# Semigroups of transformations whose characters belong to a given semigroup

Mosarof Sarkar Department of Mathematics, Central University of South Bihar,
Gaya–824236 (Bihar), India<EMAIL_ADDRESS>and  Shubh N. Singh
Department of Mathematics, Central University of South Bihar, Gaya–824236
(Bihar), India<EMAIL_ADDRESS>
###### Abstract.

Let $X$ be a nonempty set and $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ a
partition of $X$. Denote by $T(X)$ the full transformation semigroup on $X$,
and $T(X,\mathcal{P})$ the subsemigroup of $T(X)$ consisting of all
transformations that preserve $\mathcal{P}$. For every subsemigroup
$\mathbb{S}(I)$ of $T(I)$, let $T_{\mathbb{S}(I)}(X,\mathcal{P})$ be the
semigroup of all transformations $f\in T(X,\mathcal{P})$ such that
$\chi^{(f)}\in\mathbb{S}(I)$, where $\chi^{(f)}\in T(I)$ defined by
$i\chi^{(f)}=j$ whenever $X_{i}f\subseteq X_{j}$. We describe regular and
idempotent elements in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, and determine when
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a regular semigroup [inverse semigroup].
With the assumption that $\mathbb{S}(I)$ contains the identity, we
characterize Green’s relations on $T_{\mathbb{S}(I)}(X,\mathcal{P})$, describe
unit-regular elements in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, and determine
when $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a unit-regular semigroup. We apply
these general results to obtain more concrete results for $T(X,\mathcal{P})$.

###### Key words and phrases:

Transformation semigroups, Regular semigroups, Unit-regular semigroups,
Idempotents, Inverse semigroups, Green’s relations

###### 2010 Mathematics Subject Classification:

20M20, 20M17, 20M18.

##  1\. Introduction

Throughout this paper, let $X$ be a nonempty set, let
$\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a partition of $X$, and let $E$ be
the equivalence relation on $X$ induced by $\mathcal{P}$. Denote by $T(X)$ the
full transformation semigroup on $X$. The semigroup $T(X)$ and its
subsemigroups have a crucial place in the theory of semigroups, since every
semigroup is isomorphic to a subsemigroup of $T(Z)$ for some nonempty set $Z$
[12, Theorem 1.1.2]. In 1994, Pei [14] introduced the following subsemigroup
of $T(X)$.

$\displaystyle T(X,\mathcal{P})$ $\displaystyle=\\{f\in T(X)\colon\forall
X_{i}\in\mathcal{P}~{}\exists X_{j}\in\mathcal{P},~{}X_{i}f\subseteq X_{j}\\}$
$\displaystyle=\left\\{f\in T(X)\colon\forall x,y\in X,~{}(x,y)\in
E\Longrightarrow(xf,yf)\in E\right\\}.$

In that paper, Pei [14, Theorem 2.8] proved that $T(X,\mathcal{P})$ is the
semigroup of all continuous selfmaps on $X$ endowed with the topology having
$\mathcal{P}$ as a basis. Since then the semigroup $T(X,\mathcal{P})$ has been
extensively studied in the literature; see, for example, [1, 2, 15, 16, 17,
22, 25, 26, 27].

Let $f\in T(X,\mathcal{P})$. The _character_ of $f$ is the transformation
$\chi^{(f)}\colon I\to I$ defined by $i\chi^{(f)}=j$ whenever $X_{i}f\subseteq
X_{j}$. If $X$ is finite, the character of $f\in T(X,\mathcal{P})$ has been
studied using the symbol $\bar{f}$ by Araújo et al. [1], Dolinka and East [8],
and Dolinka et al. [9]. The character of $f\in T(X,\mathcal{P})$ for an
arbitrary set $X$ was first studied by Purisang and Rakbud [19]. Let
$\mathbb{S}(I)$ be a subsemigroup of $T(I)$. By using the notion of character,
Rakbud [20] introduced the following subsemigroup of $T(X,\mathcal{P})$:

$T_{\mathbb{S}(I)}(X,\mathcal{P})=\left\\{f\in
T(X,\mathcal{P})\colon\chi^{(f)}\in\mathbb{S}(I)\right\\}.$

In [21], Rakbud and Chaiya studied the semigroup
$T_{\mathbb{S}(I)}(X,\mathcal{P})$. Semigroup
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ generalizes the semigroup
$T(X,\mathcal{P})$, since $T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$. The
semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$ also generalizes both $T(X)$ and
$\mathbb{S}(I)$ in the sense that $T_{\mathbb{S}(I)}(X,\mathcal{P})=T(X)$ if
$|\mathcal{P}|=1$, and $T_{\mathbb{S}(I)}(X,\mathcal{P})\simeq\mathbb{S}(I)$
if $\mathcal{P}$ consists of singleton sets. (Note that if $\mathcal{P}$
consists of singleton sets, then $|I|=|X|$ and
$\mathbb{S}(I)\simeq\mathbb{S}(X)$.) For various proper subsemigroups
$\mathbb{S}(I)$ of $T(I)$, the semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$
has been investigated; see [5, 6, 7, 19, 23, 24, 25, 26, 27, 28].

The purpose of this paper is to study the semigroup
$T_{\mathbb{S}(I)}(X,\mathcal{P})$, where $\mathbb{S}(I)$ is a general
subsemigroup of $T(I)$. The paper is organized as follows. In the next
section, we present the definitions and notation used throughout this paper.
In Section 3, we describe regular and idempotent elements in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$. We next determine when
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a regular semigroup, and when
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse semigroup. Moreover, we prove
that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a regular semigroup if
$\mathbb{S}(I)$ is a subgroup of the symmetric group on $I$, and thus
generalize the result given by Purisang and Rakbud in [19, Theorem 3.5(1)].
With the assumption that $\mathbb{S}(I)$ contains the identity, we describe
unit-regular elements in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, and determine
when $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a unit-regular semigroup in Section
4. In Section 5, we characterize Green’s relations on
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ when $\mathbb{S}(I)$ contains the identity.
Using these characterizations, we also obtain concrete and noticeably
different descriptions of Green’s relations on $T(X,\mathcal{P})$, whose
Green’s relations are known; see [17, 18].

##  2\. Preliminaries and Notation

Let $X$ be a nonempty set. The cardinality of $X$ is denoted by $|X|$. For any
nonempty set $Y$, we write $X\setminus Y=\\{x\in X\colon x\notin Y\\}$. A
_partition_ $\mathcal{P}$ of $X$ is a collection of nonempty disjoint subsets
of $X$, called _blocks_ of $\mathcal{P}$, whose union is $X$. A _trivial
partition_ is a partition that contains only singleton blocks or a single
block. A _transversal_ of an equivalence relation $\rho$ on $X$ is a subset of
$X$ that contains exactly one element from every $\rho$-class. We denote the
identity mapping on $X$ by $\operatorname{id}_{X}$, and the symmetric group on
$X$ by $\operatorname{Sym}(X)$.

