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\begin{document}

\maketitle \vspace*{-1.5cm}

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\footnote{2010 \emph{Mathematics Subject Classification}: 20M10, 20M20.}

\footnote{\emph{Key words}: transformation semigroups, orientation-preserving, order-preserving.}

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\begin{abstract}
In this paper we introduce the notion of an orientation-preserving transformation on an arbitrary chain, as
a natural extension for infinite chains of the well known concept for finite chains introduced in 1998 by McAlister \cite{McAlister:1998} and, independently, in 1999 by Catarino and Higgins \cite{Catarino&Higgins:1999}.
We consider the monoid $\POP(X)$ of all orientation-preserving partial transformations on a finite or infinite chain $X$ and its submonoids
$\MOP(X)$ and $\POPI(X)$ of all orientation-preserving full transformations and of all orientation-preserving partial permutations on $X$, respectively.
The monoid $\PO(X)$ of all order-preserving partial transformations on $X$ and its injective counterpart $\POI(X)$ are also considered.
We study the regularity and give descriptions of the Green's relations of the monoids $\POP(X)$, $\PO(X)$, $\MOP(X)$, $\POPI(X)$ and $\POI(X)$.
\end{abstract}

\section{Introduction and preliminaries}\label{pre}

Let $X$ be an arbitrary chain (finite or infinite).

Denote by $\PT(X)$ the monoid of all partial transformations on $X$
(under composition of maps) and by $\T(X)$ its submonoid of all full transformations on $X$.
Let $\alpha\in\PT(X)$.
We say that $\alpha$ is \textit{order-preserving} if $x\le y$
implies $x\alpha\le y\alpha$, for all $x,y\in\dom(\alpha)$.
For $Y\subseteq\dom(\alpha)$, the transformation
$\alpha$ is said to be \textit{order-preserving on $Y$} if its restriction to $Y$ is order-preserving.
Denote by $\PO(X)$ the submonoid of $\PT(X)$ of all
order-preserving partial transformations, i.e.
$$
\PO(X)=\{\alpha\in\PT(X)\mid \mbox{$x\le y$ implies $x\alpha\le
y\alpha$, for all $x,y\in \dom(\alpha)$}\},
$$
and by $\MO(X)$ the submonoid of $\PO(X)$ of all order-preserving
full transformations, i.e.
$$
\MO(X)=\{\alpha\in\T(X)\mid \mbox{$x\le y$ implies $x\alpha\le
y\alpha$, for all $x,y\in X$}\} .
$$

For a finite chain $X$, it is well known, and easy to prove, that $\MO(X)$ is a regular semigroup.
The problem for an infinite chain $X$ is much more involved.
Nevertheless,  a characterization of those posets $P$
for which the semigroup of all endomorphisms of $P$ is regular was done by A\v\i zen\v stat in 1968
\cite{Aizenstat:1968} (see also the paper \cite{Adams&Gould:1989} by Adams and Gould).
A description of the regular elements of $\MO(X)$ was given in 2010 by Mora and Kemprasit \cite{Mora&Kemprasit:2010} and
the largest regular subsemigroup of $\MO(X)$ was characterized by Fernandes et al. in
\cite{Fernandes&al:2014}. In this last paper, the authors also described the Green's relations on $\MO(X)$.
For a chain $X_n$ with $n$ elements, e.g. $X_n=\{1<2<\cdots<n\}$, the monoid $\MO(X_n)$,
usually denoted by $\MO_n$, has been extensively studied since the sixties. See
\cite{Aizenstat:1962,Aizenstat:1962b,Fernandes:1997,Fernandes:2002,Fernandes&al:2010,Fernandes&Volkov:2010,Gomes&Howie:1992,
Higgins:1995,Howie:1971,Repnitskii&Vernitskii:2000,Repnitskii&Volkov:1998,Vernitskii&Volkov:1995}.

\smallskip

Let $a=(a_1,a_2,\ldots,a_t)$ be a sequence of $t$ ($t\geq0$)
elements from the chain $X_n$. We say that $a$ is \textit{cyclic}
if there exists no more than one index
$i\in\{1,\ldots,t\}$ such that $a_i>a_{i+1}$,
where $a_{t+1}$ denotes $a_1$.
An element $\alpha\in\T(X_n)$ is called an \textit{orientation-preserving}
transformation if the sequence of its
images $(1\alpha,\ldots,n\alpha)$ is cyclic.
It is a routine to check that the product
of two orientation-preserving transformations on $X_n$ is an
orientation-preserving transformation.
Denote by $\MOP_n$ the submonoid of $\T(X_n)$  whose
elements are orientation-preserving.
The notion of an orientation-preserving
transformation on a finite chain was introduced by McAlister in \cite{McAlister:1998} and,
independently, by Catarino and Higgins in \cite{Catarino&Higgins:1999}.
Several properties of the monoid $\MOP_n$ have been investigated in
these two papers. A presentation for the monoid $\MOP_n$, in terms of $2n-1$ generators, was given by Catarino in \cite{Catarino:1998}. Another presentation for $\MOP_n$, in terms of $2$ (its rank) generators, was found by
Arthur and Ru\v{s}kuc \cite{Arthur&Ruskuc:2000}.
The congruences of $\MOP_n$ were completely described by Fernandes et al. in \cite{Fernandes&Gomes&Jesus:2009}.
Semigroups of orientation-preserving transformations were also studied in several other recent papers
(e.g. see \cite{Araujo&al:2011,Dimitrova&al:2012,
Fernandes&Gomes&Jesus:2011,Fernandes&al:2016,Fernandes&Quinteiro:2011,Fernandes&Quinteiro:2012,
Fernandes&Quinteiro:2014,Zhao&Fernandes:2015}).

\smallskip

This paper is organized as follows.
In the remaining of the Section \ref{pre} we define orientation-preserving transformations on an arbitrary chain and present some basic properties,
in particular we show that the set of all orientation-preserving transformations forms a semigroup.
In Section \ref{reg} we give a criterion for the regularity of an orientation-preserving full transformation, to whose proof we dedicate all Section \ref{proof}.
Finally, in Section \ref{RGR} we give descriptions of the Green's relations for the various semigroups considered.

\medskip

Let $\alpha\in\PT(X)$.  We say that $\alpha$  is \textit{orientation-preserving} if $\alpha$ is the empty transformation or if there exists
a non-empty subset $Y$ of $\dom(\alpha)$ such that:
\begin{description}
\item (OP1) $\alpha$ is order-preserving both on $Y$ and on $\dom(\alpha)\setminus Y$;
\item (OP2) For all $a\in Y$ and $b\in\dom(\alpha)\setminus Y$, we have $a\le b$ and $a\alpha\ge b\alpha$.
\end{description}
If $\alpha\ne\emptyset$, we call to such a subset $Y$ an \textit{ideal} of $\alpha$. For $\alpha=\emptyset$, we define the \textit{ideal} of $\alpha$ as being the empty set.
Notice that $Y$ is  an order ideal of $\dom(\alpha)$ and $\dom(\alpha)\setminus Y$ is an order filter of $\dom(\alpha)$.

\begin{center}
\includegraphics[scale=.2]{pop.jpg}
\end{center}

For a finite chain, it is easy to realize that the above notion coincides with the notion of an orientation-preserving transformation introduced by McAlister \cite{McAlister:1998} and by Catarino and Higgins \cite{Catarino&Higgins:1999}.
Furthermore, it is a natural extension for infinite chains.

Denote by $\POP(X)$ the subset of $\PT(X)$ of all orientation-preserving partial transformations.
Also, denote by $\MOP(X)$ the subset of $\T(X)$ of all orientation-preserving full transformations, i.e.
$\MOP(X)=\POP(X)\cap\T(X)$. Clearly, for
$\alpha\in\PT(X)$, we have $\alpha\in\PO(X)$ if and only if $\alpha\in\POP(X)$
and $\alpha$ admits $\dom(\alpha)$ as an ideal. In particular, we have $\PO(X)\subseteq\POP(X)$ and $\MO(X)\subseteq\MOP(X)$.

Also clear is the following lemma:

\begin{lemma}\label{non-c}
Let $\alpha\in \POP(X)$ admitting
a non-empty proper subset $Y$ of $\dom(\alpha)$ as an ideal.
Then $\alpha$ is non-constant if and only if there
exist $a\in Y$ and $b\in\dom(\alpha)\setminus Y$ such that $a\alpha > b\alpha$.
\end{lemma}

Observe that any constant mapping of $\PT(X)$ is order-preserving and so it is also
orientation-preserving. Moreover, it admits
as an ideal any non-empty order ideal of its domain.
Next, we show that for non-constant mappings the situation is completely different.

\begin{proposition}
Let $\alpha$ be a non-constant transformation of $\POP(X)$.
Then $\alpha$ admits a unique ideal.
\end{proposition}
\begin{proof}
If $\alpha$ is the empty transformation then, by definition, it has only an ideal.
If $\alpha\in\PO(X)$ then $\dom(\alpha)$
is the unique ideal of $\alpha$. In fact, if $\alpha$ would admit a
non-empty proper subset $Y$ of $\dom(\alpha)$ as an ideal
then, by Lemma \ref{non-c}, there exist $a\in Y$ and $b\in \dom(\alpha)\setminus Y$
such that $a\alpha > b\alpha$. Since $a\le b$, by (OP2), then $\alpha$ would not be
order-preserving.

Next, suppose that $\alpha\not\in\PO(X)$ and
let $Y$ and $Z$ be two ideals of $\alpha$.
Then $Y$ and $Z$ are non-empty proper subsets of $\dom(\alpha)$.
By Lemma \ref{non-c}, we may consider elements $a\in Z$ and $b\in \dom(\alpha)\setminus Z$
such that $a\alpha > b\alpha$. Furthermore, $a < b$, by (OP2).

Take $x\in Y$ and, by contradiction, suppose that $x\not\in Z$. Then $a < x$ and
$a\alpha\ge x\alpha$ (since $a\in Z$ and $x\in \dom(\alpha)\setminus Z$).
From $a < x$ and $x\in Y$, it
follows that $a\in Y$, whence $a\alpha\le x\alpha$. Thus
$a\alpha=x\alpha$. As $b,x\in
\dom(\alpha)\setminus Z$, if $x\le b$ then $a\alpha=x\alpha\le
b\alpha < a\alpha$, a contradiction. Then $b < x$ and, since $x\in
Y$, it follows that $b\in Y$. Thus $a,b\in Y$ and $a < b$, whence
$a\alpha\le b\alpha$, again a contradiction. Therefore $x\in Z$
and so $Y\subseteq Z$.

Similarly, we prove that $Z\subseteq Y$, whence $Y=Z$, as
required.
\end{proof}

Another property easy to prove of an orientation-preserving transformation is the following:

\begin{proposition}\label{glued}
Let $\alpha$ be a transformation of $\POP(X)$ with
ideal $Y$. If $Y\alpha\cap(\dom(\alpha)\setminus Y)\alpha\neq\emptyset$ then
$Y\alpha\cap(\dom(\alpha)\setminus Y)\alpha=\{m\}$, for some $m\in X$.
Moreover, in this case, $Y\alpha$
has a minimum element, $(\dom(\alpha)\setminus Y)\alpha$ has a maximum element and both
of these elements coincide with $m$.
\end{proposition}
\begin{proof}
Let  $m\in Y\alpha\cap(\dom(\alpha)\setminus Y)\alpha$ and take $a\in Y$ and
$b\in \dom(\alpha)\setminus Y$ such that $a\alpha=m=b\alpha$.
Let $c\in Y\alpha$. Then $c=y\alpha$, for
some $y\in Y$. Since $b\in\dom(\alpha)\setminus Y$, we have $c=y\alpha\ge b\alpha=m$ and so
$m$ is the minimum of $Y\alpha$. Moreover, this also proves that
$Y\alpha\cap(\dom(\alpha)\setminus Y)\alpha=\{m\}$.
On the other hand, let $d\in(\dom(\alpha)\setminus Y)\alpha$.
Then $d=z\alpha$, for some $z\in\dom(\alpha)\setminus Y$.
Now, since $a\in Y$, we have
$m=a\alpha\ge z\alpha=d$ and so $m$ is also the maximum of
$(\dom(\alpha)\setminus Y)\alpha$, as required.
\end{proof}

Now, as an application of the previous proposition,
we present a property of the idempotents of $\MOP(X)$ that we will use in Section \ref{proof}.

\begin{proposition}\label{idps}
Let $\alpha$ be an idempotent of $\MOP(X)$ and let $Y$ be an ideal of $\alpha$.
Then, we have one of the following conditions:
\begin{enumerate}
\item $\alpha\in\MO(X)$;
\item $\im(\alpha)=Y\alpha\subseteq Y$, $\min\im(\alpha)$ exists and $(X\setminus Y)\alpha=\{\min\im(\alpha)\}$;
\item $\im(\alpha)=(X\setminus Y)\alpha\subseteq X\setminus Y$, $\max\im(\alpha)$ exists and $Y\alpha=\{\max\im(\alpha)\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
First, notice that, since $\alpha$ is idempotent, then $\alpha$ is order-preserving on $\im(\alpha)$.
Hence, we have $\im(\alpha)\subseteq Y$ or $\im(\alpha)\subseteq X\setminus Y$.
In fact, if $\im(\alpha)\not\subseteq Y$ and $\im(\alpha)\not\subseteq X\setminus Y$
then there exist $a,b\in X$ such that $a\alpha\not\in Y$ and $b\alpha\not\in X\setminus Y$,
whence $a\alpha\in X\setminus Y$ and $b\alpha\in Y$, from which follows
$b\alpha<a\alpha$ and $(b\alpha)\alpha\ge(a\alpha)\alpha$, i.e.
$b\alpha<a\alpha$ and $b\alpha\ge a\alpha$, which is a contradiction.
Thus, as
$Y\alpha\cup(X\setminus Y)\alpha=\im(\alpha)=\im(\alpha^2)=(\im(\alpha))\alpha$,
if $\im(\alpha)\subseteq Y$ then $\im(\alpha)=Y\alpha$ and
if $\im(\alpha)\subseteq X\setminus Y$ then $\im(\alpha)=(X\setminus Y)\alpha$.

Suppose that $\alpha\not\in\MO(X)$. Then $X\setminus Y\ne\emptyset$,
whence $Y\alpha\cap(X\setminus Y)\alpha\ne\emptyset$
and so, by Proposition \ref{glued},
we get $Y\alpha\cap(X\setminus Y)\alpha=\{m\}$, with $m=\min(Y\alpha)=\max((X\setminus Y)\alpha)$.
Therefore, either $\im(\alpha)=Y\alpha$, and so  $\min\im(\alpha)=m$ and $(X\setminus Y)\alpha=\{m\}$,
or $\im(\alpha)=(X\setminus Y)\alpha$, and so $\max\im(\alpha)=m$ and $Y\alpha=\{m\}$, as required.
\end{proof}

\medskip

Next, we show $\POP(X)$ and, consequently, $\MOP(X)$ are closed subsets of $\PT(X)$.

\begin{proposition}\label{POPsemigroup}
$\POP(X)$ is a submonoid of $\PT(X)$.
Consequently, $\MOP(X)$ is a submonoid of $\T(X)$ (and of $\POP(X)$ and of $\PT(X)$).
\end{proposition}
\begin{proof}
First, notice that the identity transformation on $X$ is order-preserving,
whence it belongs to $\POP(X)$ and to $\MOP(X)$.
Since $\MOP(X)=\POP(X)\cap\T(X)$, it suffices to show that $\POP(X)$ is closed under composition of mappings.

Let $\alpha$ and $\beta$ be transformations of $\POP(X)$ with
ideals $Y$ and $Z$, respectively.

We will show that $\alpha\beta\in\POP(X)$, by considering two
(disjoint) cases.

\smallskip

\noindent\textsc{case 1.} $Y\alpha\cap\dom(\beta)\subseteq \dom(\beta)\setminus Z$.

Let $W=(Y\cup Z\alpha^{-1})\cap\dom(\alpha\beta)$.

First, we show that $\alpha\beta$ is order-preserving on $W$. Let
$x,y\in W$ be such that $x\le y$. If $y\in Y$ then $x\in Y$,
whence $x\alpha\le y\alpha$ and
$x\alpha, y\alpha\in Y\alpha\cap\dom(\beta)\subseteq\dom(\beta)\setminus Z$,
from which it follows that $x\alpha\beta\le y\alpha\beta$.
Now, suppose that $y\not\in Y$. If $x\not\in Y$ then $x\alpha\le
y\alpha$ and $x,y\in Z\alpha^{-1}$,
whence $x\alpha, y\alpha\in Z$ and so $x\alpha\beta\le
y\alpha\beta$. On the other hand, if $x\in Y$ then
$x\alpha\in Y\alpha\cap\dom(\beta)\subseteq\dom(\beta)\setminus Z$
and, since $y\alpha\in Z$, it follows that
$x\alpha\beta\le y\alpha\beta$.

