homeomorphisms preserve topological properties. If one space has a certain toplogical property, and another space does not, then the spaces cannot be homeomorphic.
Note this technique is most useful in showing spaces are NOT homeomorphic. It not as useful if you are trying to show they ARE homeomorphic (unless the property actually characterizes a space -- having a certain property may be necessary but not sufficient to ensure two spaces are homeomorphic).