Denote the composition of mappings by juxtaposition, and compose mappings from
left to right. Let $f\colon X\to Y$ be a mapping. We write $xf$ for the image
of $x$ under $f$. For a nonempty $A\subseteq X$ (resp. $B\subseteq Y$), we put
$Af=\\{af\colon a\in A\\}$ (resp. $Bf^{-1}=\\{x\in X\colon xf\in B\\}$).
Furthermore, if $B=\\{b\\}$, then we write $bf^{-1}$ instead of
$\\{b\\}f^{-1}$. We let $D(f)=Y\setminus Xf$ and $\textnormal{d}(f)=|D(f)|$.
The _kernel_ of $f$, denoted by $\ker(f)$, is an equivalence relation on $X$
defined by $\ker(f)=\\{(a,b)\in X\times X\colon af=bf\\}$. The symbol $\pi(f)$
denotes the partition of $X$ induced by $\ker(f)$, and $T_{f}$ denotes a
transversal of $\ker(f)$. Note that $|X\setminus T_{f}|$ is independent of the
choice of transversal of $\ker(f)$ (cf. [11, p. 1356]). We let
$\textnormal{c}(f)=|X\setminus T_{f}|$.

Let $f\in T(X)$. For a nonempty subset $A$ of the domain of $f$, the
_restriction_ of $f$ to $A$ is the mapping $f_{\upharpoonright A}\colon A\to
X$ defined by $x(f_{\upharpoonright A})=xf$ for all $x\in A$. Moreover if $B$
is a subset of the codomain of $f$ such that $Af\subseteq B$, then we use the
same notation $f_{\upharpoonright A}$ for the mapping from $A$ to $B$ that
assigns $xf$ to each $x\in A$. We say that $f$ _preserves_ or _stabilizes_ the
partition $\mathcal{P}$ if for each $X_{j}\in\mathcal{P}$, there exists
$X_{k}\in\mathcal{P}$ such that $X_{j}f\subseteq X_{k}$. Thus
$T(X,\mathcal{P})$ is a subsemigroup of $T(X)$ consisting of all
transformations that preserve the partition $\mathcal{P}$.

Let $S$ be a semigroup. An element $a\in S$ is an idempotent if $a^{2}=a$. The
set of all idempotents of $S$ is denoted by $E(S)$. It is well-known that
$f\in T(X)$ is an idempotent if and only if $f$ acts as the identity map on
its image set (cf. [3, p. 6]). An element $a\in S$ is regular in $S$ if there
is $b\in S$ such that $aba=a$. Such element $b$, if exists, is called an
_inner inverse_ of $a$. The set of all inner inverses of $a$ is denoted by
$\operatorname{Pre}(a)$, and the set of all regular elements in $S$ is denoted
by $\operatorname{reg}(S)$. If $\operatorname{reg}(S)=S$, then we say that $S$
is _regular_. The semigroup $S$ is an inverse semigroup if for every $x\in S$,
there is a unique $y\in S$ such that $xyx=x$ and $yxy=y$. Equivalently, the
semigroup $S$ is an inverse semigroup if and only if $S$ is regular and its
idempotents commute (cf. [12, Theorem 5.1.1]).

Let $S$ be a semigroup with identity. Denote by $U(S)$ the set of all units in
$S$. Note that $U(T(X))=\operatorname{Sym}(X)$, and write
$U\big{(}T(X,\mathcal{P})\big{)}=S(X,\mathcal{P})$. Clearly
$S(X,\mathcal{P})=T(X,\mathcal{P})\cap\operatorname{Sym}(X)$. An element $a\in
S$ is unit-regular in $S$ if there is $u\in U(S)$ such that $aua=a$. The set
of all unit-regular elements in $S$ is denoted by $\operatorname{ureg}(S)$. If
$\operatorname{ureg}(S)=S$, then we say that $S$ is _unit-regular_.

In what follows, the letter $I$ denotes the index set of the partition
$\mathcal{P}$, and the symbol $\mathbb{S}(I)$ denotes a subsemigroup of
$T(I)$. We refer the reader to [12] for any undefined notation and terminology
of semigroup theory.

##  3\. Regular and Inverse semigroups $T_{\mathbb{S}(I)}(X,\mathcal{P})$

In this section, we give a characterization of regular elements in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ and determine when
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a regular semigroup. Using these, we
obtain known descriptions of regular elements in $T(X,\mathcal{P})$ and
regularity of $T(X,\mathcal{P})$. If $\mathbb{S}(I)$ is a subgroup of
$\operatorname{Sym}(I)$, we prove that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is
regular, and thus generalize Theorem 3.5(1) in [19]. Next, we give a
characterization of idempotents in $T_{\mathbb{S}(I)}(X,\mathcal{P})$ and then
determine when $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse semigroup.

We begin by recalling a specific case of Lemma 2.3 in [19].

######  Lemma 3.1.

[19, Lemma 2.3] We have $\chi^{(fg)}=\chi^{(f)}\chi^{(g)}$ for all $f,g\in
T(X,\mathcal{P})$.

In the next result, we observe that $\mathbb{S}(I)$ is a homomorphic image of
$T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Proposition 3.2.

The mapping $\varphi\colon T_{\mathbb{S}(I)}(X,\mathcal{P})\to\mathbb{S}(I)$
defined by $f\varphi=\chi^{(f)}$ is an epimorphism. Moreover, if
$\operatorname{id}_{X}\in T_{\mathbb{S}(I)}(X,\mathcal{P})$, then
$\operatorname{id}_{X}\varphi=\operatorname{id}_{I}$.

###### Proof.

Note from Lemma 3.1 that $\chi^{(fg)}=\chi^{(f)}\chi^{(g)}$ for all $f,g\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then we see that $\varphi\colon
T_{\mathbb{S}(I)}(X,\mathcal{P})\to\mathbb{S}(I)$ is a homomorphism, since

$(fg)\varphi=\chi^{(fg)}=\chi^{(f)}\chi^{(g)}=(f\varphi)(g\varphi)$

for all $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. It remains to prove that
$\varphi$ is surjective. For this, let $\alpha\in\mathbb{S}(I)$. Fix $x_{i}\in
X_{i}$ for each $i\in I$, and define $h\in T(X)$ by $xh=x_{i\alpha}$ whenever
$x\in X_{i}$ for some $i\in I$. Clearly $X_{i}h\subseteq X_{i\alpha}$, and so
$\chi^{(h)}=\alpha$. This gives $h\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. We
also obtain $h\varphi=\chi^{(h)}=\alpha$ as required. If
$\operatorname{id}_{X}\in T_{\mathbb{S}(I)}(X,\mathcal{P})$, then we get
$\operatorname{id}_{X}\varphi=\chi^{(\operatorname{id}_{X})}=\operatorname{id}_{I}$.
∎

The following result gives a characterization of regular elements in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Theorem 3.3.

Let $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then
$f\in\operatorname{reg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ if and
only if there exists $\alpha\in\mathbb{S}(I)$ such that:

  1. (i)

$\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$;

  2. (ii)

$X_{i}\cap Xf\subseteq X_{i\alpha}f$ for all $i\in I\chi^{(f)}$.

###### Proof.

Assume that
$f\in\operatorname{reg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. Then
there exists $g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $fgf=f$. This
yields $\chi^{(f)}\chi^{(g)}\chi^{(f)}=\chi^{(f)}$ by Lemma 3.1. Note that
$\chi^{(g)}\in\mathbb{S}(I)$, and write $\chi^{(g)}=\alpha$. Then
$\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$, and so (i) holds.

To prove (ii), let $i\in I\chi^{(f)}$ and $y\in X_{i}\cap Xf$. Then, since
$y\in Xf$, there exists $x\in X$ such that $xf=y$. Also, since $y\in X_{i}$
and $\chi^{(g)}=\alpha$, we have $yg\in X_{i\alpha}$. Therefore, since
$f=fgf$, we obtain $y=xf=(xf)gf=(yg)f\in X_{i\alpha}f$. Thus $X_{i}\cap
Xf\subseteq X_{j}f$, and so (ii) holds.