Secondly, we show that $\alpha\beta$ is order-preserving on
$\dom(\alpha\beta)\setminus W$.
Let $x,y\in \dom(\alpha\beta)\setminus W$ be such that $x\le y$.
Then $x,y\in \dom(\alpha)\setminus Y$ and $x\alpha,y\alpha\in \dom(\beta)\setminus Z$.
Hence, since $x\le y$, we have $x\alpha\le y\alpha$ and so
$x\alpha\beta\le y\alpha\beta$.

Now, we focus our attention on the condition (OP2). Let $x\in W$
and $y\in \dom(\alpha\beta)\setminus W$. Then,
as above, we have
$y\in \dom(\alpha)\setminus Y$ and $y\alpha\in \dom(\beta)\setminus Z$.

If $x\in Y$ then $x\le y$ and $x\alpha\ge y\alpha$, since $y\in \dom(\alpha)\setminus Y$.
Moreover, as $x\alpha\in Y\alpha\cap\dom(\beta)\subseteq \dom(\beta)\setminus Z$
and $y\alpha\in \dom(\beta)\setminus Z$, we
obtain $x\alpha\beta\ge y\alpha\beta$.

On the other hand, suppose that $x\not\in Y$.
Then, $x\in Z\alpha^{-1}$ and so $x\alpha\in Z$.
If $y\le x$ then $y\alpha\le x\alpha$ (since $x,y\in \dom(\alpha)\setminus Y$),
which implies that $y\alpha\in Z$, a contradiction.
Therefore $x\le y$. Moreover, since $x\alpha\in Z$ and $y\alpha\in \dom(\beta)\setminus Z$,
we also have $x\alpha\beta\ge y\alpha\beta$.

If $W=\emptyset$ then $\alpha\beta\in\PO(X)$ and so $\alpha\beta\in\POP(X)$.
Otherwise we just proved that $\alpha\beta$ admits $W$ as an ideal and so we also have
$\alpha\beta\in\POP(X)$.

\smallskip

\noindent\textsc{case 2.} $Y\alpha\cap\dom(\beta)\not\subseteq \dom(\beta)\setminus Z$.

First, notice that, there exists $a\in Y$ such that $a\alpha\in Z\cap\dom(\beta)$.
Hence, if $x\in \dom(\alpha)\setminus Y$ then $x\alpha\le a\alpha$ and so $x\alpha\in Z$.
Thus $(\dom(\alpha)\setminus Y)\alpha\subseteq Z$.

Let $V=Y\cap Z\alpha^{-1}\cap\dom(\alpha\beta)$.

We begin by showing that $\alpha\beta$ is order-preserving on $V$.
Let $x,y\in V$ be such that $x\le y$. Then $x,y\in Y$ and
$x\alpha,y\alpha\in Z$. It follows that $x\le y$ implies
$x\alpha\le y\alpha$, which in turn implies that $x\alpha\beta\le
y\alpha\beta$.

We continue by showing that $\alpha\beta$ is order-preserving on
$\dom(\alpha\beta)\setminus V$.
Let $x,y\in \dom(\alpha\beta)\setminus V$ be such that $x\le y$.
If $x\in \dom(\alpha)\setminus Y$ then $y\in \dom(\alpha)\setminus Y$ and so
$x\alpha,y\alpha\in(\dom(\alpha)\setminus Y)\alpha\subseteq Z$.
It follows that $x\le y$ implies $x\alpha\le y\alpha$,
which in turn implies that $x\alpha\beta\le y\alpha\beta$.
So, suppose that $x\in Y$.
If $y\in Y$ then, as $x,y\in \dom(\alpha\beta)\setminus V$,
we have $x,y\not\in Z\alpha^{-1}$ and so
$x\alpha,y\alpha\in \dom(\beta)\setminus Z$.
Hence, in this case,
$x\le y$ implies $x\alpha\le y\alpha$,
which in turn implies that $x\alpha\beta\le y\alpha\beta$.
On the other hand, if $y\in \dom(\alpha)\setminus Y$,
then we have $x\alpha\in\dom(\beta)\setminus Z$ and $y\alpha\in(\dom(\alpha)\setminus Y)\alpha\subseteq Z$, from which follows again $x\alpha\beta\le y\alpha\beta$.

If $V=\emptyset$ then $\alpha\beta\in\PO(X)$ and so $\alpha\beta\in\POP(X)$.
Thus, let us suppose that $V\ne\emptyset$ and prove that $\alpha\beta$ satisfies (OP2) with respect
to $V$.

Let $x\in V$ and $y\in \dom(\alpha\beta)\setminus V$.
Then $x\in Y$ and $x\alpha\in Z$.

Suppose that $y\in Y$. Then, since $y\in\dom(\alpha\beta)\setminus V$,
we have $y\not\in Z\alpha^{-1}$ and so $y\alpha\in \dom(\beta)\setminus Z$.
Hence $x\alpha\beta\ge y\alpha\beta$. On the other hand, if $y\le x$ then $y\alpha\le
x\alpha$ and so $y\alpha\in Z$, a contradiction. Thus, we also
have $x\le y$.

Now, suppose that $y\in \dom(\alpha)\setminus Y$.
Then $x\le y$ and $x\alpha\ge y\alpha$. Additionally,
$y\alpha\in(\dom(\alpha)\setminus Y)\alpha\subseteq  Z$ and so
$x\alpha\beta\ge y\alpha\beta$.

Therefore we proved that $\alpha\beta$ admits $V$ as an ideal,
whence $\alpha\beta\in\POP(X)$, as required.
\end{proof}

\medskip

Now, let $\alpha\in\POP(X)$, let $Y$ be an ideal of $\alpha$ and $A$ be any subset of $\dom(\alpha)$.
Clearly, if $A\cap Y=\emptyset$ then the restriction of $\alpha$ to $A$ is order-preserving.
On the other hand, if $A\cap Y\ne\emptyset$ then it is easy to show that $A\cap Y$ is an ideal of the restriction
of $\alpha$ to $A$ and so this transformation is also orientation-preserving. In short:

\begin{proposition}\label{rPOP}
Any restriction of an element of $\POP(X)$ also is an element of $\POP(X)$.
\end{proposition}

Another property of orientation-preserving transformations is the following:

\begin{proposition}\label{iPOP}
Let $\alpha$ be a partial permutation of $X$.
If $\alpha$ is orientation-preserving then its inverse function
$\alpha^{-1}:\im(\alpha)\longrightarrow\dom(\alpha)$ also is orientation-preserving.
In particular, if $\alpha$ is order-preserving then $\alpha^{-1}$ also is order-preserving.
\end{proposition}
\begin{proof}
Let $Y$ be an ideal of $\alpha$.
If $Y=\dom(\alpha)$ then $\alpha$ is order-preserving and it is easy to show that $\alpha^{-1}$ is also order-preserving. On the other hand, if $Y\subsetneq\dom(\alpha)$ then it is a routine matter to prove that
$(\dom(\alpha)\setminus Y)\alpha$ is an ideal of $\alpha^{-1}$.
\end{proof}

Denote by $\I(X)$ the submonoid of $\PT(X)$ of all partial permutations. This monoid is called the \textit{symmetric inverse monoid} on $X$. It follows from Proposition \ref{iPOP} that the submonoids $\POI(X)=\PO(X)\cap\I(X)$ and $\POPI(X)=\POP(X)\cap\I(X)$ of $\I(X)$ also are \textit{inverse submonoids} of $\I(X)$.

\section{Regularity} \label{reg}

The following criterion for the regularity of the elements of $\MO(X)$ was proved in 2010 by Mora and Kemprasit.

\begin{theorem}[{\cite[Theorem 2.4]{Mora&Kemprasit:2010}}] \label{regO}
Let $X$ be a chain and let $\alpha\in\MO(X)$. Then $\alpha$ is a regular element of $\MO(X)$ if and only if the following conditions hold:
\begin{enumerate}
\item If $\im(\alpha)$ has an upper bound in $X$, then $\max\im(\alpha)$ exists;
\item If $\im(\alpha)$ has a lower bound in $X$, then $\min\im(\alpha)$ exists;
\item If $x\in X\setminus\im(\alpha)$ is neither an upper bound nor a lower bound of $\im(\alpha)$,
then either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.
\end{enumerate}
\end{theorem}

Based on this theorem,  Mora and Kemprasit \cite{Mora&Kemprasit:2010} deduced several previous known results.
For instance,
that $\MO(\Z)$ is regular while $\MO(\Q)$ and $\MO(\R)$ are not regular,
being $\Z$, $\Q$ and $\R$ the sets of integers, rational numbers and real numbers, respectively,
with their usual orders.
See also \cite{Adams&Gould:1989,Aizenstat:1968,Kemprasit&Changphas:2000,Kim&Kozhukhov:2008}.

\medskip

The example below shows that $\MOP(X)$ may have more regular order-preserving elements than
$\MO(X)$.

\begin{example}\em
Consider the set $\R$ of real numbers
equipped with the usual order.
As observed above, $\MO(\R)$ is not regular.

Let $\alpha:\R\longrightarrow\R$ be the mapping defined by
$$
x\alpha=\left\{
\begin{array}{ll}
 -1/x & x\ge1\\
 -1 & x<1~,
\end{array}
\right.
$$
for $x\in\R$. Then, $\alpha\in\MO(\R)$ and $\im(\alpha)=[-1,0[$ and so,
by Theorem \ref{regO}, $\alpha$ is not regular in $\MO(\R)$.

Now, let $\beta:\R\longrightarrow\R$ be the mapping defined by
$$
x\beta=\left\{
\begin{array}{ll}
 -1/x & -1\le x<0\\
 1 & \mbox{otherwise},
\end{array}
\right.
$$
for $x\in\R$. Then $\beta\in\MOP(\R)$, with ideal
$Y=\,]-\infty,0[$. Moreover, we have $\alpha=\alpha\beta\alpha$
and $\beta=\beta\alpha\beta$. Thus $\alpha$ is regular in $\MOP(\R)$.
\end{example}

The following property gives us a necessary condition for an order-preserving transformation of $X$ to admit a non order-preserving inverse in $\MOP(X)$. As an application, we show that $\MOP(\R)$ is not regular.

\begin{proposition}\label{bound}
If $\alpha\in\MO(X)$ admits an inverse
$\beta\in\MOP(X)\setminus\MO(X)$ then $\im(\alpha)$ is bounded.
\end{proposition}
\begin{proof}
Let $Y$ be an ideal of $\beta$. Then $Y$ is a nonempty proper
subset of $X$, since $\beta\in\MOP(X)\setminus\MO(X)$.

First, suppose that $(X\setminus Y)\beta\alpha\cap
Y\neq\emptyset$. Then, there exist $a\in Y$ and $b\in X\setminus
Y$ such that $b\beta\alpha=a$. Hence,
$b\beta=b\beta\alpha\beta=a\beta$ and so $m=b\beta=a\beta\in
(X\setminus Y)\beta\cap Y\beta$. By Proposition \ref{glued}, it
follows that $(X\setminus Y)\beta\cap Y\beta=\{m\}$ and
$\max((X\setminus Y)\beta)=m=\min(Y\beta)$. Thus, $\beta$ is
constant in $]-\infty,a]\cup(X\setminus Y)$ with value $m$. In
fact, given $y\in ]-\infty,a]$, we have $y\beta\le a\beta=m$,
whence $y\beta=m$. On the other hand, for $x\in X\setminus Y$, we
have $x\beta\le m$ and so $x\beta\alpha\le
m\alpha=b\beta\alpha=a$. Hence, $x\beta=(x\beta\alpha)\beta=m$. In
addition, since $\beta$ is order-preserving on $Y$, it follow that
$\im(\beta)\subseteq[m,+\infty[$, whence $\im(\beta\alpha)\subseteq
[m\alpha,+\infty[=[a,+\infty[$. On the other hand, if
$x\beta\alpha\in X\setminus Y$, for some $x\in X$, then
$m=(x\beta\alpha)\beta=x\beta$, whence $a=m\alpha=x\beta\alpha\in
X\setminus Y$, which is a contradiction. Therefore
$\im(\beta\alpha)\subseteq Y$ and thus $\im(\beta\alpha)\subseteq
Y\cap[a,+\infty[$. Now, as $Y$ is upper bounded by any element of
$X\setminus Y$, we have $\im(\beta\alpha)\subseteq [a,b[$.

Next, we admit the other case, i.e. $(X\setminus
Y)\beta\alpha\subseteq X\setminus Y$. Then, we also have
$Y\beta\alpha\subseteq X\setminus Y$. In fact, being $y\in Y$ and
$x\in X\setminus Y$, we have $x\beta\le y\beta$ and so
$x\beta\alpha\le y\beta\alpha$. Since $x\beta\alpha\in X\setminus
Y$ and $X\setminus Y$ is an order filter of $X$, then
$y\beta\alpha\in X\setminus Y$, as required. Thus
$\im(\beta\alpha)\subseteq X\setminus Y$. It follows that
$X\beta=(X\beta\alpha)\beta\subseteq (X\setminus Y)\beta$, whence
$Y\beta\cap(X\setminus Y)\beta=Y\beta\cap
X\beta=Y\beta\neq\emptyset$ and so, by Proposition \ref{glued},
there exists $m\in X$ such that $Y\beta=\{m\}$, with
$\max(X\beta)=\max((X\setminus Y)\beta)=m\,(=\min(Y\beta))$. Then
$\im(\beta)\subseteq \,]-\infty, m]$ and so
$\im(\beta\alpha)\subseteq \,]-\infty, m\alpha]$. Now, as
$X\setminus Y$ is lower bounded by any element of $Y$, we have
$\im(\beta\alpha)\subseteq \,]y,m\alpha]$, for any $y\in Y$, which
finishes the proof.
\end{proof}

\begin{example}\em
Let $\R$ be the set of real numbers equipped with the usual order.
Then $\MOP(\R)$ is not a regular semigroup. Furthermore,
it contains non regular order-preserving transformations.

In fact, let $\alpha:\R\longrightarrow\R$ be the mapping defined by
$x\alpha=-e^{-x}, $ for $x\in\R$. Then $\alpha\in\MO(\R)$ and
$\im(\alpha)=\,]-\infty,0[$.
Hence, by Theorem \ref{regO},
$\alpha$ is not regular in $\MO(\R)$
and, by Proposition \ref{bound}, $\alpha$ is not regular in $\MOP(\R)$ too.
\end{example}

\medskip

Next, we show that for partial transformations, such as for partial permutations, we only have regular elements.

Let $\alpha\in\POP(X)\setminus\{\emptyset\}$. Let $Y$ be an ideal of $\alpha$.
For each $x\in Y\alpha$, choose $z_x\in x\alpha^{-1}\cap Y$.
For each $x\in\im(\alpha)\setminus(Y\alpha)$, choose $z_x\in x\alpha^{-1}$ (observe that, in this case $z_x\in \dom(\alpha)\setminus Y$).
Let $D=\{z_x\mid x\in\im(\alpha)\}$. Then
$$
\begin{array}{rrclcrrcl}
\zeta: & \im(\alpha) &\longrightarrow & D & \quad\text{and}\quad & \alpha|_D: & D & \longrightarrow & \im(\alpha) \\
          & x & \longmapsto & z_x && & x & \longmapsto & x\alpha
\end{array}
$$
are mutually inverse bijections. By Proposition \ref{rPOP}, $\alpha|_D\in\POP(X)$. In addition, $\alpha|_D$ is a partial permutation of $X$ and so, by Proposition \ref{iPOP}, $\zeta\in\POP(X)$. More precisely, $\alpha|_D,\zeta\in\POPI(X)$.
Moreover, if $\alpha\in\PO(X)$ then $\alpha|_D,\zeta\in\POI(X)$.