Conversely, assume that the given conditions hold. To prove
$f\in\operatorname{reg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$, we
will construct $g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $fgf=f$. Let
$i\in I\chi^{(f)}$. For each $x\in X_{i}\cap Xf$, by (ii) there exists
$x^{\prime}\in X_{i\alpha}$ such that $x=x^{\prime}f$. For each $j\in I$, we
fix $x_{j}\in X_{j}$. Define $g\in T(X)$ by:

$\displaystyle xg=\begin{cases}x^{\prime}&\text{if $x\in X_{i}\cap Xf$, where
$i\in I\chi^{(f)}$};\\\ x_{i\alpha}&\text{if $x\in X_{i}\setminus Xf$, where
$i\in I\chi^{(f)}$};\\\ x_{j\alpha}&\text{if $x\in X_{j}$, where $j\in
I\setminus I\chi^{(f)}$}.\end{cases}$

Clearly $\chi^{(g)}=\alpha$. Since $\alpha\in\mathbb{S}(I)$, it follows that
$g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. It is routine to verify that $fgf=f$,
and hence
$f\in\operatorname{reg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. ∎

Using [13, Theorem 3.1], Pei [17, Corollary 2.3] obtained the following result
that characterizes regular elements in $T(X,\mathcal{P})$. Using Theorem 3.3,
we give an alternative proof of the same result.

######  Proposition 3.4.

Let $f\in T(X,\mathcal{P})$. Then
$f\in\operatorname{reg}\big{(}T(X,\mathcal{P})\big{)}$ if and only if for
every $i\in I$, there exists $j\in I$ such that $X_{i}\cap Xf\subseteq
X_{j}f$.

###### Proof.

Assume that $f\in\operatorname{reg}\big{(}T(X,\mathcal{P})\big{)}$, and let
$i\in I$. If $i\in I\setminus I\chi^{(f)}$, then we get $X_{i}\cap
Xf=\varnothing$, and so $X_{i}\cap Xf\subseteq X_{j}f$ for all $j\in I$.
Assume that $i\in I\chi^{(f)}$. By Theorem 3.3, since
$T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, there exists $\alpha\in T(I)$ such
that $X_{i}\cap Xf\subseteq X_{i\alpha}f$. Thus we conclude that $X_{i}\cap
Xf\subseteq X_{j}f$ for some $j\in I$.

Conversely, assume that the given condition holds. To prove
$f\in\operatorname{reg}\big{(}T(X,\mathcal{P})\big{)}$, since
$T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, we will construct $\alpha\in T(I)$
such that each condition of Theorem 3.3 holds. For each $i\in I$, by
hypothesis we fix $j_{i}\in I$ such that $X_{i}\cap Xf\subseteq X_{j_{i}}f$.
Consider $\alpha\in T(I)$ defined by $i\alpha=j_{i}$. We now show that
$\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$. Let $k\in I$ and write
$k\chi^{(f)}=\ell$. Then

$k\left(\chi^{(f)}\alpha\chi^{(f)}\right)=\left(k\chi^{(f)}\right)\alpha\chi^{(f)}=(\ell\alpha)\chi^{(f)}=j_{\ell}\chi^{(f)}=\ell=k\chi^{(f)},$

and so $\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$. Thus the condition (i) of
Theorem 3.3 holds.

Next, let $i\in I\chi^{(f)}$. Then $i\alpha=j_{i}$ by the definition of
$\alpha$. Therefore, by hypothesis and the choice of $j_{i}$, we get
$X_{i}\cap Xf\subseteq X_{i\alpha}f$. Therefore the condition (ii) of Theorem
3.3 holds.

Thus, since $T_{T(I)}(X,\mathcal{P})=T(X,\mathcal{P})$, we conclude from
Theorem 3.3 that $f\in\operatorname{reg}\big{(}T(X,\mathcal{P})\big{)}$. ∎

Note that the regularity of $\mathbb{S}(I)$ may not imply the regularity of
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ (cf. [20, p. 631]). In the next result, we
determine when $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a regular semigroup.

######  Theorem 3.5.

The semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular if and only if:

  1. (i)

$\mathbb{S}(I)$ is regular;

  2. (ii)

for each $\alpha\in\mathbb{S}(I)$ if there exists $i\in I$ such that
$|i\alpha^{-1}|\geq 2$, then $|X_{i}|=1$.

###### Proof.

Assume that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular. Then (i) is true by
Proposition 3.2 and the fact that any homomorphic image of a regular semigroup
is regular (cf. [12, Lemma 2.4.4]).

To prove (ii), let $\alpha\in\mathbb{S}(I)$. If $\mathcal{P}$ is trivial, then
there is nothing to prove. Assume that $\mathcal{P}$ is nontrivial. Suppose to
the contrary that $|X_{j}|\geq 2$ for some $j\in I$ with $|j\alpha^{-1}|\geq
2$. Choose distinct $y_{1},y_{2}\in X_{j}$. Fix $x_{t}\in X_{t}$ for each
$t\in I\alpha\setminus\\{j\\}$ if $I\alpha\setminus\\{j\\}\neq\varnothing$. We
also fix $\ell\in j\alpha^{-1}$. Define $f\in T(X)$ by:

$\displaystyle xf=\begin{cases}y_{1}&\text{if $x\in X_{\ell}$};\\\
y_{2}&\text{if $x\in X_{k}$, where $k\in j\alpha^{-1}\setminus\\{\ell\\}$};\\\
x_{i\alpha}&\text{if $x\in X_{i}$, where $i\in I\setminus
j\alpha^{-1}$}.\end{cases}$

Clearly $\chi^{(f)}=\alpha$. Since $\alpha\in\mathbb{S}(I)$, it follows that
$f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Therefore
$f\in\operatorname{reg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ by
hypothesis. By (i), let $\beta\in\operatorname{Pre}(\alpha)$. Note that
$\alpha\beta\alpha=\alpha$ and $j\in I\alpha$, which gives $j\beta\in
j\alpha^{-1}$. Therefore either $X_{j\beta}f=\\{y_{1}\\}$ or
$X_{j\beta}f=\\{y_{2}\\}$ by the definition of $f$. However, we have
$X_{j}\cap Xf=\\{y_{1},y_{2}\\}$, and so $X_{j}\cap Xf\not\subseteq
X_{j\beta}f$. This is a contradiction by Theorem 3.3, since
$f\in\operatorname{reg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. Thus
(ii) holds.

Conversely, assume that the given conditions hold and let $f\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$. To prove the desired result, we will show
by Theorem 3.3 that
$f\in\operatorname{reg}(T_{\mathbb{S}(I)}(X,\mathcal{P}))$. Since $f\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$, we have $\chi^{(f)}\in\mathbb{S}(I)$. Then
by (i), there exists $\beta\in\mathbb{S}(I)$ such that
$\chi^{(f)}\beta\chi^{(f)}=\chi^{(f)}$.

Next, let $i\in I\chi^{(f)}$. Then there are two cases:

Case 1: Suppose $|i\left(\chi^{(f)}\right)^{-1}|=1$. Write
$i\left(\chi^{(f)}\right)^{-1}=\\{j\\}$. Clearly $X_{i}\cap Xf=X_{j}f$. Since
$\chi^{(f)}\beta\chi^{(f)}=\chi^{(f)}$ and $j\chi^{(f)}=i$, we see that
$i\beta=j$. Therefore $X_{i}\cap Xf\subseteq X_{i\beta}f$.

Case 2: Suppose $|i\left(\chi^{(f)}\right)^{-1}|\geq 2$. Then by $\rm(ii)$, we
get $|X_{i}|=1$. Write $i\beta=k$. Then, since
$\chi^{(f)}\beta\chi^{(f)}=\chi^{(f)}$, we see that $k\chi^{(f)}=i$. It
follows that $X_{k}f=X_{i}$ and $X_{i}\cap Xf=X_{i}$. Therefore $X_{i}\cap
Xf\subseteq X_{i\beta}f$.

Thus, in either case, we have $X_{i}\cap Xf\subseteq X_{i\beta}f$. Hence we
conclude from Theorem 3.3 that
$f\in\operatorname{reg}(T_{\mathbb{S}(I)}(X,\mathcal{P}))$. ∎

If $\mathbb{S}(I)=\operatorname{Sym}(I)$, Purisang and Rakbud [19, Theorem
3.5(1)] proved that the semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is
regular. Using Theorem 3.5, we prove in the following result that
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular for every subgroup
$\mathbb{S}(I)$ of $\operatorname{Sym}(I)$, and thus generalizes Theorem
3.5(1) in [19].

######  Corollary 3.6.

If $\mathbb{S}(I)$ is a subgroup of $\operatorname{Sym}(I)$, then
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular.

###### Proof.

If $\mathbb{S}(I)$ is a subgroup of $\operatorname{Sym}(I)$, then it is
obvious that $\mathbb{S}(I)$ is regular. Also for each
$\alpha\in\mathbb{S}(I)$, we get $|i\alpha^{-1}|=1$ for all $i\in I$.
Therefore the semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular by
Theorem 3.5. ∎

Using Theorem 3.5, we obtain the following result, which was first appeared in
[17, Proposition 2.4].

######  Proposition 3.7.

The semigroup $T(X,\mathcal{P})$ is regular if and only if $\mathcal{P}$ is
trivial.

###### Proof.

Assume that $T(X,\mathcal{P})$ is regular. If $|X|\leq 2$, then there is
nothing to prove. Assume that $|X|\geq 3$. Suppose to the contrary that
$\mathcal{P}$ is not trivial. Then $|\mathcal{P}|\geq 2$ and $|X_{j}|\geq 2$
for some $j\in I$. Consider the constant mapping $\alpha\in T(I)$ defined by
$i\alpha=j$. Since $|\mathcal{P}|\geq 2$, we get $|j\alpha^{-1}|\geq 2$.
Therefore, since $T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, the semigroup
$T(X,\mathcal{P})$ is not regular by Theorem 3.5, a contradiction.

Conversely, assume that $\mathcal{P}$ is trivial. Then
$T(X,\mathcal{P})=T(X)$, and hence $T(X,\mathcal{P})$ is regular by [12,
Exercise 15, p. 63]. ∎

Recall that a _block mapping_ is a mapping whose both domain and codomain are
blocks of a partition of a nonempty set (cf. [25, Definition 5.1]). Since
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a subsemigroup of $T(X,\mathcal{P})$,
the following result is immediate from [25, Lemma 5.2].

######  Lemma 3.8.

Let $f\in T(X)$. Then $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if
there exists a unique indexed family
$B(f,I)=\\{f_{\upharpoonright{X_{i}}}\colon i\in I\\}$ of block mappings
induced by $f$.

Note that an element $f\in T(X)$ is an idempotent if and only if $f$ acts as
the identity on its image set (cf. [3, p. 6]). If $X$ is finite, the
idempotents of $T(X,\mathcal{P})$ are characterized in [8, Proposition 3.1]
and [9, Proposition 3.1] for a uniform partition $\mathcal{P}$ and a non-
uniform partition $\mathcal{P}$, respectively.

The following result describes idempotents of
$T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Proposition 3.9.

Let $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $f\in
E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ if and only if:

  1. (i)

$\chi^{(f)}\in E(\mathbb{S}(I))$;

  2. (ii)

$f_{\upharpoonright{X_{i}}}\in E(T(X_{i}))$ for all $i\in I\chi^{(f)}$;

  3. (iii)

$X_{i}f\subseteq X_{i\chi^{(f)}}f$ for all $i\in I\setminus I\chi^{(f)}$.

###### Proof.

Assume that $f\in E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. Then
$\chi^{(f)}\in\mathbb{S}(I)$ and $f^{2}=f$.

  1. (i)

It is clear that $\chi^{(f)}\in E(\mathbb{S}(I))$, since
$\left(\chi^{(f)}\right)^{2}=\chi^{(f^{2})}=\chi^{(f)}$ by Lemma 3.1.

  2. (ii)

Let $i\in I\chi^{(f)}$, and let $y\in X_{i}f_{\upharpoonright{X_{i}}}$. Then
$i\chi^{(f)}=i$ by $\rm(i)$, and $y\in Xf$. Therefore, since $f$ is
idempotent, we obtain $yf_{\upharpoonright{X_{i}}}=yf=y$. Hence
$f_{\upharpoonright{X_{i}}}\in E(T(X_{i}))$.

  3. (iii)

It is straightforward, since $f$ stabilizes $\mathcal{P}$ and $f^{2}=f$.

Conversely, assume that the given conditions hold. To prove $f\in
E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$, let $y\in Xf$. Then $y\in
X_{i}$ for some $i\in I\chi^{(f)}$. Therefore $i\chi^{(f)}=i$ by (i), and
$f_{\upharpoonright{X_{i}}}\in E(T(X_{i}))$ by (ii). By (iii), we then see
that $y\in X_{i}f_{\upharpoonright{X_{i}}}$. Thus we obtain
$yf=yf_{\upharpoonright{X_{i}}}=y$. Hence $f\in
E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. ∎

Recall that a semigroup is an inverse semigroup if and only if it is regular
and its idempotents commute (cf. [12, Theorem 5.1.1]).

The following result determines when $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an
inverse semigroup.

######  Theorem 3.10.

The semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse semigroup if
and only if:

  1. (i)

$\mathbb{S}(I)$ is an inverse semigroup;

  2. (ii)

for each $\alpha\in E(\mathbb{S}(I))$, we have $|X_{i}|=1$ for all $i\in
I\alpha$.

###### Proof.

Assume that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse semigroup. Then
(i) is true by Proposition 3.2 and the fact that any homomorphic image of an
inverse semigroup is an inverse semigroup (cf. [12, Theorem 5.1.4]).