\smallskip

Let $x\in\dom(\alpha)$. Then $x\alpha\zeta\alpha=(x\alpha)\zeta\alpha=z_{x\alpha}\alpha=x\alpha$.
Hence $\alpha=\alpha\zeta\alpha$. This proves the following result.

\begin{theorem}
$\PO(X)$ and $\POP(X)$ are regular monoids.
\end{theorem}

In the case of full orientation-preserving transformations, the situation is much more complex.
In fact, we have the following criterion for the regularity of the elements of $\MOP(X)$:

\begin{theorem}\label{regOP}
Let $X$ be a chain and let $\alpha\in\MOP(X)$. Then $\alpha$ is a regular element of $\MOP(X)$ if and only if the following conditions are satisfied:
\begin{enumerate}
\item If $\im(\alpha)$ has an upper bound or a lower bound in $X$, then $\max\im(\alpha)$ exists or $\min\im(\alpha)$ exists;
\item If $x\in X\setminus\im(\alpha)$ is neither an upper bound nor a lower bound of $\im(\alpha)$,
then either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.
\end{enumerate}
\end{theorem}

This criterion is similar to that of full order-preserving transformations (Mora and Kemprasit's Theorem quoted above). However, its proof,
that we will present in Section \ref{proof}, is much more involved.

\smallskip

As an application of this result, we have the following two examples.

\begin{example}\em
Consider the set of integers $\Z$ with the usual order.
Since any non-empty subset of $\Z$ with upper bounds has a maximum and any non-empty subset of $\Z$ with lower bounds has a minimum,
by Theorem \ref{regOP}, it is easy to deduce that $\MOP(\Z)$ is a regular semigroup.
\end{example}

\begin{example}\em
Let $\Q$ be the set of rational numbers
equipped with its usual order.

Let $\alpha:\Q\longrightarrow\Q$ be the mapping defined by
$$
x\alpha=\left\{
\begin{array}{ll}
 1-\frac{1}{x+1} & x\ge0\\
-1-\frac{1}{x-1}  & x<0~,
\end{array}
\right.
$$
for $x\in\Q$. Then, $\alpha\in\MOP(\Q)$ (notice that, in fact, we also have that $\alpha\in\MO(\Q)$) and $\im(\alpha)=]-1,1[$.
Hence, by Theorem \ref{regOP}, $\alpha$ is not regular
in $\MOP(\Q)$.

Therefore, the semigroup $\MOP(\Q)$ is not regular.
\end{example}

\medskip

In \cite{Fernandes&al:2014} Fernandes et al. showed that the product of any two regular elements of $\MO(X)$ is a regular element of $\MO(X)$.
We finish this section leaving as an open problem the corresponding question for $\MOP(X)$.
\begin{question}
Is the set $\reg(\MOP(X))$ of all regular elements of $\MOP(X)$ a subsemigroup of $\MOP(X)$?
\end{question}

\section{The proof of Theorem \ref{regOP}} \label{proof}

This section consists of a series of lemmas. All together demonstrate Theorem \ref{regOP}.

\medskip

Let $\alpha$ be a regular element of $\MOP(X)$ and let $\beta\in\MOP(X)$ be such that $\alpha=\alpha\beta\alpha$.
Then $\beta\alpha$ is an idempotent of $\MOP(X)$ and so, by Proposition \ref{idps}, $\beta\alpha\in\MO(X)$
or $\min\im(\beta\alpha)$ exists or $\max\im(\beta\alpha)$ exists.
If $\beta\alpha\in\MO(X)$ and  $\im(\beta\alpha)$ has an upper bound or a lower bound in $X$ then,
since $\beta\alpha$ is a regular element of $\MO(X)$, by Theorem \ref{regO},
$\max\im(\beta\alpha)$ exists or $\min\im(\beta\alpha)$ exists.
On the other hand, $\im(\alpha)=\im(\alpha\beta\alpha)\subseteq\im(\beta\alpha)\subseteq\im(\alpha)$
and so $\im(\alpha)=\im(\beta\alpha)$.
Therefore, it follows immediately:

\begin{lemma}
Under the above conditions,
if $\im(\alpha)$ has an upper bound or a lower bound in $X$, then $\max\im(\alpha)$ exists or $\min\im(\alpha)$ exists.
\end{lemma}

Our next lemma completes the proof of the direct implication of Theorem \ref{regOP}.

\begin{lemma}
Under the above conditions,
if $x\in X\setminus\im(\alpha)$ is neither an upper bound nor a lower bound of $\im(\alpha)$,
then either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.
\end{lemma}
\begin{proof}
Let $x\in X\setminus\im(\alpha)$ be such that $x$ is neither an upper bound nor a lower bound of $\im(\alpha)$.
As $\im(\alpha)=\im(\beta\alpha)$, then $x\in X\setminus\im(\beta\alpha)$ and
$x$ is neither an upper bound nor a lower bound of $\im(\beta\alpha)$.

If $\beta\alpha\in\MO(X)$ then, since $\beta\alpha$ is a regular element of $\MO(X)$, by Theorem \ref{regO},
either $\max\{t\in\im(\beta\alpha)\mid t<x\}$ or $\min\{t\in\im(\beta\alpha)\mid t>x\}$ exists, i.e.
either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.

Thus, suppose that $\beta\alpha\not\in\MO(X)$ and let $Y$ be the ideal of $\beta\alpha$. Then,
as $\beta\alpha$ is an idempotent of $\MOP(X)$,
by Proposition \ref{idps}, we have two possible cases.

\smallskip

\noindent\textsc{case 1.} $\im(\beta\alpha)=Y\beta\alpha\subseteq Y$.

Fix $z\in X\setminus Y$ (notice that $X\setminus Y\ne\emptyset$) and
define a transformation $\gamma$ of $X$ by $x\gamma=x\beta\alpha$,
if $x\in Y$, and $x\gamma=z$, if $x\in X\setminus Y$.
Clearly, $\gamma\in\MO(X)$ and $\gamma$ is an idempotent. In particular, $\gamma$ is a regular element of $\MO(X)$.
Moreover, $\im(\gamma)=Y\beta\alpha\cup\{z\}=\im(\beta\alpha)\cup\{z\}$.

As $\im(\beta\alpha)\subseteq\im(\gamma)$ and
$x$ is neither an upper bound nor a lower bound of $\im(\beta\alpha)$,
then we also have that $x$ is neither an upper bound nor a lower bound of $\im(\gamma)$.
On the other hand,
since $z$ is an upper bound of $\im(\gamma)$ (notice that $\im(\beta\alpha)\subseteq Y$ and $z\in X\setminus Y$),
then $x<z$, whence as also $x\in X\setminus\im(\beta\alpha)$, we obtain $x\in X\setminus\im(\gamma)$.
Thus, by Theorem \ref{regO}, either $\max\{t\in\im(\gamma)\mid t<x\}$ or $\min\{t\in\im(\gamma)\mid t>x\}$ exists.

Since $\{t\in\im(\alpha)\mid t<x\}=\{t\in\im(\beta\alpha)\mid t<x\}=\{t\in\im(\gamma)\mid t<x\}$,
if $\max\{t\in\im(\gamma)\mid t<x\}$ exists then $\max\{t\in\im(\alpha)\mid t<x\}$ also exists.

On the other hand, since $x<z$, then
$\{t\in\im(\gamma)\mid t>x\}=\{t\in\im(\beta\alpha)\mid t>x\}\cup\{z\}$. Hence, as $t<z$, for all $t\in\im(\beta\alpha)$,
if $\min\{t\in\im(\gamma)\mid t>x\}$ exists then $\min\{t\in\im(\beta\alpha)\mid t>x\}=\min\{t\in\im(\alpha)\mid t>x\}$ also exists
(and coincide with each other).

Thus, in this case, we showed that either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.

\smallskip

\noindent\textsc{case 2.} $\im(\beta\alpha)=(X\setminus Y)\beta\alpha\subseteq X\setminus Y$.

We proceed similarly to \textsc{case 1}.
For a fixed element $y\in Y$,
define a transformation $\lambda$ of $X$ by $x\lambda=x\beta\alpha$,
if $x\in X\setminus Y$, and $x\lambda=y$, if $x\in Y$.
Clearly, $\lambda\in\MO(X)$ and $\lambda^2=\lambda$ and so, in particular, $\lambda$ is a regular element of $\MO(X)$.
In addition, we have $\im(\lambda)=(X\setminus Y)\beta\alpha\cup\{y\}=\im(\beta\alpha)\cup\{y\}$.

Since $\im(\beta\alpha)\subseteq X\setminus Y$ and $y\in Y$, then $y$ is a lower bound of $\im(\lambda)$,
whence $y<x$ and so $x\in X\setminus\im(\lambda)$. Moreover, as $\im(\beta\alpha)\subseteq\im(\lambda)$
then $x$ is neither an upper bound nor a lower bound of $\im(\lambda)$.
Thus, by Theorem \ref{regO}, either $\max\{t\in\im(\lambda)\mid t<x\}$ or $\min\{t\in\im(\lambda)\mid t>x\}$ exists.

Since $y<x$, then $\{t\in\im(\lambda)\mid t<x\}=\{t\in\im(\beta\alpha)\mid t<x\}\cup\{y\}$ and so, as $y<t$, for all $t\in\im(\beta\alpha)$,
if $\max\{t\in\im(\lambda)\mid t<x\}$ exists then $\max\{t\in\im(\beta\alpha)\mid t<x\}=\max\{t\in\im(\alpha)\mid t<x\}$ also exists
(and they are the same).

On the other hand, as $\{t\in\im(\alpha)\mid t>x\}=\{t\in\im(\beta\alpha)\mid t>x\}=\{t\in\im(\lambda)\mid t>x\}$,
if $\min\{t\in\im(\lambda)\mid t>x\}$ exists then $\min\{t\in\im(\alpha)\mid t>x\}$ also exists.

Thus, also in this case, we proved that either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.
\end{proof}

\medskip

For the rest of this section, we will prove the converse implication of Theorem \ref{regOP}.

\smallskip

Now, let $\alpha$ be any element of $\MOP(X)$ and let $Y$ be an ideal of $\alpha$.

\begin{lemma}\label{a}
Let $x\in X$.
\begin{enumerate}
\item If $d=\max\{t\in\im(\alpha)\mid t<x\}$ exists and $b_1<x$, for some $b_1\in Y\alpha$, then $d\in Y\alpha$;
\item If $d=\min\{t\in\im(\alpha)\mid t>x\}$ exists and $b_1<x$, for some $b_1\in Y\alpha$, then $d\in Y\alpha$;
\item If $c=\max\{t\in\im(\alpha)\mid t<x\}$ exists and $x<b_2$, for some $b_2\in (X\setminus Y)\alpha$, then $c\in (X\setminus Y)\alpha$;
\item If $c=\min\{t\in\im(\alpha)\mid t>x\}$ exists and $x<b_2$, for some $b_2\in (X\setminus Y)\alpha$, then $c\in (X\setminus Y)\alpha$.
\end{enumerate}
\end{lemma}
\begin{proof} Recall that, by definition, $x_1\ge x_2$, for all  $x_1\in Y\alpha$ and $x_2\in (X\setminus Y)\alpha$.
Hence, $b_1\in Y\alpha$, $d\in\im(\alpha)$ and $b_1\le d$ imply $d\in Y\alpha$ and, analogously,
$b_2\in (X\setminus Y)\alpha$, $c\in\im(\alpha)$ and $c\le b_2$ imply $c\in (X\setminus Y)\alpha$.

\smallskip

1. Since $b_1<x$ and $b_1\in\im(\alpha)$, then $b_1\le d$. As $b_1\in Y\alpha$ and $d\in\im(\alpha)$, it follows that $d\in Y\alpha$.

\smallskip

2. We have $b_1<x<d$, whence $b_1<d$. As $d\in\im(\alpha)$ and  $b_1\in Y\alpha$, then $d\in Y\alpha$.

\smallskip

3. We have $c<x<b_2$, whence $c<b_2$. As $c\in\im(\alpha)$ and $b_2\in (X\setminus Y)\alpha$, then $c\in (X\setminus Y)\alpha$.

\smallskip

4. Since $b_2>x$ and $b_2\in\im(\alpha)$, then $c\le b_2$. As $b_2\in (X\setminus Y)\alpha$ and $c\in\im(\alpha)$,
it follows that $c\in (X\setminus Y)\alpha$.
\end{proof}

\smallskip

Next, as in Section \ref{reg}, for each $x\in Y\alpha$, choose $z_x\in x\alpha^{-1}\cap Y$ and,
for each $x\in\im(\alpha)\setminus(Y\alpha)$, choose $z_x\in x\alpha^{-1}$ (notice that $z_x\in X\setminus Y$).
Recall that $\zeta:\im(\alpha)\longrightarrow X$, $x \longmapsto z_x$, is an injective mapping and $z_x\alpha=x$, for all $x\in\im(\alpha)$.
Moreover, it is easy to show that $\zeta$ is order-preserving both on $Y\alpha$ and on $(X\setminus Y)\alpha\setminus(Y\alpha)$ and,
for $x\in Y\alpha$ and $y\in (X\setminus Y)\alpha\setminus(Y\alpha)$, we have $x>y$ and $z_x<z_y$. In other words:

\begin{lemma}\label{aaa}
The mapping $\zeta:\im(\alpha)\longrightarrow X$ is either order-preserving or $(X\setminus Y)\alpha\setminus(Y\alpha)$ is an ideal of $\zeta$.
\end{lemma}

And, as an immediate consequence, we have:

\begin{lemma}\label{aa}
\begin{enumerate}
\item Suppose that $\alpha\in\MO(X)$.
\begin{enumerate}
\item If $\im(\alpha)$ has a maximum then $z_x\le z_{\max\im(\alpha)}$, for all $x\in\im(\alpha)$;
\item If $\im(\alpha)$ has a minimum then $z_{\min\im(\alpha)}\le z_x$, for all $x\in\im(\alpha)$.
\end{enumerate}

\item Suppose that $\alpha\not\in\MO(X)$ and $Y\alpha\cap(X\setminus Y)\alpha=\emptyset$.
\begin{enumerate}
\item If $\im(\alpha)$ has a maximum then $\max\im(\alpha)=\max(Y\alpha)$ and
$z_d\le z_{\max\im(\alpha)}<z_c$, for all $d\in Y\alpha$ and for all $c\in(X\setminus Y)\alpha$.
\item If $\im(\alpha)$ has a minimum then $\min\im(\alpha)=\min((X\setminus Y)\alpha)$ and
$z_d<z_{\min\im(\alpha)}\le z_c$, for all $d\in Y\alpha$ and for all $c\in(X\setminus Y)\alpha$.
\end{enumerate}

\item Suppose that $Y\alpha\cap(X\setminus Y)\alpha=\{m\}$, for some $m\in X$.
\begin{enumerate}
\item If $\im(\alpha)$ has a maximum then $\max\im(\alpha)=\max(Y\alpha)$,
$z_d\le z_{\max\im(\alpha)}$, for all $d\in Y\alpha$, and $z_{\max\im(\alpha)}<z_c$, for all $c\in(X\setminus Y)\alpha\setminus\{m\}$.
\item If $\im(\alpha)$ has a minimum then $\min\im(\alpha)=\min((X\setminus Y)\alpha)$.
In addition,
\begin{enumerate}
\item if $\min\im(\alpha)\neq m$ then $z_d<z_{\min\im(\alpha)}\le z_c$, for all $d\in Y\alpha$ and for all $c\in(X\setminus Y)\alpha\setminus\{m\}$;
\item if $\min\im(\alpha)=m$ (i.e. $(X\setminus Y)\alpha=\{m\}$) then $z_{\min\im(\alpha)}\le z_x$, for all $x\in\im(\alpha)$.
\end{enumerate}
\end{enumerate}
\end{enumerate}
\end{lemma}

\smallskip

Now, let us suppose that $\alpha$ satisfies conditions 1 and 2 of Theorem \ref{regOP}, i.e.
\begin{enumerate}
\item if $\im(\alpha)$ has an upper bound or a lower bound in $X$, then $\max\im(\alpha)$ exists or $\min\im(\alpha)$ exists;
\item if $x\in X\setminus\im(\alpha)$ is neither an upper bound nor a lower bound of $\im(\alpha)$,
then either $\max\{t\in\im(\alpha)\mid t<x\}$ or $\min\{t\in\im(\alpha)\mid t>x\}$ exists.
\end{enumerate}
These conditions allow us to extend $\zeta$ to a full transformations on $X$, as follows. We define $\beta\in\T(X)$ by:
\begin{enumerate}
\item If $x\in\im(\alpha)$, then $x\beta=z_x$;

\item If $x\in X\setminus\im(\alpha)$ is an upper bound or a lower bound of $\im(\alpha)$, then
$$
x\beta=\left\{
\begin{array}{ll}
z_{\max\im(\alpha)} & \mbox{if $\im(\alpha)$ has a maximum}\\
z_{\min\im(\alpha)} & \mbox{otherwise}
\end{array}\right.
$$
(notice that if $\max\im(\alpha)$ does not exist then $\min\im(\alpha)$ has to exist);

\item  If $x\in X\setminus\im(\alpha)$ is neither an upper bound nor a lower bound of $\im(\alpha)$, then
$$
x\beta=\left\{
\begin{array}{ll}
z_{\max\{t\in\im(\alpha)\mid t<x\}} & \mbox{if $\{t\in\im(\alpha)\mid t<x\}$ has a maximum}\\
z_{\min\{t\in\im(\alpha)\mid t>x\}} & \mbox{otherwise}
\end{array}\right.
$$
(notice that if $\max\{t\in\im(\alpha)\mid t<x\}$ does not exist then $\min\{t\in\im(\alpha)\mid t>x\}$ has to exist).
\end{enumerate}

Let $x\in X$. Then $x\alpha\in\im(\alpha)$ and so $x(\alpha\beta\alpha)=((x\alpha)\beta)\alpha=z_{x\alpha}\alpha=x\alpha$.
Thus, we showed:

\begin{lemma}\label{r}
$\alpha=\alpha\beta\alpha$.
\end{lemma}

Our objective now is to prove that $\beta\in\MOP(X)$.