To prove (ii), let $\alpha\in E(\mathbb{S}(I))$. If $|X|=1$, then there is
nothing to prove. Assume that $|X|\geq 2$. Suppose to the contrary that there
exists $j\in I\alpha$ such that $|X_{j}|\geq 2$. Since $\alpha\in
E(\mathbb{S}(I))$ and $j\in I\alpha$, we get $j\alpha=j$. Now, choose distinct
$y,z\in X_{j}$. If $I\setminus\\{j\\}\neq\varnothing$, then we fix $x_{i}\in
X_{i}$ for each $i\in I\setminus\\{j\\}$. Define $f,g\in T(X)$ by:

$\displaystyle xf=\begin{cases}y&\text{if $x\in X_{i}$, where $i\in
j\alpha^{-1}$};\\\ x_{i\alpha}&\text{if $x\in X_{i}$, where $i\in I\setminus
j\alpha^{-1}$}.\end{cases}$

and

$\displaystyle xg=\begin{cases}z&\text{if $x\in X_{i}$, where $i\in
j\alpha^{-1}$};\\\ x_{i\alpha}&\text{if $x\in X_{i}$, where $i\in I\setminus
j\alpha^{-1}$}.\end{cases}$

Clearly $\chi^{(f)}=\alpha=\chi^{(g)}$. Since $\alpha\in E(\mathbb{S}(I))$, it
follows that $\chi^{(f)},\chi^{(g)}\in E(\mathbb{S}(I))$. Now, we observe from
the definition of $f$ that $f_{\upharpoonright{X_{i}}}\in E(T(X_{i}))$ for all
$i\in I\alpha$, and
$X_{i}f_{\upharpoonright{X_{i}}}=X_{i\alpha}f_{\upharpoonright{X_{i\alpha}}}$
for all $i\in I\setminus I\alpha$. Therefore $f\in
E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ by Proposition 3.9.
Similarly, we can obtain $g\in
E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. However, since $y(fg)\neq
y(gf)$, we have $fg\neq gf$. This contradicts our assumption that
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse semigroup. Thus $|X_{i}|=1$
for all $i\in I\alpha$, and hence (ii) holds.

Conversely, assume that the given conditions hold. To prove that
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse semigroup, we will show that
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular and its idempotents commute.

First, we show that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is regular. By (i), it
is clear that $\mathbb{S}(I)$ is regular. Now, let $\alpha\in\mathbb{S}(I)$.
Since $\mathbb{S}(I)$ is regular, there exists $\beta\in E(\mathbb{S}(I))$
such that $I\alpha=I\beta$ (cf. [12, Theorem 2.3.2 and Exercise 16(a) on p.
63]). Then by (ii), we have $|X_{i}|=1$ for all $i\in I\beta$. It follows that
$|X_{i}|=1$ for all $i\in I\alpha$. Thus $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is
regular by Theorem 3.5.

To prove that idempotents of $T_{\mathbb{S}(I)}(X,\mathcal{P})$ commute, let
$f,g\in E\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ and let $x\in X$.
Then $\chi^{(f)},\chi^{(g)}\in E(\mathbb{S}(I))$ by Proposition 3.9, and $x\in
X_{i}$ for some $i\in I$. Write $i\chi^{(fg)}=j$. Observe from (i) and Lemma
3.1 that
$i\chi^{(gf)}=i(\chi^{(g)}\chi^{(f)})=i(\chi^{(f)}\chi^{(g)})=i\chi^{(fg)}=j$.
Note also from Lemma 3.1 that $I\chi^{(fg)}=I(\chi^{(f)}\chi^{(g)})\subseteq
I\chi^{(g)}$, and so $j\in I\chi^{(g)}$. By (ii), we then get $X_{j}=\\{y\\}$
for some $y\in X$. Therefore we obtain $x(fg)=y=x(gf)$. Hence idempotents of
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ commute.

Thus we conclude that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is an inverse
semigroup.

∎

##  4\. Unit-regularity of $T_{\mathbb{S}(I)}(X,\mathcal{P})$

Throughout this section, assume that $\mathbb{S}(I)$ contains the identity
$\operatorname{id}_{I}$. We describe unit-regular elements in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then we determine when
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a unit-regular semigroup. Using this, we
determine when $T(X,\mathcal{P})$ is a unit-regular semigroup.

Observe that $\mathbb{S}(I)$ contains the identity $\operatorname{id}_{I}$ if
and only if $T_{\mathbb{S}(I)}(X,\mathcal{P})$ contains the identity
$\operatorname{id}_{X}$. Note that
$U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}\subseteq
U\big{(}T(X,\mathcal{P})\big{)}=S(X,\mathcal{P})$, and so
$U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}=T_{\mathbb{S}(I)}(X,\mathcal{P})\cap
S(X,\mathcal{P})$.

We require the next three results from [25, 27] to prove Theorem 4.4, which
describes unit-regular elements in $T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Lemma 4.1.

[25, Lemma 3.6(i)] If $f\in S(X,\mathcal{P})$, then $X_{i}f\in\mathcal{P}$ for
all $i\in I$.

######  Lemma 4.2.

[25, Theorem 5.8] Let $f\in T(X,\mathcal{P})$. Then $f\in S(X,\mathcal{P})$ if
and only if:

  1. (i)

every element of $B(f,I)$ is bijective;

  2. (ii)

$\chi^{(f)}$ is bijective.

######  Lemma 4.3.

[27, Lemma 3.1] Let $f\colon X\to Y$ and $g\colon Y\to X$ be mappings. If
$fgf=f$, then $X(fg)$ is a transversal of the equivalence relation $\ker(f)$.

######  Theorem 4.4.

Let $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then
$f\in\operatorname{ureg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ if and
only if there exists $\alpha\in U(\mathbb{S}(I))$ such that:

  1. (i)

$\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$;

  2. (ii)

$X_{i}\cap Xf\subseteq X_{i\alpha}f$ for all $i\in I\chi^{(f)}$;

  3. (iii)

$|X_{i}|=|X_{i\alpha}|$ for all $i\in I$;

  4. (iv)

$\textnormal{c}(f_{\upharpoonright{X_{i\alpha}}})=\textnormal{d}(f_{\upharpoonright{X_{i\alpha}}})$
for all $i\in I\chi^{(f)}$.

###### Proof.

Assume that
$f\in\operatorname{ureg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. Then
there exists $g\in U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ such that
$fgf=f$. This gives $\chi^{(f)}\chi^{(g)}\chi^{(f)}=\chi^{(f)}$ by Lemma 3.1.
Note that $\chi^{(g)}\in U(\mathbb{S}(I))$, and write $\chi^{(g)}=\alpha$.

  1. (i)

It is clear from the above.

  2. (ii)

Let $i\in I\chi^{(f)}$ and $y\in X_{i}\cap Xf$. Then, since $y\in Xf$, there
exists $x\in X$ such that $xf=y$. Also, since $y\in X_{i}$ and
$\chi^{(g)}=\alpha$, we see that $yg\in X_{i\alpha}$. Therefore, since
$f=fgf$, we obtain $y=xf=(xf)gf=(yg)f\in X_{i\alpha}f$. Hence $X_{i}\cap
Xf\subseteq X_{i\alpha}f$.

  3. (iii)

Let $i\in I$. Note that $g\in U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$
and $\chi^{(g)}=\alpha$. Then, since
$U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}\subseteq S(X,\mathcal{P})$,
we get $X_{i}g=X_{i\alpha}$ by Lemma 4.1, which implies
$|X_{i}|=|X_{i\alpha}|$.

  4. (iv)

Let $i\in I\chi^{(f)}$, and write $i\alpha=j$. Recall that $g\in
U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ and $\chi^{(g)}=\alpha$.
Since $U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}\subseteq
S(X,\mathcal{P})$, it follows from Lemma 4.1 that $X_{i}g=X_{j}$. This yields
$X_{i}g_{\upharpoonright{X_{i}}}=X_{j}$. By (ii), we have
$X_{j}f_{\upharpoonright{X_{j}}}\subseteq X_{i}$. Then, since $fgf=f$, we
obtain
$f_{\upharpoonright{X_{j}}}g_{\upharpoonright{X_{i}}}f_{\upharpoonright{X_{j}}}=f_{\upharpoonright{X_{j}}}$.
It follows that $X_{j}(f_{\upharpoonright{X_{j}}}g_{\upharpoonright{X_{i}}})$
is a transversal of $\ker(f_{\upharpoonright{X_{j}}})$ by Lemma 4.3. Write
$T_{f_{\upharpoonright{X_{j}}}}=X_{j}(f_{\upharpoonright{X_{j}}}g_{\upharpoonright{X_{i}}})$.
Since $g\in S(X,\mathcal{P})$, it follows that $g_{\upharpoonright{X_{i}}}$ is
bijective by Theorem 4.2. Therefore $(X_{i}\setminus
X_{j}f_{\upharpoonright{X_{j}}})g_{\upharpoonright{X_{i}}}=X_{j}\setminus
T_{f_{\upharpoonright{X_{j}}}}$, and hence
$\textnormal{d}(f_{\upharpoonright{X_{j}}})=\textnormal{c}(f_{\upharpoonright{X_{j}}})$.

Conversely, assume that the given conditions hold. To prove
$f\in\operatorname{ureg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$, we
will construct $g\in U(T_{\mathbb{S}(I)}(X,\mathcal{P}))$ such that $fgf=f$.
Let $i\in I\chi^{(f)}$, and write $i\alpha=j$. Choose a transversal
$T_{f_{\upharpoonright{X_{j}}}}$ of $\ker(f_{\upharpoonright{X_{j}}})$, and
define a mapping $h_{i}^{\prime}\colon X_{j}f_{\upharpoonright{X_{j}}}\to
T_{f_{\upharpoonright{X_{j}}}}$ by $xh_{i}^{\prime}=a_{x}$ if and only if
$x(f_{\upharpoonright{X_{j}}})^{-1}\cap
T_{f_{\upharpoonright{X_{j}}}}=\\{a_{x}\\}$. It is easy to verify that
$h_{i}^{\prime}$ is a bijection. By (iv), there exists a bijection
$h_{i}^{\prime\prime}\colon D(f_{\upharpoonright{X_{j}}})\to
C(f_{\upharpoonright{X_{j}}})$, where
$C(f_{\upharpoonright{X_{j}}})=X_{j}\setminus T_{f_{\upharpoonright{X_{j}}}}$.
By (i), we see that $j\chi^{(f)}=i$. Thus we may write
$X_{i}=X_{j}f_{\upharpoonright{X_{j}}}\cup D(f_{\upharpoonright{X_{j}}})$ and
$X_{j}=T_{f_{\upharpoonright{X_{j}}}}\cup C(f_{\upharpoonright{X_{j}}})$.
Define a mapping $h_{i}\colon X_{i}\to X_{j}$ by:

$\displaystyle xh_{i}=\begin{cases}xh_{i}^{\prime}&\text{if $x\in
X_{j}f_{\upharpoonright{X_{j}}}$};\\\ xh_{i}^{\prime\prime}&\text{if $x\in
D(f_{\upharpoonright{X_{j}}})$}.\end{cases}$

Clearly $h_{i}$ is a bijection. Now for each $j\in I\setminus I\chi^{(f)}$, by
(iii) there is a bijection $g_{j}\colon X_{j}\to X_{j\alpha}$. Finally, define
$g\in T(X)$ by:

$\displaystyle xg=\begin{cases}xh_{i}&\text{if $x\in X_{i}$, where $i\in
I\chi^{(f)}$};\\\ xg_{j}&\text{if $x\in X_{j}$, where $j\in I\setminus
I\chi^{(f)}$}.\end{cases}$

Clearly $\chi^{(g)}=\alpha$. Since $\alpha\in U(\mathbb{S}(I))$, it follows
that $g$ is a bijection. Therefore we have $g\in
U\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. It is routine to verify that
$fgf=f$, and hence
$f\in\operatorname{ureg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$. ∎

We require the following result to prove Theorem 4.6, which determines unit-
regularity of $T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Lemma 4.5.

Let $X$ and $Y$ be nonempty sets. Then there is a mapping $f\colon X\to Y$
such that $\textnormal{c}(f)\neq\textnormal{d}(f)$ if any of the following
conditions hold:

  1. (i)

$|X|\neq|Y|$;

  2. (ii)

$X$ is infinite;

  3. (iii)

$Y$ is infinite.

###### Proof.

  1. (i)

If $|X|<|Y|$, then there exists an injection $f\colon X\to Y$. Therefore
$\textnormal{c}(f)=0$ and $|Xf|=|X|$. Since $|X|<|Y|$, we have $|Xf|<|Y|$.
Therefore $\textnormal{d}(f)\neq 0$, and hence
$\textnormal{c}(f)\neq\textnormal{d}(f)$. If $|X|>|Y|$, we may dually prove
that $\textnormal{c}(f)\neq\textnormal{d}(f)$.

  2. (ii)

If $|X|\neq|Y|$, then we are done by (i). Assume that $|X|=|Y|$. Since $X$ is
infinite, there exists an injection $f\colon X\to Y$ that is not a surjection.
Therefore $\textnormal{c}(f)=0$ but $\textnormal{d}(f)\neq 0$. Hence
$\textnormal{c}(f)\neq\textnormal{d}(f)$.

  3. (iii)

It is similar to (ii).

∎

######  Theorem 4.6.

The semigroup $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is unit-regular if and only
if:

  1. (i)

$\mathbb{S}(I)$ is unit-regular;

  2. (ii)

for each $\alpha\in U(\mathbb{S}(I))$, we have $|X_{i}|=|X_{i\alpha}|$ for all
$i\in I$;

  3. (iii)

$|X_{i}|$ is finite for all $i\in I$;

  4. (iv)

for each $\beta\in\mathbb{S}(I)$ if there exists $i\in I$ such that
$|i\beta^{-1}|\geq 2$, then $|X_{i}|=1$.

###### Proof.

Assume that $T_{\mathbb{S}(I)}(X,\mathcal{P})$ is unit-regular.

  1. (i)

The semigroup $\mathbb{S}(I)$ is unit-regular by Proposition 3.2 and the fact
that any homomorphic image of a unit-regular semigroup is unit-regular (cf.
[10, Proposition 2.7]).

  2. (ii)

Let $\alpha\in U(\mathbb{S}(I))$. Then $\alpha\in\operatorname{Sym}(I)$.
Suppose to the contrary that there exists $j\in I$ such that
$|X_{j}|\neq|X_{j\alpha}|$. Then by Lemma 4.5, there exists a mapping $h\colon
X_{j}\to X_{j\alpha}$ such that $\textnormal{c}(h)\neq\textnormal{d}(h)$. For
each $i\in I\setminus\\{j\alpha\\}$, we fix $x_{i}\in X_{i}$. Define $f\in
T(X)$ by:

$\displaystyle xf=\begin{cases}xh&\text{if $x\in X_{j}$};\\\
x_{i\alpha}&\text{if $x\in X_{i}$, where $i\in
I\setminus\\{j\\}$}.\end{cases}$

Clearly $\chi^{(f)}=\alpha$, and so $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$.
Now, let $j\alpha=k$ and $\alpha^{-1}=\beta$. Clearly $\beta\in
U(\mathbb{S}(I))$ and $k\beta=j$. Note that $k\in I\chi^{(f)}$ and
$f_{\upharpoonright{X_{k\beta}}}=f_{\upharpoonright{X_{j}}}=h$. Since
$\textnormal{c}(h)\neq\textnormal{d}(h)$, it follows that
$\textnormal{c}(f_{\upharpoonright{X_{k\beta}}})\neq\textnormal{d}(f_{\upharpoonright{X_{k\beta}}})$.
This means $f$ does not satisfy the condition $\rm(iv)$ of Theorem 4.4, and so
$f\notin\operatorname{ureg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ by
Theorem 4.4. This is a contradiction, because
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is unit-regular. Hence
$|X_{i}|=|X_{i\alpha}|$ for all $i\in I$.