\smallskip

Let $C$ be a subset of $X$. We call \textit{convex closure} of $C$ to the set $\cl{C}=\bigcup_{x,y\in C, x\le y}[x,y]$. It is clear that $\cl{C}$ is a \textit{convex} subset of $X$ (in the sense that, for all $x,y\in\cl{C}$ and for all $z\in X$, $x\le z\le y$ implies $z\in\cl{C}$).

\smallskip

\begin{lemma}\label{x}
The transformation $\beta$ is order-preserving on $\cl{Y\alpha}$. Moreover, if $x\in \cl{Y\alpha}$ then $x\beta=z_d\in Y$, for some $d\in Y\alpha$.
\end{lemma}
\begin{proof}
Let $x\in \cl{Y\alpha}$. If $x\in Y\alpha$ then $x\beta=z_x\in Y$, by definition. Suppose that $x\not\in Y\alpha$.
Then $b_1<x<b_2$, for some $b_1,b_2\in Y\alpha$, and so $x\in X\setminus\im(\alpha)$ (since $x>b_1$, with $b_1\in Y\alpha$)
and $x$ is neither an upper bound nor a lower bound of $\im(\alpha)$.  Hence
$x\beta=z_d$, with
$$
d=\left\{
\begin{array}{ll}
\max\{t\in\im(\alpha)\mid t<x\} & \mbox{if $\{t\in\im(\alpha)\mid t<x\}$ has a maximum}\\
\min\{t\in\im(\alpha)\mid t>x\} & \mbox{otherwise.}
\end{array}\right.
$$
As $b_1<x$ and $b_1\in Y\alpha$, then $d\in Y\alpha$, by Lemma \ref{a}, and so $z_d\in Y$.

Let $x,y\in\cl{Y\alpha}$ be such that $x<y$.

If $x,y\in Y\alpha$, then $x\beta=z_x=x\zeta\le y\zeta=z_y=y\beta$, since $\zeta$ is order-preserving on $Y\alpha$ (we will use   this fact below several times without explicit mention).

Next, suppose that $x\in Y\alpha$ and $y\not\in Y\alpha$. Then $x\beta=z_x$ and $y\beta=z_d$,
with $d=\max\{t\in\im(\alpha)\mid t<y\}$ or $d=\min\{t\in\im(\alpha)\mid t>y\}$.
If $d=\max\{t\in\im(\alpha)\mid t<y\}$, as $x<y$, then $x\le d$.
On the other hand, if $d=\min\{t\in\im(\alpha)\mid t>y\}$ then $d>y>x$, whence $x<d$.
Thus, in both cases, we have $x\beta=z_x=x\zeta\le d\zeta=z_d=y\beta$.

Now, we consider the case $x\not\in Y\alpha$ and $y\in Y\alpha$. Then $y\beta=z_y$ and $x\beta=z_d$,
with  $d=\max\{t\in\im(\alpha)\mid t<x\}$ or $d=\min\{t\in\im(\alpha)\mid t>x\}$.
If $d=\max\{t\in\im(\alpha)\mid t<x\}$ then $d<x<y$, whence $d<y$.
On the other hand, if $d=\min\{t\in\im(\alpha)\mid t>x\}$, as $y>x$, then $d\le y$.
So, in both cases, we get $x\beta=z_d=d\zeta\le y\zeta=z_y=y\beta$.

Finally, we suppose that $x,y\not\in Y\alpha$.
Then $x\beta=z_c$, with $c=\max\{t\in\im(\alpha)\mid t<x\}$ or $c=\min\{t\in\im(\alpha)\mid t>x\}$,
and $y\beta=z_d$, with  $d=\max\{t\in\im(\alpha)\mid t<y\}$ or $d=\min\{t\in\im(\alpha)\mid t>y\}$.
If $c=\max\{t\in\im(\alpha)\mid t<x\}$ and $d=\max\{t\in\im(\alpha)\mid t<y\}$ then $c<x<y$, whence $c<y$ and so $c\le d$.
If $c=\max\{t\in\im(\alpha)\mid t<x\}$ and $d=\min\{t\in\im(\alpha)\mid t>y\}$ then $c<x<y<d$ and so $c<d$.
Next, suppose that $c=\min\{t\in\im(\alpha)\mid t>x\}$ and $d=\max\{t\in\im(\alpha)\mid t<y\}$.
Notice that, in this case, by definition of $\beta$, $\max\{t\in\im(\alpha)\mid t<x\}$ does not exist.
If $d<x$ then there exists $e\in\im(\alpha)$ such that $d<e<x$, whence $e<y$ and so $e\le d$,
which is a contradiction. Therefore, $d>x$ and so $c\le d$.
If $c=\min\{t\in\im(\alpha)\mid t>x\}$ and $d=\min\{t\in\im(\alpha)\mid t>y\}$ then $d>y>x$, whence $d>x$ and so $c\le d$.
Thus, in all cases, we have $x\beta=z_c=c\zeta\le d\zeta=z_d=y\beta$, as required.
\end{proof}

\begin{lemma}\label{c}
\begin{enumerate}
\item If $Y\alpha\cap(X\setminus Y)\alpha=\emptyset$, then $\beta$ is order-preserving on $W=\cl{(X\setminus Y)\alpha}$.

\item If $Y\alpha\cap(X\setminus Y)\alpha=\{m\}$, for some $m\in X$,  then $\beta$ is order-preserving on $W=\cl{(X\setminus Y)\alpha\setminus\{m\}}$.

\item[] Moreover, in both cases, if $x\in W$ then $x\beta=z_c\in X\setminus Y$, for some $c\in (X\setminus Y)\alpha\setminus (Y\alpha)$.

\end{enumerate}
\end{lemma}
\begin{proof}
First, notice that, in the case of 2, we have $\max((X\setminus Y)\alpha)=m=\min(Y\alpha)$ and so $x<m$, for all $x\in W$.
In particular, $m\not\in W$.

Let $x\in W$.
If $x\in (X\setminus Y)\alpha$ then $x\not\in Y\alpha$ and so $x\beta=z_x\in X\setminus Y$, by definition.
Suppose that $x\not\in(X\setminus Y)\alpha$.
Then $b_1<x<b_2$, for some $b_1,b_2\in (X\setminus Y)\alpha$ (and $b_2<m$, in the case of 2), and so $x\in X\setminus\im(\alpha)$
(since $x<b_2$, with $b_2\in (X\setminus Y)\alpha$)
and $x$ is neither an upper bound nor a lower bound of $\im(\alpha)$.  Hence
$x\beta=z_c$, with
$$
c=\left\{
\begin{array}{ll}
\max\{t\in\im(\alpha)\mid t<x\} & \mbox{if $\{t\in\im(\alpha)\mid t<x\}$ has a maximum}\\
\min\{t\in\im(\alpha)\mid t>x\} & \mbox{otherwise.}
\end{array}\right.
$$
As $x<b_2$ and $b_2\in (X\setminus Y)\alpha$, then $c\in(X\setminus Y)\alpha$, by Lemma \ref{a}. In the case of 1, it is immediate that $c\not\in Y\alpha$. In the case of 2, if $c\in Y\alpha$ then $c=m$. In this case, if $m=\max\{t\in\im(\alpha)\mid t<x\}$ then $m<x<b_2<m$, which is a contradiction, and if $m=\min\{t\in\im(\alpha)\mid t>x\}$, as $b_2>x$, then $m<b_2$, which is again a contradiction.
Hence, we also have $c\not\in Y\alpha$. Thus, in both cases, $c\in(X\setminus Y)\alpha\setminus(Y\alpha)$ and so $z_c\in X\setminus Y$.

Let $x,y\in W$ be such that $x<y$. By using the fact that $\zeta$ is also order-preserving on $(X\setminus Y)\alpha\setminus (Y\alpha)$ and replacing,  in the corresponding part of the proof of Lemma \ref{x}, each instance of $Y\alpha$ by $(X\setminus Y)\alpha\setminus (Y\alpha)$,
we show that $x\beta\le y\beta$, as required.
\end{proof}

\begin{lemma}\label{b}
If $Y\alpha\cap(X\setminus Y)\alpha=\{m\}$, for some $m\in X$,
and $(X\setminus Y)\alpha\setminus\{m\}\ne\emptyset$ then
$$
x\beta=\left\{
\begin{array}{lll}
z_{\max((X\setminus Y)\alpha\setminus\{m\})} &\in X\setminus Y& \mbox{if $(X\setminus Y)\alpha\setminus\{m\}$ has a maximum}\\
z_{m} &\in Y & \mbox{otherwise},
\end{array}\right.
$$
for $x\in X$ such that $c<x<m$, for all $c\in(X\setminus Y)\alpha\setminus\{m\}$.
\end{lemma}
\begin{proof}
Let $x\in X$ be such that $c<x<m$, for all $c\in(X\setminus Y)\alpha\setminus\{m\}$.
Then $\{t\in\im(\alpha)\mid t<x\}=(X\setminus Y)\alpha\setminus\{m\}$ and $m=\min\{t\in\im(\alpha)\mid t>x\}$ and so the result follows.
\end{proof}

Define
$$
\lb(\alpha)=\{x\in X\setminus\im(\alpha)\mid\mbox{$x$ is a lower bound of $\im(\alpha)$}\}
$$
and
$$
\ub(\alpha)=\{x\in X\setminus\im(\alpha)\mid\mbox{$x$ is an upper bound of $\im(\alpha)$}\}.
$$

\begin{lemma}\label{d}
If $Y\alpha\cap(X\setminus Y)\alpha\neq\emptyset$ then $\beta\in\MOP(X)$.
\end{lemma}
\begin{proof}
Let $m\in X$ be such that $Y\alpha\cap(X\setminus Y)\alpha=\{m\}$.
Recall that $\max((X\setminus Y)\alpha)=m=\min(Y\alpha)$.

\smallskip

Firstly, we prove that $\beta$ is order-preserving on $\lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$.

Let $x,y \in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$ be such that $x<y$. As $\beta$ is constant in $\lb(\alpha)$, by definition, and order-preserving on $\cl{(X\setminus Y)\alpha\setminus\{m\}}$, by Lemma \ref{c}, it suffices to consider the case $x\in\lb(\alpha)$ and $y\in \cl{(X\setminus Y)\alpha\setminus\{m\}}$.
In this case, $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ and, by Lemma \ref{c}, $y\beta=z_c$,
for some $c\in (X\setminus Y)\alpha\setminus\{m\}$.
Then, by Lemma \ref{aa}, $x\beta\le z_c=y\beta$.

\smallskip

In second place,
if $(X\setminus Y)\alpha\ne\{m\}$ or $\im(\alpha)$ has a maximum,
we prove that $\beta$ is order-preserving on $\cl{Y\alpha}\cup\ub(\alpha)$.

Let $x,y\in \cl{Y\alpha}\cup\ub(\alpha)$ be such that $x<y$. As $\beta$ is constant in $\ub(\alpha)$, by definition, and order-preserving on $\cl{Y\alpha}$, by Lemma \ref{x}, it suffices to consider the case $x\in\cl{Y\alpha}$ and $y\in \ub(\alpha)$.
In this case, $x\beta=z_d$, for some $d\in Y\alpha$, by Lemma \ref{x}, and $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$. Then, by Lemma \ref{aa}, $x\beta=z_d\le y\beta$.

\smallskip

Thirdly, take $x\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$ and $y\in \cl{Y\alpha}\cup\ub(\alpha)$.
It is clear that $x<y$. Moreover,
if $(X\setminus Y)\alpha\ne\{m\}$ or $\im(\alpha)$ has a maximum,
we also have $x\beta\ge y\beta$. In fact, if $x\in\lb(\alpha)$ and $y\in\ub(\alpha)$ then $x\beta=y\beta$.
If  $x\in\lb(\alpha)$ and $y\in \cl{Y\alpha}$ then
$x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ and $y\beta=z_d$,
for some $d\in Y\alpha$, by Lemma \ref{x},
whence $x\beta\ge z_d=y\beta$, by Lemma \ref{aa}.
If $x\in \cl{(X\setminus Y)\alpha\setminus\{m\}}$ and $y\in\ub(\alpha)$ then $x\beta=z_c$,
for some $c\in (X\setminus Y)\alpha\setminus\{m\}$, by Lemma \ref{c}, and $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$,
whence $x\beta=z_c\ge y\beta$, by Lemma \ref{aa}.
Finally, if $x\in \cl{(X\setminus Y)\alpha\setminus\{m\}}$ and $y\in \cl{Y\alpha}$ then, by Lemmas \ref{c} and \ref{x},
$x\beta=z_c\in X\setminus Y$, for some $c\in (X\setminus Y)\alpha\setminus\{m\}$, and
$y\beta=z_d\in Y$, for some $d\in Y\alpha$, whence $x\beta=z_c>z_d=y\beta$.

\smallskip

Next, we will consider four cases.

Let $A=\{x\in X\mid\mbox{$t<x<m$, for all $t\in(X\setminus Y)\alpha\setminus\{m\}$}\}$.

\smallskip

\noindent{\sc case 1}. $(X\setminus Y)\alpha\setminus\{m\}$ has a maximum.

Let $a=\max((X\setminus Y)\alpha\setminus\{m\})$.

Let $Z=]-\infty,m[$. Notice that, $Z=\lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}\cup A$ (and,
in this case, $A=]a,m[$) and $X\setminus Z=\cl{Y\alpha}\cup\ub(\alpha)$.

If $y\in A$ then $y\beta=z_a\in X\setminus Y$, by Lemma \ref{b}, and so $\beta$ is constant in $A$.
Let $x,y\in Z$ be such that $x<y$. As $\beta$ is constant in $A$ and order-preserving on $\lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$, in order to prove that $\beta$ is order-preserving on $Z$, it suffices to consider the case
$x\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$ and $y\in A$.
Then $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ or $x\beta=z_c$,
for some $c\in (X\setminus Y)\alpha\setminus\{m\}$.
If $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ then, by Lemma \ref{aa}, $x\beta\le z_a=y\beta$.
If $x\beta=z_c$,
for some $c\in (X\setminus Y)\alpha\setminus\{m\}$, then $c\le a$ and so $x\beta=z_c\le z_a=y\beta$, by Lemma \ref{c}.
Hence, we proved that $\beta$ is order-preserving on $Z$.

Above, we already proved that $\beta$ is also order-preserving on $X\setminus Z$.

Let $x\in Z$ and $y\in X\setminus Z$. Clearly, $x<y$. If $x\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$ then, as proved above, we have $x\beta\ge y\beta$. So, suppose that $x\in A$.
If $y\in \cl{Y\alpha}$ then $y\beta=z_d\in Y$,
for some $d\in Y\alpha$, by Lemma \ref{x}, whence $x\beta=z_a>z_d=y\beta$.
If $y\in\ub(\alpha)$ then $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$ and so,
by Lemma \ref{aa}, we have $x\beta=z_a\ge y\beta$.

Thus, in this case, we showed that $Z$ is an ideal of $\beta$ and so $\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 2}. $(X\setminus Y)\alpha\setminus\{m\}\ne\emptyset$ and does not have a maximum.

Let $Z=\lb(\alpha)\cup\cl{(X\setminus Y)\alpha\setminus\{m\}}$. Then $X\setminus Z=A\cup\cl{Y\alpha}\cup\ub(\alpha)$.

We proved above that $\beta$ is order-preserving on $Z$.