  3. (iii)

Suppose to the contrary that there exists $j\in I$ such that $|X_{j}|$ is
infinite. Take $\alpha\in U(\mathbb{S}(I))$, and write $j\alpha=\ell$. Since
$|X_{j}|$ is infinite, by Lemma 4.5 there is a mapping $h\colon X_{j}\to
X_{\ell}$ such that $\textnormal{c}(h)\neq\textnormal{d}(h)$. For each $i\in
I\setminus\\{j\alpha\\}$, we fix $x_{i}\in X_{i}$. Define $f\in T(X)$ by:

$\displaystyle xf=\begin{cases}xh&\text{if $x\in X_{j}$};\\\
x_{i\alpha}&\text{if $x\in X_{i}$, where $i\in
I\setminus\\{j\\}$}.\end{cases}$

Clearly $\chi^{(f)}=\alpha$, and so $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$.
Write $\alpha^{-1}=\beta$. Clearly $\beta\in U(\mathbb{S}(I))$ and
$\ell\beta=j$. Note that $\ell\in I\chi^{(f)}$ and
$f_{\upharpoonright{X_{\ell\beta}}}=f_{\upharpoonright{X_{j}}}=h$. Since
$\textnormal{c}(h)\neq\textnormal{d}(h)$, it follows that
$\textnormal{c}(f_{\upharpoonright{X_{\ell\beta}}})\neq\textnormal{d}(f_{\upharpoonright{X_{\ell\beta}}})$.
This means $f$ does not satisfy the condition $\rm(iv)$ of Theorem 4.4, and so
$f\notin\operatorname{ureg}\big{(}T_{\mathbb{S}(I)}(X,\mathcal{P})\big{)}$ by
Theorem 4.4. This is a contradiction, because
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is unit-regular. Hence $|X_{i}|$ is finite
for all $i\in I$.

  4. (iv)

It is directly followed from Theorem 3.5, since
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ is a regular semigroup.

Conversely, assume that the given conditions hold, and let $f\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$. To prove the desired result, we will show
that $f\in\operatorname{ureg}(T_{\mathbb{S}(I)}(X,\mathcal{P}))$ by Theorem
4.4. Since $f\in T_{\mathbb{S}(I)}(X,\mathcal{P})$, we have
$\chi^{(f)}\in\mathbb{S}(I)$. By (i), there exists $\alpha\in
U(\mathbb{S}(I))$ such that $\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$. By (ii),
we have $|X_{i}|=|X_{i\alpha}|$ for all $i\in I$.

Next, let $i\in I\chi^{(f)}$. Since $\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$,
we see that $X_{i\alpha}f\subseteq X_{i}$. That means
$f_{\upharpoonright{X_{i\alpha}}}$ is a block mapping from $X_{i\alpha}$ to
$X_{i}$. Note from $\rm(ii)$ and (iii) that $|X_{i}|=|X_{i\alpha}|$ and
$X_{i}$ is finite, respectively. Therefore
$\textnormal{c}(f_{\upharpoonright{X_{i\alpha}}})=\textnormal{d}(f_{\upharpoonright{X_{i\alpha}}})$.

Finally, since $i\in I\chi^{(f)}$, there are two cases:

Case 1: Suppose $|i(\chi^{(f)})^{-1}|=1$. Since
$\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$, we see that $(i\alpha)\chi^{(f)}=i$
and so $i\left(\chi^{(f)}\right)^{-1}=\\{i\alpha\\}$. This gives
$X_{i\alpha}f=X_{i}\cap Xf$, and so $X_{i}\cap Xf\subseteq X_{i\alpha}f$.

Case 2: Suppose $|i(\chi^{(f)})^{-1}|\geq 2$. Then by (iv), we get
$|X_{i}|=1$. Therefore, since $i\in I\chi^{(f)}$, we obtain $X_{i}\cap
Xf=X_{i}$. Since $\chi^{(f)}\alpha\chi^{(f)}=\chi^{(f)}$, we see that
$(i\alpha)\chi^{(f)}=i$. Therefore $X_{i\alpha}f=X_{i}$, and so $X_{i}\cap
Xf\subseteq X_{i\alpha}f$.

In either case, we have $X_{i}\cap Xf\subseteq X_{i\alpha}f$. Thus $f$
satisfies each condition of Theorem 4.4, and hence
$f\in\operatorname{ureg}(T_{\mathbb{S}(I)}(X,\mathcal{P}))$ by Theorem 4.4. ∎

The authors [27, Theorem 4.7] recently described unit-regular elements in
$T(X,\mathcal{P})$. We require the following result to prove Proposition 4.8,
which determines unit-regularity of $T(X,\mathcal{P})$.

######  Proposition 4.7.

[4, Proposition 5] The semigroup $T(X)$ is unit-regular if and only if $X$ is
finite.

######  Proposition 4.8.

The semigroup $T(X,\mathcal{P})$ is unit-regular if and only if $X$ is finite
and $\mathcal{P}$ is trivial.

###### Proof.

Assume that $T(X,\mathcal{P})$ is unit-regular. Then, since
$T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, the semigroup $T(I)$ is unit-
regular by Theorem 4.6(i). Therefore $I$ is finite by Proposition 4.7. It
follows from Theorem 4.6(iii) that $X$ is finite. Next, since
$T(X,\mathcal{P})$ is regular, the partition $\mathcal{P}$ is trivial by
Proposition 3.7.

Conversely, assume that the given conditions hold. Since $\mathcal{P}$ is
trivial, it follows that $T(X,\mathcal{P})=T(X)$. Moreover, since $X$ is
finite, the semigroup $T(X,\mathcal{P})$ is unit-regular by Proposition 4.7. ∎

##  5\. Green’s Relations on $T_{\mathbb{S}(I)}(X,\mathcal{P})$

Throughout this section, assume that the semigroup $\mathbb{S}(I)$ contains
the identity $\operatorname{id}_{I}$. We characterize Green’s relations on
$T_{\mathbb{S}(I)}(X,\mathcal{P})$. Using these characterizations, we obtain
more concrete and noticeably different descriptions of Green’s relations on
$T(X,\mathcal{P})$, whose Green’s relations are known; see [17, 18]. We refer
the reader to [12, Chapter 2] for the basic definitions and notation related
to Green’s relations on a semigroup.

First, we recall the following well-known characterizations of Green’s
relations $\mathcal{L}$, $\mathcal{R}$, $\mathcal{D}$, and $\mathcal{J}$ on
$T(X)$ (cf. [3, p. 52], [12, p. 63, Exercise 16]).

######  Theorem 5.1.

Let $f,g\in T(X)$. Then:

  1. (1)

$L_{f}\leq L_{g}$ if and only if $Xf\subseteq Xg$. Subsequently,
$(f,g)\in\mathcal{L}$ if and only if $Xf=Xg$;

  2. (2)

$R_{f}\leq R_{g}$ if and only if $\ker(g)\subseteq\ker(f)$. Subsequently,
$(f,g)\in\mathcal{R}$ if and only if $\ker(f)=\ker(g)$;

  3. (3)

$J_{f}\leq J_{g}$ if and only if $|Xf|\leq|Xg|$. Subsequently,
$(f,g)\in\mathcal{J}$ if and only if $|Xf|=|Xg|$;

  4. (4)

$\mathcal{D}=\mathcal{J}$.