If $x\in A$ then $x\beta=z_m\in Y$, by Lemma \ref{b}, and so $\beta$ is constant in $A$.
Let $x,y\in X\setminus Z$ be such that $x<y$. As $\beta$ is constant in $A$
and order-preserving on $\cl{Y\alpha}\cup\ub(\alpha)$,
in order to prove that $\beta$ is order-preserving on $X\setminus Z$, it suffices to consider the case
$x\in A$ and $y\in \cl{Y\alpha}\cup\ub(\alpha)$.
If $y\in \cl{Y\alpha}$ then $y\beta=z_d$, for some $d\in Y\alpha$, by Lemma \ref{x},
and so, as $m\le d$, we have $x\beta=z_m\le z_d=y\beta$.
If $y\in\ub(\alpha)$ then $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$,
whence $x\beta=z_m\le y\beta$, by Lemma \ref{aa}.
Thus, $\beta$ is order-preserving on $X\setminus Z$.

Let $x\in Z$ and $y\in X\setminus Z$. Clearly, $x<y$.
If $y\in \cl{Y\alpha}\cup\ub(\alpha)$ then, as proved above, we have $x\beta\ge y\beta$. So, suppose that $y\in A$.
If $x\in\lb(\alpha)$ then $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ and so,
by Lemma \ref{aa}, we have $x\beta\ge z_m=y\beta$.
If $x\in \cl{(X\setminus Y)\alpha\setminus\{m\}}$ then $x\beta=z_c\in X\setminus Y$, for some $c\in (X\setminus Y)\alpha\setminus\{m\}$, by Lemma \ref{c}, whence $x\beta=z_c>z_m=y\beta$.

Thus, also in this case, we showed that $Z$ is an ideal of $\beta$ and so $\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 3}. $(X\setminus Y)\alpha=\{m\}$ and $\im(\alpha)$ has a maximum.

Let $Z=\lb(\alpha)$. Then $X\setminus Z=\cl{Y\alpha}\cup\ub(\alpha)$.

In this case, if $Z=\emptyset$ then, as proved above, $\beta$ is order-preserving on $\cl{Y\alpha}\cup\ub(\alpha)=X$,
whence $\beta\in\MO(X)$ and so $\beta\in\MOP(X)$.
On the other hand, if $Z\ne\emptyset$ then, as proved above, $Z$ is an ideal of $\beta$ and so again we obtain
$\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 4}. $(X\setminus Y)\alpha=\{m\}$ and $\im(\alpha)$ does not have a maximum.

Let $Z=\lb(\alpha)\cup\cl{Y\alpha}$. Then $X\setminus Z=\ub(\alpha)$.

Notice that, in this case, we have $\min\im(\alpha)=m$ and $x\beta=z_m$, for all $x\in\lb(\alpha)\cup\ub(\alpha)$.
Moreover, by Lemma \ref{aa}, we have $z_m\le z_x$, for all $x\in\im(\alpha)$ and, by Lemma \ref{x}, $\beta$ is order-preserving on $\cl{Y\alpha}$. All this together allow us to easily deduce, in this last case, that $Z$ is an ideal of $\beta$
and thus $\beta\in\MOP(X)$, as required.
\end{proof}

Now, suppose that $Y\alpha\cap(X\setminus Y)\alpha=\emptyset$, $X\setminus Y\neq\emptyset$
and there exists $x\in X$ such that $c<x<d$, for all $c\in(X\setminus Y)\alpha$ and for all $d\in Y\alpha$.
Since $\{t\in\im(\alpha)\mid t<x\}=(X\setminus Y)\alpha$ and $\{t\in\im(\alpha)\mid t>x\}=Y\alpha$ (for any such $x$), then
$(X\setminus Y)\alpha$ has a maximum or $Y\alpha$ has a minimum. Hence, we clearly have:

\begin{lemma}\label{bb}
If $Y\alpha\cap(X\setminus Y)\alpha=\emptyset$ and  $X\setminus Y\ne\emptyset$, then
$$
x\beta=\left\{
\begin{array}{lll}
z_{\max((X\setminus Y)\alpha)} &\in X\setminus Y& \mbox{if $(X\setminus Y)\alpha$ has a maximum}\\
z_{\min(Y\alpha)} &\in Y & \mbox{otherwise},
\end{array}\right.
$$
for $x\in X$ such that $c<x<d$, for all $c\in(X\setminus Y)\alpha$ and for all $d\in Y\alpha$.
\end{lemma}

With the following lemma we conclude the proof of Theorem \ref{regOP}.

\begin{lemma}\label{e}
If $Y\alpha\cap(X\setminus Y)\alpha=\emptyset$ then $\beta\in\MOP(X)$.
\end{lemma}
\begin{proof}
Firstly, we suppose that $\alpha\in\MO(X)$ and consider three cases.

\smallskip

\noindent{\sc case 1.1}. $\lb(\alpha)\ne\emptyset$ and $\im(\alpha)$ has a maximum.

Let $Z=\lb(\alpha)$. Then $X\setminus Z=\cl{\im(\alpha)}\cup\ub(\alpha)$.

By definition, $\beta$ is constant on $Z$. In fact, $x\beta=z_{\max\im(\alpha)}$, for all $x\in\lb(\alpha)\cup\ub(\alpha)$.
Since $\im(\alpha)$ has a maximum, then combining Lemmas \ref{aa} and \ref{x},
we deduce that $\beta$ is order-preserving on $\cl{\im(\alpha)}\cup\ub(\alpha)$.
On the other hand, if $x\in Z$ and $y\in X\setminus Z$ then, clearly, $x<y$ and, by Lemmas \ref{aa} and \ref{x}, we have $x\beta\ge y\beta$.
Thus, $Z$ is an ideal of $\beta$ and so $\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 1.2}. $\lb(\alpha)\ne\emptyset$ and $\im(\alpha)$ does not have a maximum.

Let $Z=\lb(\alpha)\cup\cl{\im(\alpha)}$. Then $X\setminus Z=\ub(\alpha)$.

In this case, $\min\im(\alpha)$ must exist and we have $x\beta=z_{\min\im(\alpha)}$, for all $x\in\lb(\alpha)\cup\ub(\alpha)$.
Then, by using Lemmas \ref{aa} and \ref{x}, it is easy to conclude that $Z$ is an ideal of $\beta$ and so $\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 1.3}. $\lb(\alpha)=\emptyset$.

If $\im(\alpha)$ has a maximum then, as in {\sc case 1.1}, by Lemmas \ref{aa} and \ref{x},
we obtain that $\beta$ is order-preserving on $\cl{\im(\alpha)}\cup\ub(\alpha)$, i.e. $\beta\in\MO(X)$ and so $\beta\in\MOP(X)$.
On the other hand, if $\im(\alpha)$ does not have a maximum, again by Lemmas \ref{aa} and \ref{x},
it is easy to conclude that $Z=\cl{\im(\alpha)}$ is an ideal of $\beta$ (notice that $\ub(\alpha)=\emptyset$ or $\im(\alpha)$ has a minimum) and so, also in this case, $\beta\in\MOP(X)$.

\medskip

Secondly, we suppose that $\alpha\not\in\MO(X)$. Then, we have $X\setminus Y\ne\emptyset$.

\smallskip

Within this assumption, we begin by proving that $\beta$ is order-preserving on $\lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$.

Let $x,y\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$ be such that $x<y$.
Since $\beta$ is constant on $\lb(\alpha)$ and, by Lemma \ref{c}, $\beta$ is order-preserving on $\cl{(X\setminus Y)\alpha}$,
it suffices to consider $x\in\lb(\alpha)$ and $y\in\cl{(X\setminus Y)\alpha}$.
In this case, we have $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ and, by Lemma \ref{c}, $y\beta=z_c$,
for some $c\in (X\setminus Y)\alpha$.
Then, by Lemma \ref{aa}, it follows that $x\beta\le z_c=y\beta$.

\smallskip

We continue by showing that $\beta$ is order-preserving on $\cl{Y\alpha}\cup\ub(\alpha)$.

Let $x,y\in \cl{Y\alpha}\cup\ub(\alpha)$ be such that $x<y$. Once again,
by Lemma \ref{x}, we have that $\beta$ is order-preserving on $\cl{Y\alpha}$ and, by definition,
$\beta$ is constant on $\ub(\alpha)$. So, it suffices to consider $x\in \cl{Y\alpha}$ and $y\in\ub(\alpha)$.
In this case, we have $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$ and, by Lemma \ref{x}, $x\beta=z_d$,
for some $d\in Y\alpha$. Then, by Lemma \ref{aa}, it follows that $x\beta=z_d\le y\beta$.

\smallskip

Next, take $x\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$ and $y\in \cl{Y\alpha}\cup\ub(\alpha)$. Clearly, $x<y$.
Let us prove that we also have $x\beta\ge y\beta$.
If $x\in\lb(\alpha)$ and $y\in\cl{Y\alpha}$ then $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ and, by Lemma \ref{x}, $y\beta=z_d$,
for some $d\in Y\alpha$, whence $x\beta\ge z_d=y\beta$, by Lemma \ref{aa}.
If $x\in\lb(\alpha)$ and $y\in \ub(\alpha)$ then $x\beta=y\beta$.
If $x\in\cl{(X\setminus Y)\alpha}$ and $y\in\cl{Y\alpha}$ then, by Lemmas \ref{c} and \ref{x},
$x\beta=z_c\in X\setminus Y$, for some $c\in (X\setminus Y)\alpha$, and $y\beta=z_d\in Y$, for some $d\in Y\alpha$,
whence $x\beta=z_c>z_d=y\beta$.
Finally, if  $x\in\cl{(X\setminus Y)\alpha}$ and $y\in \ub(\alpha)$ then, by Lemma \ref{c}, $x\beta=z_c$,
for some $c\in (X\setminus Y)\alpha$, and $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$,
whence $x\beta=z_c\ge y\beta$, by Lemma \ref{aa}.

\smallskip

Now, let $B=\{x\in X\mid\mbox{$c<x<d$, for all $c\in(X\setminus Y)\alpha$ and for all $d\in Y\alpha$}\}$.
We will consider three cases.

\smallskip

\noindent{\sc case 2.1}. $B=\emptyset$.

In this case, being $Z=\lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$, we just proved that $Z$ is an ideal of $\beta$
(notice that $X\setminus Z= \cl{Y\alpha}\cup\ub(\alpha)$) and so $\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 2.2}. $B\ne\emptyset$ and $(X\setminus Y)\alpha$ has a maximum.

Notice that, for $x\in B$, we have $x\beta=z_{\max((X\setminus Y)\alpha)}\in X\setminus Y$, by Lemma \ref{bb}.

Take $Z=\lb(\alpha)\cup\cl{(X\setminus Y)\alpha}\cup B$ and let us prove that $Z$ is an ideal of $\beta$.

In order to prove that $\beta$ is order-preserving on $Z$, let $x,y\in Z$ be such that $x<y$.
Since we already proved that $\beta$ is order-preserving on $\lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$,
it suffices to consider $x\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$ and $y\in B$.
Then $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ or, by Lemma \ref{c},
$x\beta=z_c\in X\setminus Y$, for some $c\in (X\setminus Y)\alpha$, and
$y\beta=z_{\max((X\setminus Y)\alpha)}\in X\setminus Y$.
If  $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ then, by Lemma \ref{aa},
we have $x\beta\le z_{\max((X\setminus Y)\alpha)}=y\beta$.
On the other hand, if $x\beta=z_c$, for some $c\in (X\setminus Y)\alpha$, then $c\le \max((X\setminus Y)\alpha)$ and so,
as  $\zeta$ is order-preserving on $(X\setminus Y)\alpha$, we have
$x\beta=z_c=c\zeta\le (\max((X\setminus Y)\alpha))\zeta=z_{\max((X\setminus Y)\alpha)}=y\beta$.
Thus, we proved that $\beta$ is order-preserving on $Z$.

Since $X\setminus Z=\cl{Y\alpha}\cup\ub(\alpha)$, we have already proved that $\beta$ is order-preserving on $X\setminus Z$.

Let $x\in Z$ and $y\in X\setminus Z$. Clearly, $x<y$. If $x\in \lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$,
we have proved above that $x\beta\ge y\beta$. So, suppose that $x\in B$. Then
$x\beta=z_{\max((X\setminus Y)\alpha)}\in X\setminus Y$.
If $y\in \ub(\alpha)$ then $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$ and so, by Lemma \ref{aa},
we have $y\beta\le z_{\max((X\setminus Y)\alpha)}=x\beta$.
On the other hand, if $y\in\cl{Y\alpha}$ then, by Lemma \ref{x}, $y\beta=z_d\in Y$, for some $d\in Y\alpha$, and so
$y\beta=z_d < z_{\max((X\setminus Y)\alpha)}=x\beta$.

Thus, we have proved that $Z$ is an ideal of $\beta$ and so $\beta\in\MOP(X)$.

\smallskip

\noindent{\sc case 2.3}. $B\ne\emptyset$ and $(X\setminus Y)\alpha$ does not have a maximum.

In this case, $\min(Y\alpha)$ exists and, for $x\in B$, by Lemma \ref{bb}, we have $x\beta=z_{\min(Y\alpha)} \in Y$.

Let $Z=\lb(\alpha)\cup\cl{(X\setminus Y)\alpha}$. We aim to show that $Z$ is an ideal of $\beta$.

We already proved that $\beta$ is order-preserving on $Z$.

We proceed by showing that $\beta$ is order-preserving on $X\setminus Z=B\cup\cl{Y\alpha}\cup\ub(\alpha)$.
Let $x,y\in X\setminus Z$ be such that $x<y$.
Since we already proved that $\beta$ is order-preserving on $\cl{Y\alpha}\cup\ub(\alpha)$,
it suffices to consider $x\in B$ and $y\in \cl{Y\alpha}\cup\ub(\alpha)$.
If $y\in \ub(\alpha)$ then $y\beta=z_{\max\im(\alpha)}$ or $y\beta=z_{\min\im(\alpha)}$ and so, by Lemma \ref{aa},
we have $x\beta=z_{\min(Y\alpha)} \le y\beta$.
On the other hand, if $y\in \cl{Y\alpha}$ then $y\beta=z_d$, for some $d\in Y\alpha$,
whence $\min(Y\alpha)\le d$ and so, as $\zeta$ is order-preserving on $Y\alpha$, we have
$x\beta= z_{\min(Y\alpha)} =(\min(Y\alpha))\zeta\le d\zeta=z_d=y\beta$.
Hence, $\beta$ is order-preserving on $X\setminus Z$.

Finally, let $x\in Z$ and $y\in X\setminus Z$. It is clear that $x<y$.
If $y\in \cl{Y\alpha}\cup\ub(\alpha)$ then we already proved that $x\beta\ge y\beta$.
So, let us suppose that $y\in B$. Then $y\beta=z_{\min(Y\alpha)} \in Y$.
If $x\in \lb(\alpha)$ then $x\beta=z_{\max\im(\alpha)}$ or $x\beta=z_{\min\im(\alpha)}$ and so, by Lemma \ref{aa},
we have $y\beta=z_{\min(Y\alpha)} \le x\beta$.
On the other hand, if $x\in\cl{(X\setminus Y)\alpha}$ then, by Lemma \ref{c}, we have
$x\beta=z_c\in X\setminus Y$, for some $c\in (X\setminus Y)\alpha$,  and so
$y\beta=z_{\min(Y\alpha)} < z_c=x\beta$.

Thus, also in this case, we have proved that $Z$ is an ideal of $\beta$ and so $\beta\in\MOP(X)$.

The proof of the lemma is now complete.
\end{proof}

\section{Green's relations}\label{RGR}

A description of the Green's relations in $\MO(X)$ was given by Fernandes et al. in \cite{Fernandes&al:2014}.
In this section we characterize Green's relations in $\MOP(X)$, $\PO(X)$, $\POP(X)$, $\POI(X)$ and $\POPI(X)$.
We begin by focusing our attention in the semigroup $\MOP(X)$, which is, in fact, the most challenging case.

\smallskip

First, recall the following description of the Green's relations in $\T(X)$.
Let $\alpha,\beta\in\T(X)$. Then, in $\T(X)$, we have:
\begin{description}
\item $\alpha\GRL\beta$ if and only if $\im(\alpha)=\im(\beta)$;
\item $\alpha\GRR\beta$ if and only if $\ker(\alpha)=\ker(\beta)$;
\item $\alpha\GRD\beta$ if and only if $|\im(\alpha)|=|\im(\beta)|$;
\item $\GRJ=\GRD$.
\end{description}
See \cite{Howie:1995}.

\medskip

Let $\alpha,\beta\in\MOP(X)$. If $\alpha\GRL\beta$ in $\MOP(X)$ then $\alpha\GRL\beta$ in $\T(X)$
and so $\im(\alpha)=\im(\beta)$.  Next, we aim to show that the converse is also true.
We start by proving a series of three lemmas.

First, notice that, if $\alpha\in\POP(X)$ admits $Y$ as an ideal then $x\leq y$,
for all $x\in (X\setminus Y)\alpha$ and $y\in Y\alpha$.

\begin{lemma}\label{lema1L}
Let $\alpha,\beta\in\MOP(X)$ be such that $\im(\alpha)=\im(\beta)$.
If $A$ and $B$ are ideals of $\alpha$ and $\beta$, respectively,
then $A\alpha\subseteq B\beta$ or $B\beta\subseteq A\alpha$.
\end{lemma}
\begin{proof}
Suppose that $A\alpha\nsubseteq B\beta$.
Then, there exists $z\in A\alpha$ such that $z\not\in B\beta$.
As $z\in A\alpha\subseteq\im(\alpha)=\im(\beta)=B\beta\cup (X\setminus B)\beta$
and $z\not\in B\beta$, then $z\in(X\setminus B)\beta$. It follows that
$z<y$, for all $y\in B\beta$.

Let $y\in B\beta$. Since $B\beta\subseteq\im(\beta)=\im(\alpha)=A\alpha\cup (X\setminus A)\alpha$,
then $y\in A\alpha$ or $y\in(X\setminus A)\alpha$.
If $y\in(X\setminus A)\alpha$ then $y\leq z$, a contradiction.
Hence $y\in A\alpha$ and so $B\beta\subseteq A\alpha$, as required.
\end{proof}

\begin{lemma}\label{lema2L}
Let $\alpha,\beta\in\MOP(X)$ be such that $\im(\alpha)=\im(\beta)$.
If $A$ and $B$ are ideals of $\alpha$ and $\beta$, respectively,
and $A\alpha\subsetneqq B\beta$ then $(X\setminus B)\beta\subseteq (X\setminus A)\alpha$.
\end{lemma}
\begin{proof}
Let $z\in B\beta\setminus A\alpha$.
Then, as $z\in B\beta\subseteq\im(\beta)=\im(\alpha)=A\alpha\cup(X\setminus A)\alpha$, we can
conclude that $z\in(X\setminus A)\alpha$.
Let $y\in(X\setminus B)\beta$. Then $y\leq z$. If $y\in A\alpha$, as
$z\in(X\setminus A)\alpha$, then $z\leq y$ and so
$y=z\not\in A\alpha$, a contradiction. Hence $y\in(X\setminus A)\alpha$,
since $(X\setminus B)\beta\subseteq\im(\beta)=\im(\alpha)=A\alpha\cup(X\setminus A)\alpha$.
Thus  $(X\setminus B)\beta\subseteq (X\setminus A)\alpha$, as required.
\end{proof}

\begin{lemma}\label{lema3L}
Let $\alpha,\beta\in\MOP(X)$ be such that $\im(\alpha)=\im(\beta)$.
If $A$ and $B$ are ideals of $\alpha$ and $\beta$, respectively,
and $A\alpha=B\beta$ then:
\begin{enumerate}
    \item $A\alpha\cap(X\setminus A)\alpha=\{m_\alpha\}$ and
    $B\beta\cap(X\setminus B)\beta=\{m_\beta\}$ implies $m_\alpha=m_\beta$ and
    $(X\setminus A)\alpha=(X\setminus B)\beta$;
    \item $A\alpha\cap(X\setminus A)\alpha=\{m_\alpha\}$ and
    $B\beta\cap(X\setminus B)\beta=\emptyset$ implies $(X\setminus A)\alpha=(X\setminus B)\beta\cup\{m_\alpha\}$;
    \item $A\alpha\cap(X\setminus A)\alpha=\emptyset$ and
    $B\beta\cap(X\setminus B)\beta=\{m_\beta\}$ implies $(X\setminus B)\beta=(X\setminus A)\alpha\cup\{m_\beta\}$;
    \item $A\alpha\cap(X\setminus A)\alpha=\emptyset$ and
    $B\beta\cap(X\setminus B)\beta=\emptyset$ implies $(X\setminus A)\alpha=(X\setminus B)\beta$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let us admit the hypothesis of 1. Then, by Proposition \ref{glued},
we have $m_\alpha=\min(A\alpha)=\min(B\beta)=m_\beta$.
On the other hand,
$(X\setminus A)\alpha=(\im(\alpha)\setminus A\alpha)\ \cup \
\{m_\alpha\}=(\im(\beta)\setminus B\beta)\ \cup \
\{m_\beta\}=(X\setminus B)\beta$.

Properties 2, 3, and 4 are clear.
\end{proof}

Now, we prove the description of the Green relation $\GRL$ on $\MOP(X)$ announced above.

\begin{proposition}\label{prop1}
Let $\alpha,\beta\in\MOP(X)$. Then $\alpha\GRL\beta$ in $\MOP(X)$ if and only if $\im(\alpha)=\im(\beta)$.
\end{proposition}
\begin{proof}
It remains to prove that $\im(\alpha)=\im(\beta)$ implies $\alpha\GRL\beta$.
Therefore, suppose that $\im(\alpha)=\im(\beta)$ and
let $A$ and $B$ be ideals of $\alpha$ and $\beta$, respectively.

For each $y\in\im(\beta)=\im(\alpha)$,
choose $z_y\in y\beta^{-1}$
such that if $y\in B\beta$ then $z_y\in B$.

Define a transformation $\gamma\in\T(X)$ by $x\gamma=z_{x\alpha}$, for all $x\in X$.
Then, $x\gamma\beta=z_{x\alpha}\beta=x\alpha$, for all $x\in X$, i.e. $\alpha=\gamma\beta$.
Now, our aim is to prove that $\gamma\in\MOP(X)$. With this in mind, we consider two cases.

\smallskip

\noindent\textsc{case 1.} $A\alpha\subseteq B\beta$.

Let $C=X\setminus\{x\in X\setminus A\mid x\alpha\in B\beta\}$.

We begin by proving that $C$ satisfies (OP1) for $\gamma$.

Let $x_1,x_2\in C$ be such that $x_1\leq x_2$.
Then either $x_i\in A$ or $x_i\in X\setminus A$ and $x_i\alpha\not\in B\beta$, for $i=1,2$.

First, suppose that $x_1,x_2\in A$. Then
$x_1\alpha,x_2\alpha\in A\alpha\subseteq B\beta$,
from which it follows that $z_{x_1\alpha},z_{x_2\alpha}\in B$.
Moreover, $x_1\alpha\leq x_2\alpha$.
If $z_{x_2\alpha}<z_{x_1\alpha}$ then
$x_2\alpha=z_{x_2\alpha}\beta\le z_{x_1\alpha}\beta=x_1\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$,
which is a contradiction.
Hence $x_1\gamma=z_{x_1\alpha}\leq z_{x_2\alpha}=x_2\gamma$.

Secondly, suppose that $x_1,x_2\in X\setminus A$.
Then $x_1\alpha\leq x_2\alpha$. Additionally,  $x_1\alpha, x_2\alpha\not\in B\beta$
from which follows $z_{x_1\alpha},z_{x_2\alpha}\in X\setminus B$.
As above, if $z_{x_2\alpha}<z_{x_1\alpha}$ then
$x_2\alpha=z_{x_2\alpha}\beta\le z_{x_1\alpha}\beta=x_1\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$,
which is a contradiction.
Hence $x_1\gamma=z_{x_1\alpha}\leq z_{x_2\alpha}=x_2\gamma$.

Finally, suppose that $x_1\in A$ and $x_2\in X\setminus A$.
Then, as above, we may deduce that $z_{x_1\alpha}\in B$ and
$z_{x_2\alpha}\in X\setminus B$. Hence
$x_1\gamma=z_{x_1\alpha}<z_{x_2\alpha}=x_2\gamma$.

Notice that, since $x_1\le x_2$, we cannot have $x_1\in X\setminus A$
and $x_2\in A$.

Now, let $x_1,x_2\in X\setminus C$ be such that $x_1\leq x_2$.
Then $x_1,x_2\in X\setminus A$ and $x_1\alpha,x_2\alpha\in B\beta$.
Hence $x_1\alpha\leq x_2\alpha$ and $z_{x_1\alpha},z_{x_2\alpha}\in B$.
Once again, if $z_{x_2\alpha}<z_{x_1\alpha}$ then
$x_2\alpha=z_{x_2\alpha}\beta\le z_{x_1\alpha}\beta=x_1\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$,
which is a contradiction. Hence $x_1\gamma=z_{x_1\alpha}\leq z_{x_2\alpha}=x_2\gamma$.

Thus, we proved that $C$ satisfies (OP1) for $\gamma$.

Next, we prove that $C$ satisfies (OP2) for $\gamma$.

Let $x_1\in C$ and $x_2\in X\setminus C$.
Then $x_2\in X\setminus A$ and $x_2\alpha\in B\beta$.
Regarding $x_1$, we have two cases.

First, suppose that $x_1\in A$.
Then, we obtain  $x_1\le x_2$ and $x_1\alpha\geq x_2\alpha$.
On the other hand, we also have $x_1\alpha\in A\alpha\subseteq B\beta$.
Thus $z_{x_1\alpha},z_{x_2\alpha}\in B$.
If $z_{x_1\alpha}<z_{x_2\alpha}$ then
$x_1\alpha=z_{x_1\alpha}\beta\leq z_{x_2\alpha}\beta=x_2\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$,
which is a contradiction.
Hence $x_1\gamma=z_{x_1\alpha}\geq z_{x_2\alpha}=x_2\gamma$.

Lastly, suppose that $x_1\in X\setminus A$.
Then $x_1\alpha\not\in B\beta$ and so $x_1\alpha\in (X\setminus B)\beta$ and $z_{x_1\alpha}\in X\setminus B$.

On the other hand, as $x_2\alpha\in B\beta$, then $z_{x_2\alpha}\in B$. Hence
$x_1\alpha\le x_2\alpha$ and $x_2\gamma=z_{x_2\alpha}< z_{x_1\alpha}=x_1\gamma$.

If $x_2\le x_1$ then $x_2\alpha\le x_1\alpha$, since $x_1,x_2\in X\setminus A$, and so $x_1\alpha=x_2\alpha$,
from which follows that $z_{x_1\alpha}= z_{x_2\alpha}$, a contradiction. Thus, we also have $x_1< x_2$.
This finishes the proof that $C$ satisfies (OP2) for $\gamma$.

Therefore, $C$ is an ideal of $\gamma$ (notice that, as $A\subseteq C$, we have $C\ne\emptyset$) and so $\gamma\in\MOP(X)$.

\smallskip

\noindent\textsc{case 2.} $B\beta\subsetneqq A\alpha$.

Notice that, by Lemma \ref{lema2L}, we have $(X\setminus A)\alpha\subseteq (X\setminus B)\beta$.

Let $C=\{x\in A\mid x\alpha\in A\alpha\setminus B\beta\}$.

Observe that $C\ne\emptyset$, otherwise we would have $A\alpha\subseteq B\beta$, a contradiction.

We start by proving that $C$ satisfies (OP1) for $\gamma$.

Let $x_1,x_2\in C$ be such that $x_1\leq x_2$.
Then $x_1,x_2\in A$ and so $x_1\leq x_2$ implies $x_1\alpha\leq x_2\alpha$.
Moreover,  $x_1\alpha,x_2\alpha\in A\alpha\setminus B\beta$, from which follows
$z_{x_1\alpha},z_{x_2\alpha}\in X\setminus B$.
If $z_{x_2\alpha}<z_{x_1\alpha}$ then $x_2\alpha=z_{x_2\alpha}\beta\leq z_{x_1\alpha}\beta=x_1\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$, which is a contradiction.
Thus $x_1\gamma=z_{x_1\alpha}\le z_{x_2\alpha}=x_2\gamma$.

Next, let $x_1,x_2\in X\setminus C$ be such that $x_1\leq x_2$.
Then, either $x_i\in X\setminus A$ or $x_i\in A$ and $x_i\alpha\in B\beta$,
for $i=1,2$.

First, suppose that $x_1,x_2\in A$.
Then $x_1\alpha\leq x_2\alpha$. Additionally,
$x_1\alpha,x_2\alpha\in B\beta$ and so $z_{x_1\alpha},z_{x_2\alpha}\in B$.
Once again,
if $z_{x_2\alpha}<z_{x_1\alpha}$ then $x_2\alpha=z_{x_2\alpha}\beta\leq z_{x_1\alpha}\beta=x_1\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$, which is a contradiction.
Thus $x_1\gamma=z_{x_1\alpha}\le z_{x_2\alpha}=x_2\gamma$.

Secondly, suppose that $x_1\in A$ and $x_2\in X\setminus A$.
Then, as above,  $x_1\alpha\in B\beta$ and so $z_{x_1\alpha}\in B$.
Moreover, $x_2\alpha\in(X\setminus A)\alpha\subseteq (X\setminus B)\beta$.
Let us suppose that we also have $x_2\alpha\in B\beta$.
Then, by Proposition \ref{glued}, we get $x_2\alpha=\max((X\setminus B)\beta)=\min(B\beta)$.
Since $B\beta\subsetneqq A\alpha$, there exists $y\in A\alpha$ such that $y\not\in B\beta$.
Hence $y\in(X\setminus B)\beta$ and so $y\leq x_2\alpha$.
On the other hand,
as $x_2\alpha\in(X\setminus A)\alpha$ and $y\in A\alpha$, we obtain $x_2\alpha\leq y$.
Then $y=x_2\alpha\in B\beta$, which is a contradiction. Thus $x_2\alpha\not\in B\beta$ and
so $z_{x_2\alpha}\in X\setminus B$. Therefore,
$x_1\gamma=z_{x_1\alpha}<z_{x_2\alpha}=x_2\gamma$.

Again, notice that, since $x_1\le x_2$, we cannot have $x_1\in X\setminus A$
and $x_2\in A$. Therefore, finally, we suppose that $x_1,x_2\in X\setminus A$.
Then, we have $x_1\alpha, x_2\alpha\in(X\setminus A)\alpha\subseteq (X\setminus B)\beta$
and, as above, we may deduce that $x_1\alpha, x_2\alpha\not\in B\beta$, from which follows that
$z_{x_1\alpha},z_{x_2\alpha}\in X\setminus B$.
Moreover, we have $x_1\alpha\le x_2\alpha$.
So, yet again,
if $z_{x_2\alpha}<z_{x_1\alpha}$ then $x_2\alpha=z_{x_2\alpha}\beta\leq z_{x_1\alpha}\beta=x_1\alpha$,
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$, which is a contradiction.
Thus $x_1\gamma=z_{x_1\alpha}\le z_{x_2\alpha}=x_2\gamma$.

Thus, we proved that $C$ satisfies (OP1) for $\gamma$.

It remains to prove that $C$ satisfies (OP2) for $\gamma$.

Let $x_1\in C$ and $x_2\in X\setminus C$.
Then $x_1\in A$ and $x_1\alpha\not\in B\beta$ and so $z_{x_1\alpha}\in X\setminus B$.
On the other hand, either $x_2\in X\setminus A$ or $x_2\in A$ and $x_2\alpha\in B\beta$.

First, suppose that $x_2\in X\setminus A$. As $x_1\in A$, we have $x_1\le x_2$ and $x_1\alpha\ge x_2\alpha$.
On the other hand,  $x_2\alpha\in(X\setminus A)\alpha\subseteq (X\setminus B)\beta$
and, once again, we may deduce that $x_2\alpha\not\in B\beta$, from which follows that
$z_{x_2\alpha}\in X\setminus B$.
If $z_{x_1\alpha}<z_{x_2\alpha}$ then $x_1\alpha=z_{x_1\alpha}\beta\le z_{x_2\alpha}\beta=x_2\alpha$
(since $z_{x_1\alpha}, z_{x_2\alpha}\in X\setminus B$),
whence $x_1\alpha=x_2\alpha$ and so $z_{x_1\alpha}=z_{x_2\alpha}$, which is a contradiction.
Thus $x_1\gamma=z_{x_1\alpha}\ge z_{x_2\alpha}=x_2\gamma$.

Finally, we suppose that $x_2\in A$ and so  $x_2\alpha\in B\beta$.
As $x_1\alpha\not\in B\beta$, we must have $x_1\alpha\le x_2\alpha$.
If $x_1>x_2$ then $x_1\alpha\ge x_2\alpha$ (since $x_1,x_2\in A$)
and so $x_1\alpha=x_2\alpha\in B\beta$, which is a contradiction.
Hence $x_1\le x_2$.
On the other hand, from $x_1\alpha\not\in B\beta$ and $x_2\alpha\in B\beta$,
it follows that $z_{x_1\alpha}\in X\setminus B$ and $z_{x_2\alpha}\in B$ and so
$x_1\gamma=z_{x_1\alpha}> z_{x_2\alpha}=x_2\gamma$.

Therefore, $C$ is an ideal of $\gamma$ and so, also in this case, we have $\gamma\in\MOP(X)$.

\smallskip

So, we have showed that $\alpha=\gamma\beta$, with $\gamma\in\MOP(X)$.
Analogously, we can prove that $\beta=\lambda\alpha$, for some $\lambda\in\MOP(X)$.
Therefore, $\alpha\GRL\beta$, as required.
\end{proof}

\medskip

In order to describe the Green's relation $\GRR$ in $\MOP(X)$,
we need to introduce the following concepts.

Let $A$ and $B$ be two subsets of $X$ and let $\theta:A\longrightarrow B$ be a mapping.
We say that $\theta$ is \textit{completable} in $\MOP(X)$ if there exists $\gamma\in\MOP(X)$ such that
$x\gamma=x\theta$, for all $x\in A$.
To such a transformation $\gamma$ (not necessarily unique) we call a \textit{complete extension}
of $\theta$ in $\MOP(X)$.
If $\theta:A\longrightarrow B$ is a bijection, we say that $\theta$ is \textit{bicompletable} in $\MOP(X)$
if both $\theta$ and its inverse $\theta^{-1}:B\longrightarrow A$ are completable in $\MOP(X)$.
Let $\theta:A\longrightarrow B$ be an injective mapping.
By the \textit{inverse} $\theta^{-1}$ of $\theta$ we mean the inverse mapping $\theta^{-1}:\im(\theta)\longrightarrow A$
of the bijection $\theta:A\longrightarrow \im(\theta)$.
Thus, we say that an injective mapping $\theta:A\longrightarrow B$ admits a completable inverse in $\MOP(X)$
if its inverse $\theta^{-1}:\im(\theta)\longrightarrow A$ is completable in $\MOP(X)$.

\smallskip

Let $\alpha,\beta\in\PT(X)$ be such that $\ker\alpha=\ker\beta$.
Let $\theta\subseteq \im(\alpha)\times\im(\beta)$ be the relation defined by
$$
(a,b)\in\theta\,\text{ if and only if }\,a\alpha^{-1}=b\beta^{-1},\,\text{ for all
}\,(a,b)\in\im(\alpha)\times\im(\beta).
$$
Then, it is a routine matter to show that
$\theta:\im(\alpha)\longrightarrow \im(\beta)$ is a bijection such that
$\alpha=\beta\theta^{-1}$ and $\beta=\alpha\theta$.
Under these conditions, we say that $\theta$ is the \textit{canonical bijection} from $\im(\alpha)$ into $\im(\beta)$.
Notice that, given $a\in\im(\alpha)$, we have $a\theta=x\beta$, for any $x\in a\alpha^{-1}$.

\smallskip

Now, we can present our description of the relation $\GRR$ in $\MOP(X)$.

\begin{proposition}\label{prop3}
Let $\alpha,\beta\in\MOP(X)$. Then
$\alpha\GRR\beta$ in $\MOP(X)$ if and only if
$\ker(\alpha)=\ker(\beta)$ and the canonical bijection
$\theta:\im(\alpha)\longrightarrow \im(\beta)$ is bicompletable in
$\MOP(X)$.
\end{proposition}
\begin{proof}
First, suppose that $\alpha\GRR\beta$ in $\MOP(X)$ and
let $\gamma,\lambda\in\MOP(X)$ be such that $\alpha=\beta\gamma$ and
$\beta=\alpha\lambda$. As $\alpha\GRR\beta$ in $\MOP(X)$, we also have
$\alpha\GRR\beta$ in $\T(X)$ and so $\ker(\alpha)=\ker(\beta)$.
Let $\theta:\im(\alpha)\longrightarrow \im(\beta)$ the canonical
bijection.
Let $a\in\im(\alpha)$. Then $a\alpha^{-1}=(a\theta)\beta^{-1}$ and,
by taking $x\in a\alpha^{-1}$, we have $a\lambda=x\alpha\lambda=x\beta=a\theta$.
Hence $\lambda$ is a complete extension of $\theta$.
Similarly, we may show that $\gamma$ is a complete extension in $\MOP(X)$ of $\theta^{-1}$.
Therefore $\theta$ is bicompletable  in $\MOP(X)$.

Conversely, suppose that $\ker(\alpha)=\ker(\beta)$ and the
canonical bijection $\theta:\im(\alpha)\longrightarrow \im(\beta)$
is bicompletable in $\MOP(X)$.
Let $\lambda$ and $\gamma$
be complete extensions in $\MOP(X)$ of $\theta$ and $\theta^{-1}$, respectively.
Since $\beta=\alpha\theta$, $\alpha=\beta\theta^{-1}$, $\dom(\theta)=\im(\alpha)$ and
$\dom(\theta^{-1})=\im(\beta)$, we have $\beta=\alpha\lambda$ and $\alpha=\beta\gamma$,
whence $\alpha\GRR\beta$, as required.
\end{proof}

From Proposition \ref{prop1} and Proposition \ref{prop3} it follows immediately:

\begin{corollary}\label{Hrelation}
Let $\alpha,\beta\in\MOP(X)$. Then
$\alpha\GRH\beta$ in $\MOP(X)$ if and only if
$\im(\alpha)=\im(\beta)$, $\ker(\alpha)=\ker(\beta)$ and the
canonical bijection $\theta:\im(\alpha)\longrightarrow \im(\beta)$
is bicompletable in $\MOP(X)$.
\end{corollary}

Regarding the relation $\GRD$, we have:

\begin{proposition}\label{prop4}
Let $\alpha,\beta\in\MOP(X)$. Then
$\alpha\GRD\beta$ in $\MOP(X)$ if and only if there exists a
bijective mapping  $\theta:\im(\alpha)\longrightarrow \im(\beta)$ which is bicompletable in $\MOP(X)$.
\end{proposition}
\begin{proof}
Suppose that $\alpha\GRD\beta$. Then, there exists
$\gamma\in\MOP(X)$ such that $\alpha\GRR\gamma$ and
$\gamma\GRL\beta$. By Proposition \ref{prop3}, we have that
$\ker(\alpha)=\ker(\gamma)$ and the canonical bijection
$\theta:\im(\alpha)\longrightarrow \im(\gamma)$ is bicompletable
in $\MOP(X)$. On the other hand, by Proposition \ref{prop1}, we
have $\im(\gamma)=\im(\beta)$. Thus, we obtain a bijection
$\theta:\im(\alpha)\longrightarrow \im(\gamma)=\im(\beta)$ which
is bicompletable in $\MOP(X)$.

Conversely, suppose that there exists a bijection
$\theta:\im(\alpha)\longrightarrow\im(\beta)$ which is bicompletable in
$\MOP(X)$. Let $\xi$ be a complete extension in $\MOP(X)$ of $\theta$.
Let $\gamma=\alpha\xi$. Then $\gamma\in\MOP(X)$.
Moreover, $\gamma=\alpha\theta$ and $\im(\gamma)=(\im(\alpha))\theta=\im(\beta)$.
Hence, by Proposition \ref{prop1}, we have
$\gamma\GRL\beta$.
On the other hand, let $x,y\in X$. Since $\theta$ is a bijection, we have
$$
x\alpha=y\alpha\,\Leftrightarrow\,(x\alpha)\theta=(y\alpha)\theta\,\Leftrightarrow\,x\gamma=y\gamma.
$$
Then $\ker(\alpha)=\ker(\gamma)$.
Let us consider the canonical bijection $\tau:\im(\alpha)\longrightarrow\im(\gamma)$.
Next, we prove that $\tau$ is bicompletable in $\MOP(X)$ by showing that $\tau=\theta$.
Let $a\in\im(\alpha)$ and take $x\in a\alpha^{-1}$. Then,
$a\alpha^{-1}=(a\tau)\gamma^{-1}$ and
$a\tau=x\gamma=x\alpha\theta=a\theta$.
Thus, $\tau=\theta$ and so $\tau$ is bicompletable in $\MOP(X)$,
whence $\alpha\GRR\gamma$, by Proposition \ref{prop3}.

Therefore $\alpha\GRD\beta$, as required.
\end{proof}

To finish the study of the Green's relations in $\MOP(X)$, we give the following description of $\GRJ$.

\begin{proposition}\label{prop5}
Let $\alpha,\beta\in\MOP(X)$. Then $\alpha\GRJ\beta$ in $\MOP(X)$ if and only if there exist
injective mappings $\theta:\im(\alpha)\longrightarrow \im(\beta)$ and
$\tau:\im(\beta)\longrightarrow \im(\alpha)$ admitting completable
inverses in $\MOP(X)$.
\end{proposition}

\begin{proof}
Suppose that $\alpha\GRJ\beta$. Then, em particular, there exist
$\lambda,\gamma\in\MOP(X)$ such that
$\alpha=\lambda\beta\gamma$.

For each $a\in\im(\alpha)=(\im(\lambda\beta))\gamma$,
choose $w_a\in\im(\lambda\beta)\cap a\gamma^{-1}$.
Then, define a mapping $\theta:\im(\alpha)\longrightarrow X$ by $a\theta=w_a$,
for all $a\in\im(\alpha)$. Notice that $\im(\theta)\subseteq\im(\beta)$.

Next, we prove that $\theta$ is injective.
Let $a,b\in\im(\alpha)$ be such that $a\theta=b\theta$.
Then $w_a=w_b$ and, as $w_a\in a\gamma^{-1}$ and $w_b\in b\gamma^{-1}$, we obtain
$a=w_a\gamma=w_b\gamma=b$. Hence, $\theta$ is an injective mapping.

Finally, we show that $\theta:\im(\alpha)\longrightarrow \im(\beta)$ admits a completable inverse in
$\MOP(X)$. Let $w\in\dom(\theta^{-1})=\im(\theta)$ and take $a=w\theta^{-1}\in\im(\alpha)$.
Then $w=a\theta=w_a$ and so $w\theta^{-1}=a=w_a\gamma=w\gamma$.
Thus $\gamma$ is a complete extension of $ \theta^{-1}$ in $\MOP(X)$.

Similarly,
by taking $\delta,\xi\in\MOP(X)$ such that $\beta=\delta\alpha\xi$,
we can construct an injective mapping
$\tau:\im(\beta)\longrightarrow\im(\alpha)$
that admits a completable inverse in $\MOP(X)$.

\smallskip

Conversely, let $\theta:\im(\alpha)\longrightarrow\im(\beta)$ be an injective
mapping such that its inverse $\theta^{-1}:\im(\theta)\longrightarrow\im(\alpha)$ is completable in
$\MOP(X)$. Let $\gamma\in\MOP(X)$ be a complete extension of $\theta^{-1}$.
Notice that, by Proposition \ref{rPOP}, we have $\theta^{-1}\in\POP(X)$ and so, by Proposition \ref{iPOP}, it follows that $\theta\in\POP(X)$.

Let $B$ be an ideal of $\beta$ and consider $C=(X\setminus B)\beta$.
For each $c\in C$, choose $z_c\in c\beta^{-1}\cap (X\setminus B)$.
For each $c\in\im(\beta)\setminus C$, choose $z_c\in c\beta^{-1}\cap B$
(notice that $\im(\beta)\setminus C\subseteq B\beta$).
Then, define a mapping $\bar\beta\in\PT(X)$, with $\dom(\bar\beta)=\im(\beta)$,
by $c\bar\beta=z_c$, for all $c\in\im(\beta)$.
Hence, it is a routine matter to show that, if $C\ne\emptyset$ then $\bar\beta$ admits $C$ as an ideal,
and, if $C=\emptyset$ then $\bar\beta$ is order-preserving.
Thus, in both cases, we have $\bar\beta\in\POP(X)$.

Now, let $\lambda=\alpha\theta\bar\beta$. Then, as $\alpha,\theta,\bar\beta\in\POP(X)$,
we have $\lambda\in\POP(X)$. Additionally, $\dom(\lambda)=X$ and so $\lambda\in\MOP(X)$.

Next, take $x\in X$.
Then
$$
x\lambda\beta\gamma=
x\alpha\theta\bar\beta\beta\gamma=
z_{x\alpha\theta}\beta\gamma=x\alpha\theta\gamma=x\alpha\theta\theta^{-1}=x\alpha.
$$
Thus $\alpha=\lambda\beta\gamma$.

Similarly, by considering an injective mapping $\tau:\im(\beta)\longrightarrow \im(\alpha)$
that admits a completable inverse in $\MOP(X)$, we can construct transformations $\delta,\xi\in\MOP(X)$ such that
$\beta=\delta\alpha\xi$. Therefore $\alpha\GRJ\beta$, as required.
\end{proof}

\medskip

Now, we recall the following descriptions of the Green's relations in the semigroups $\PT(X)$ and $\I(X)$ (see \cite{Howie:1995}).
Let $S\in\{\PT(X),\I(X)\}$ and let $\alpha,\beta\in S$. Then, in S, we have:
\begin{description}
\item $\alpha\GRL\beta$ if and only if $\im(\alpha)=\im(\beta)$;
\item $\alpha\GRR\beta$ if and only if $\ker(\alpha)=\ker(\beta)$;
\item $\alpha\GRD\beta$ if and only if $|\im(\alpha)|=|\im(\beta)|$;
\item $\GRJ=\GRD$.
\end{description}

Notice that, if $\alpha,\beta\in\I(X)$ then $\ker(\alpha)=\ker(\beta)$ if and only if $\dom(\alpha)=\dom(\beta)$.

\smallskip

The situation regarding the Green's relations of the partial, and the partial injective, counterparts of $\MO(X)$ and $\MOP(X)$,
such as for the regularity, is much more simple than for these semigroups.

In fact, beginning with the Green's relations $\GRR$ and $\GRL$, as an immediate consequence of the regularity of each of the semigroups, we have:

\begin{proposition}\label{rellr}
Let $S\in\{\PO(X),\POP(X),\POI(X),\POPI(X)\}$.
Let  $\alpha,\beta\in S$. Then, in $S$, we have:
\begin{enumerate}
\item $\alpha\GRL\beta$ if and only if $\im(\alpha)=\im(\beta)$;
\item $\alpha\GRR\beta$ if and only if $\ker(\alpha)=\ker(\beta)$ \\
(i.e. $\alpha\GRR\beta$ if and only if $\dom(\alpha)=\dom(\beta)$, if $S\in\{\POI(X),\POPI(X)\}$);
\item $\alpha\GRH\beta$ if and only if $\im(\alpha)=\im(\beta)$ and $\ker(\alpha)=\ker(\beta)$ \\
(i.e. $\alpha\GRH\beta$ if and only if $\dom(\alpha)=\dom(\beta)$ and $\im(\alpha)=\im(\beta)$, if $S\in\{\POI(X),\POPI(X)\}$).
\end{enumerate}
\end{proposition}

\smallskip

In order to characterize the Green's relation $\GRD$, we first prove the following lemma.

\begin{lemma}\label{cabij}
Let $\alpha,\beta\in\POP(X)$ be such that $\ker(\alpha)=\ker(\beta)$ and let $\theta:\im(\alpha)\longrightarrow \im(\beta)$ be the canonical bijection from $\im(\alpha)$ into $\im(\beta)$.
Then $\theta$ is an orientation-preserving bijection. Moreover, if $\alpha,\beta\in\PO(X)$ then $\theta$ is an order-isomorphism.
\end{lemma}
\begin{proof}
First, observe that $\dom(\alpha)=\dom(\beta)$, since $\ker(\alpha)=\ker(\beta)$.
Let $Y$ be an ideal of $\alpha$ and $Z$ be an ideal of $\beta$. As $Y$ and $Z$ are order ideals of $\dom(\alpha)$, then $Y\subseteq Z$ or $Z\subseteq Y$. Without loss of generality, we suppose that $Y\subseteq Z$. Let $W=(Z\setminus Y)\alpha$.

\smallskip

We begin by showing that $\theta$ is order-preserving on $W$.
Let $a_1,a_2\in W$ be such that $a_1<a_2$.
Then, there exist $x_1,x_2\in Z\setminus Y$ such that $x_1\in a_1\alpha^{-1}$ and $x_2\in a_2\alpha^{-1}$.
As $Z\setminus Y\subseteq \dom(\alpha)\setminus Y$, if $x_2<x_1$ then $a_2=x_2\alpha\le x_1\alpha=a_1$, which is a contradiction.
Hence, $x_1\le x_2$ and, as $Z\setminus Y\subseteq Z$, we have $a_1\theta=x_1\beta\le x_2\beta=a_2\theta$.

\smallskip

Next, we prove that $\theta$ is order-preserving on $\im(\alpha)\setminus W$.
Let $a_1,a_2\in \im(\alpha)\setminus W$ be such that $a_1<a_2$.
Take $x_1,x_2\in \dom(\alpha)$ such that $x_1\in a_1\alpha^{-1}$ and $x_2\in a_2\alpha^{-1}$. Clearly, $x_1,x_2\not\in Z\setminus Y$ and so
$x_1,x_2\in Y\cup (\dom(\alpha)\setminus Z)$.
Suppose that $x_1,x_2\in Y$. If $x_2<x_1$ then $a_2=x_2\alpha\le x_1\alpha=a_1$, which is a contradiction.
Hence, $x_1\le x_2$ and, as $Y\subseteq Z$, we have $a_1\theta=x_1\beta\le x_2\beta=a_2\theta$.
Secondly, consider that $x_1\in Y$ and $x_2\in\dom(\alpha)\setminus Z$.
As $\dom(\alpha)\setminus Z\subseteq\dom(\alpha)\setminus Y$, we have $x_1\in Y$ and $x_2\in\dom(\alpha)\setminus Y$, whence $a_1=x_1\alpha\ge x_2\alpha=a_2$, which is a contradiction. So, this case does not occur.
In third place,  if $x_1\in\dom(\alpha)\setminus Z$ and $x_2\in Y$ then, as $Y\subseteq Z$,
$x_1\in\dom(\beta)\setminus Z$ and $x_2\in Z$ and so $a_1\theta=x_1\beta\le x_2\beta=a_2\theta$.
Finally, suppose that $x_1,x_2\in\dom(\alpha)\setminus Z$. As $\dom(\alpha)\setminus Z\subseteq\dom(\alpha)\setminus Y$,
if $x_2<x_1$ then  $a_2=x_2\alpha\le x_1\alpha=a_1$, which is a contradiction.
Hence, $x_1,x_2\in\dom(\beta)\setminus Z$ and $x_1\le x_2$ , from which follows that $a_1\theta=x_1\beta\le x_2\beta=a_2\theta$.

\smallskip

Now, let $a_1\in W$ and $a_2\in \im(\alpha)\setminus W$. We aim to show that $a_1\le a_2$ and $a_1\theta\ge a_2\theta$.
Take $x_1\in Z\setminus Y$ and $x_2\in Y\cup (\dom(\alpha)\setminus Z)$ such that $x_1\in a_1\alpha^{-1}$ and $x_2\in a_2\alpha^{-1}$.
First, suppose that $x_2\in Y$. As $x_1\in Z\setminus Y\subseteq \dom(\alpha)\setminus Y$, we have $x_2<x_1$ and
$a_2=x_2\alpha\ge x_1\alpha=a_1$.
On the other hand, $x_1\in Z\setminus Y\subseteq Z$, $x_2\in Y\subseteq Z$ and $x_2< x_1$ imply $a_2\theta=x_2\beta\le x_1\beta=a_1\theta$.
Secondly, suppose that $x_2\in \dom(\alpha)\setminus Z$.
Then $x_1\in Z\setminus Y\subseteq Z$ and $x_2\in \dom(\beta)\setminus Z$, whence $x_1<x_2$ and $a_1\theta=x_1\beta\ge x_2\beta=a_2\theta$.
Furthermore, $x_1\in Z\setminus Y\subseteq \dom(\alpha)\setminus Y$, $x_2\in \dom(\alpha)\setminus Z\subseteq\dom(\alpha)\setminus Y$ and
$x_1<x_2$ imply $a_1=x_1\alpha\le x_2\alpha=a_2$.

\smallskip

Thus, if $W\ne\emptyset$, we proved that $W$ is an ideal of $\theta$. On the other hand, if $W=\emptyset$ then we proved that $\theta$ is order-preserving (on $\im(\alpha)$). In both cases,  we have $\theta\in\POPI(X)$. Moreover, if $\alpha,\beta\in\PO(X)$ then $W=\emptyset$ and so
$\theta\in\POI(X)$, as required.
\end{proof}

\begin{proposition}\label{reld}
Let $S\in\{\PO(X),\POI(X)\}$ [respectively, $S\in\{\POP(X),\POPI(X)\}$].
Let  $\alpha,\beta\in S$. Then $\alpha\GRD\beta$ in $S$ if and only if there exists an order-isomorphism
[respectively, an orientation-preserving bijection] $\theta:\im(\alpha)\longrightarrow \im(\beta)$.
\end{proposition}
\begin{proof}
Suppose that $\alpha\GRD\beta$. Then, there exists
$\gamma\in S$ such that $\alpha\GRR\gamma$ and
$\gamma\GRL\beta$. By Proposition \ref{rellr}, we have that
$\ker(\alpha)=\ker(\gamma)$ and $\im(\gamma)=\im(\beta)$.
Then, by Lemma \ref{cabij}, the canonical bijection $\theta:\im(\alpha)\longrightarrow \im(\gamma)=\im(\beta)$ from $\im(\alpha)$ into $\im(\gamma)$
is an order-isomorphism [respectively, an orientation-preserving bijection].

Conversely, suppose that there exists an order-isomorphism
[respectively, an orientation-preserving bijection] $\theta:\im(\alpha)\longrightarrow \im(\beta)$.
Let $\gamma=\alpha\theta$. Then $\gamma\in S$.
Moreover, $\im(\gamma)=(\im(\alpha))\theta=\im(\beta)$.
On the other hand, $\dom(\gamma)=\dom(\theta)\alpha^{-1}=\im(\alpha)\alpha^{-1}=\dom(\alpha)$ and,
since $\theta$ is a bijection, we have
$
x\alpha=y\alpha\,\Leftrightarrow\,(x\alpha)\theta=(y\alpha)\theta\,\Leftrightarrow\,x\gamma=y\gamma,
$
for all $x,y\in\dom(\alpha)=\dom(\gamma)$.
Then $\ker(\alpha)=\ker(\gamma)$.
Hence, by Proposition \ref{rellr}, we have
$\gamma\GRL\beta$ and $\alpha\GRR\gamma$
and so $\alpha\GRD\beta$, as required.
\end{proof}

Finally, for $\GRJ$, we have:

\begin{proposition}\label{relj}
Let $S\in\{\PO(X),\POI(X)\}$ [respectively, $S\in\{\POP(X),\POPI(X)\}$].
Let  $\alpha,\beta\in S$. Then $\alpha\GRJ\beta$ in $S$ if and only if there exist order-preserving
[respectively, orientation-preserving] injections $\theta:\im(\alpha)\longrightarrow \im(\beta)$ and
$\tau:\im(\beta)\longrightarrow \im(\alpha)$.
\end{proposition}
\begin{proof}
First, suppose that $\alpha\GRJ\beta$.
Let $\lambda,\gamma\in S$ be such that $\alpha=\lambda\beta\gamma$.

For each $a\in\im(\alpha)=(\im(\lambda\beta))\gamma$,
choose $w_a\in\im(\lambda\beta)\cap a\gamma^{-1}$.
Then, define a mapping $\theta:\im(\alpha)\longrightarrow X$ by $a\theta=w_a$,
for all $a\in\im(\alpha)$. Notice that $\im(\theta)\subseteq\im(\beta)$.

Next, we prove that $\theta$ is injective.
Let $a,b\in\im(\alpha)$ be such that $a\theta=b\theta$.
Then $w_a=w_b$ and, as $w_a\in a\gamma^{-1}$ and $w_b\in b\gamma^{-1}$, we obtain
$a=w_a\gamma=w_b\gamma=b$. Hence, $\theta$ is an injective mapping.

Let us consider the inverse mapping $\theta^{-1}:\im(\theta)\longrightarrow\im(\alpha)$ of $\theta$.
Let $w\in\dom(\theta^{-1})=\im(\theta)$ and take $a=w\theta^{-1}\in\im(\alpha)$.
Then $w=a\theta=w_a$ and so $w\theta^{-1}=a=w_a\gamma=w\gamma$.
Therefore, $\theta^{-1}$ is a restriction of $\gamma$ and so $\theta^{-1}$ is an order-preserving transformation
[respectively, orientation-preserving transformation, by Proposition \ref{rPOP}].
Thus, as $\theta:\im(\alpha)\longrightarrow\im(\theta)$ is the inverse of $\theta^{-1}$, by Proposition \ref{iPOP}, $\theta$ also is an order-preserving
[respectively, orientation-preserving] transformation.

Similarly,
by taking $\delta,\xi\in\MOP(X)$ such that $\beta=\delta\alpha\xi$,
we can construct an order-preserving
[respectively, orientation-preserving] injection $\tau:\im(\beta)\longrightarrow \im(\alpha)$.

\smallskip

Conversely, let $\theta:\im(\alpha)\longrightarrow\im(\beta)$ be an injective
order-preserving [respectively, orientation-preserving] transformation.
Then, by Proposition \ref{iPOP}, $\theta^{-1}:\im(\theta)\longrightarrow\im(\alpha)$ also is an order-preserving
[respectively, orientation-preserving] transformation.

Let $B=\dom(\beta)$ [respectively, $B$ be an ideal of $\beta$] and consider $C=(\dom(\beta)\setminus B)\beta$.
For each $c\in C$, choose $z_c\in c\beta^{-1}\cap (\dom(\beta)\setminus B)$.
For each $c\in\im(\beta)\setminus C$, choose $z_c\in c\beta^{-1}\cap B$
(notice that $\im(\beta)\setminus C\subseteq B\beta$).
Then, define a mapping $\bar\beta\in\PT(X)$, with $\dom(\bar\beta)=\im(\beta)$,
by $c\bar\beta=z_c$, for all $c\in\im(\beta)$.
Hence, it is a routine matter to show that $\bar\beta$ is order-preserving
[respectively, if $C\ne\emptyset$ then $\bar\beta$ admits $C$ as an ideal,
and, if $C=\emptyset$ then $\bar\beta$ is order-preserving.
Thus, in both cases, we have that $\bar\beta$ is orientation-preserving].

Now, let $\lambda=\alpha\theta\bar\beta$. Then, as $\alpha,\theta,\bar\beta\in S$,
we have $\lambda\in S$.  Take $x\in \dom(\alpha)$.
Then
$$
x\lambda\beta\theta^{-1}=
x\alpha\theta\bar\beta\beta\theta^{-1}=
z_{x\alpha\theta}\beta\theta^{-1}=x\alpha\theta\theta^{-1}=x\alpha.
$$
Thus $\alpha=\lambda\beta\theta^{-1}$. Notice that $\theta^{-1}\in S$.

Similarly, by considering an injective order-preserving [respectively, orientation-preserving] transformation $\tau:\im(\beta)\longrightarrow \im(\alpha)$, we can construct transformations $\delta,\xi\in S$ such that
$\beta=\delta\alpha\xi$. Therefore $\alpha\GRJ\beta$, as required.
\end{proof}

\medskip

We finish this paper by showing that, such as for $\MO(X)$ (see \cite{Fernandes&al:2014}),
for any
$$
S\in\{\MOP(X), \PO(X), \POP(X), \POI(X), \POPI(X)\},
$$
we may have $\GRD\subsetneq\GRJ$ in $S$.

\smallskip

In the following lemma and examples, we consider the set of real numbers $\R$ equipped with the usual order.

\begin{lemma}\label{exes}
Let $a,b,c,d\in\R$ be such that $a<b$ and $c<d$.
Then, there exists no orientation-preserving bijection from the interval $I=]a,b[$ into the interval $J=[c,d]$.
\end{lemma}
\begin{proof}
By contradiction, suppose there exists an orientation-preserving bijection $\theta:I\longrightarrow J$.
Let $Y$ be an ideal of $\theta$. If $Y=I$ then $\theta$ would be an order-preserving bijection from $I$ into $J$, which is not possible
(for instance, because $J$ has a minimum and $I$ does not).
Hence, $Y$ is a proper order ideal of $I$ and $I\setminus Y$ is a non-empty order filter of $I$.
Since $I$ has no minimum or maximum, then $Y$ has no minimum and $I\setminus Y$ has no maximum.
Hence, as $\theta$ is order-preserving both in $Y$ and in $I\setminus Y$,
$Y\theta$ also has no minimum and $(I\setminus Y)\theta$ also has no maximum.
On the other hand, since each element of $(I\setminus Y)\theta$ is a lower bound of $Y\theta$ and
each element of $Y\theta$ is an upper bound of $(I\setminus Y)\theta$,
we have that $(I\setminus Y)\theta$ has a supremum $c'\in I$, $Y\theta$ has an infimum $d'\in I$ and $c'\le d'$.
Since $(I\setminus Y)\theta$ has no maximum, $c'\not\in (I\setminus Y)\theta$ and so,
as $J=I\theta=(I\setminus Y)\theta\cup Y\theta$, we have $c'\in Y\theta$, whence $d'\le c'$.
Therefore, $d'=c'$ and so $d'$ is the minimum of $Y\theta$, which is a contradiction.
Thus, there exists no orientation-preserving bijection $\theta:I\longrightarrow J$, as required.
\end{proof}

\smallskip

\begin{example}\em
Let  $\alpha,\beta\in\T(\R)$ be defined by $x\alpha=\arctan(x)$, for $x\in\R$, and
$$
x\beta=\left\{
\begin{array}{cl}
-1 & \mbox{if $x<-1$}\\
x & \mbox{if $-1\le x\le1$}\\
1& \mbox{if $x>1$.}
\end{array}
\right.
$$
Then $\alpha,\beta\in\MO(\R)$ and so, in particular, $\alpha,\beta\in\MOP(\R)$.
In \cite{Fernandes&al:2014} the authors have showed that  $(\alpha,\beta)\not\in\GRD$ and $(\alpha,\beta)\in\GRJ$ in $\MO(\R)$.
Hence, as $\MO(\R)$ is a subsemigroup of $\MOP(\R)$, we may immediately deduce that also $(\alpha,\beta)\in\GRJ$ in $\MOP(\R)$.
On the other hand, $\im(\alpha)=\left]-\pi/2,\pi/2\right[$ and $\im(\beta)=\left[-1,1\right]$ and so, by the above lemma,
there exists no orientation-preserving bijection $\theta:\im(\alpha)\longrightarrow\im(\beta)$. Therefore,
in view of Proposition \ref{rPOP}, there exists no (bi)completable in $\MOP(\R)$ bijection $\theta:\im(\alpha)\longrightarrow\im(\beta)$.
Thus, $(\alpha,\beta)\not\in\GRD$ in $\MOP(\R)$ and so we have $\GRD\subsetneq\GRJ$ in $\MOP(\R)$.
\end{example}

\smallskip

Now, observe that $\POI(X)$ is a subsemigroup of $S$, for any $S\in\{\PO(X),\POP(X),\POPI(X)\}$.
Furthermore, $S$ is a subsemigroup of $\POP(X)$, for any $S\in\{\PO(X),\POI(X),\POPI(X)\}$.

\smallskip

\begin{example}\em
Let $a,b,c,d\in\R$ be such that $a<b$ and $c<d$. Let $\alpha$ and $\beta$ be the partial identities on the intervals $]a,b[$ and $[c,d]$, respectively.  Then $\alpha,\beta\in\POI(\R)$ and so $\alpha,\beta\in S$, for any $S\in\{\PO(\R),\POP(\R),\POPI(\R)\}$.
Moreover, $\im(\alpha)=]a,b[$ and $\im(\beta)=[c,d]$.

By Lemma \ref{exes} there exists no orientation-preserving bijection from $\im(\alpha)$ into $\im(\beta)$ and so, by Proposition \ref{reld},
$(\alpha,\beta)\not\in\GRD$ in $\POP(\R)$.
Therefore, also $(\alpha,\beta)\not\in\GRD$ in $S$,  for any $S\in\{\PO(\R),\POI(\R),\POPI(\R)\}$.

On the other hand, it is clear that there exist order-preserving injections $\theta:]a,b[\longrightarrow [c,d]$ and
$\tau:[c,d]\longrightarrow ]a,b[$
(for instance, $x\theta=\frac{1}{3(b-a)}((d-c)x+2bc-2ad+bd-ac)$, for $x\in ]a,b[$, and
$x\tau=\frac{1}{3(d-c)}((b-a)x+2ad-2bc+bd-ac)$, for $x\in [c,d]$) and so,
by Proposition  \ref{relj}, $(\alpha,\beta)\in\GRJ$ in $\POI(\R)$.
Therefore, also $(\alpha,\beta)\in\GRJ$ in $S$,  for any $S\in\{\PO(\R),\POP(\R),\POPI(\R)\}$.
Thus $\GRD\subsetneq\GRJ$ in $S$, for any $S\in\{\PO(\R),\POP(\R),\POI(X),\POPI(\R)\}$.
\end{example}