We require the following result to obtain a characterization of the relation
$\mathcal{L}$ on $T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Lemma 5.2.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $L_{f}\leq L_{g}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if there exists
$\alpha\in\mathbb{S}(I)$ such that $\chi^{(f)}=\alpha\chi^{(g)}$ and
$X_{i}f\subseteq X_{i\alpha}g$ for all $i\in I$.

###### Proof.

Assume that $L_{f}\leq L_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then
$f=hg$ for some $h\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. This gives
$\chi^{(f)}=\chi^{(hg)}=\chi^{(h)}\chi^{(g)}$ by Lemma 3.1. Write
$\chi^{(h)}=\alpha$, and thus $\chi^{(f)}=\alpha\chi^{(g)}$. To prove the
remaining part, let $i\in I$. Notice that $X_{i}h\subseteq X_{i\alpha}$, and
therefore $X_{i}f=X_{i}(hg)=(X_{i}h)g\subseteq X_{i\alpha}g$.

Conversely, assume that the given conditions hold. To prove $L_{f}\leq L_{g}$
in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we will construct $h\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $f=hg$. For each $i\in I$ and for
each $x\in X_{i}$, by hypothesis we fix $y\in X_{i\alpha}$ such that $xf=yg$.
Define $h\in T(X)$ by $xh=y$ whenever $x\in X_{i}$ for some $i\in I$. It is
clear by the definition of $h$ that $\chi^{(h)}=\alpha$, and so $h\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$. To prove $f=hg$, let $x\in X$. Then $x\in
X_{i}$ for some $i\in I$. Therefore $x(hg)=(xh)g=yg=xf$, which yields $f=hg$.
Hence $L_{f}\leq L_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. ∎

######  Theorem 5.3.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{L}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if there exist
$\alpha,\beta\in\mathbb{S}(I)$ such that:

  1. (i)

$\chi^{(f)}=\alpha\chi^{(g)}$ and $\chi^{(g)}=\beta\chi^{(f)}$;

  2. (ii)

$X_{i}f\subseteq X_{i\alpha}g$ and $X_{i}g\subseteq X_{i\beta}f$ for all $i\in
I$.

###### Proof.

It follows directly from Lemma 5.2. ∎

In 2005, Pei [17, Theorem 3.2] gave a characterization of the relation
$\mathcal{L}$ on $T(X,\mathcal{P})$. Using Theorem 5.3, we also provide
another characterization of the relation $\mathcal{L}$ on $T(X,\mathcal{P})$
as follows.

######  Proposition 5.4.

Let $f,g\in T(X,\mathcal{P})$. Then $(f,g)\in\mathcal{L}$ in
$T(X,\mathcal{P})$ if and only if for each $i\in I$, there exist $j,k\in I$
such that $X_{i}f\subseteq X_{j}g$ and $X_{i}g\subseteq X_{k}f$.

###### Proof.

Assume that $(f,g)\in\mathcal{L}$ in $T(X,\mathcal{P})$, and let $i\in I$. By
Theorem 5.3, since $T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, there exist
$\alpha,\beta\in T(I)$ such that $X_{i}f\subseteq X_{i\alpha}g$ and
$X_{i}g\subseteq X_{i\beta}f$. Write $i\alpha=j$ and $i\beta=k$. Thus
$X_{i}f\subseteq X_{j}g$ and $X_{i}g\subseteq X_{k}f$.

Conversely, assume that the given conditions hold. For each $i\in I$, by
hypothesis we choose $j_{i},k_{i}\in I$ such that $X_{i}f\subseteq X_{j_{i}}g$
and $X_{i}g\subseteq X_{k_{i}}f$. Define $\alpha,\beta\in T(I)$ by
$i\alpha=j_{i}$ and $i\beta=k_{i}$. Then we have $X_{i}f\subseteq
X_{i\alpha}g$ and $X_{i}g\subseteq X_{i\beta}f$ for all $i\in I$. Next, we
prove that $\chi^{(f)}=\alpha\chi^{(g)}$. For this, let $i\in I$. Then
$X_{i}f\subseteq X_{j_{i}}g$, and so $i\chi^{(f)}=j_{i}\chi^{(g)}$. Therefore
we obtain
$i\left(\alpha\chi^{(g)}\right)=(i\alpha)\chi^{(g)}=j_{i}\chi^{(g)}=i\chi^{(f)}$,
which yields $\chi^{(f)}=\alpha\chi^{(g)}$. Similarly, we may prove that
$\chi^{(g)}=\beta\chi^{(f)}$ by using $X_{i}g\subseteq X_{k_{i}}f$. Thus,
since $T_{T(I)}(X,\mathcal{P})=T(X,\mathcal{P})$, we conclude from Theorem 5.3
that $(f,g)\in\mathcal{L}$ in $T(X,\mathcal{P})$. ∎

To obtain a characterization of the relation $\mathcal{R}$ on
$T_{\mathbb{S}(I)}(X,\mathcal{P})$, we require the following definition and
Lemma 5.5.

Let $f,g\in T(X)$. We say that $\pi(f)$ refines $\pi(g)$, denoted by
$\pi(f)\preceq\pi(g)$, if $\ker(f)\subseteq\ker(g)$. We write $\pi(f)=\pi(g)$
if $\pi(f)\preceq\pi(g)$ and $\pi(g)\preceq\pi(f)$.

######  Lemma 5.5.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $R_{f}\leq R_{g}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if $R_{\chi^{(f)}}\leq
R_{\chi^{(g)}}$ in $\mathbb{S}(I)$ and $\pi(g)\preceq\pi(f)$.

###### Proof.

Assume that $R_{f}\leq R_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then
$f=gh$ for some $h\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. This gives
$\chi^{(f)}=\chi^{(gh)}=\chi^{(g)}\chi^{(h)}$ by Lemma 3.1. Since
$\chi^{(f)},\chi^{(g)},\chi^{(h)}\in\mathbb{S}(I)$, it follows that
$R_{\chi^{(f)}}\leq R_{\chi^{(g)}}$ in $\mathbb{S}(I)$. Next, since
$T_{\mathbb{S}(I)}(X,\mathcal{P})\subseteq T(X)$, we have $f,g,h\in T(X)$.
Therefore $\ker(g)\subseteq\ker(f)$ by Theorem 5.1(2), and hence
$\pi(g)\preceq\pi(f)$.

Conversely, assume that the given conditions hold. To prove $R_{f}\leq R_{g}$
in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we will construct $h\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $f=gh$. Since $R_{\chi^{(f)}}\leq
R_{\chi^{(g)}}$ in $\mathbb{S}(I)$, we have $\chi^{(f)}=\chi^{(g)}\beta$ for
some $\beta\in\mathbb{S}(I)$. Now, for each $i\in I$, fix $x_{i}\in X_{i}$;
for each $x\in Xg$, fix $x^{\prime}\in xg^{-1}$. Define $h\in T(X)$ by:

$\displaystyle xh=\begin{cases}x^{\prime}f&\text{if $x\in X_{i}\cap Xg$};\\\
x_{i\beta}&\text{if $x\in X_{i}\setminus Xg$.}\end{cases}$

It is clear that $h$ is well-defined, since $\pi(g)\preceq\pi(f)$. Since
$\chi^{(f)}=\chi^{(g)}\beta$, we see that $\chi^{(h)}=\beta$, and so $h\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$. To prove $f=gh$, let $x\in X$. Then $x\in
X_{i}$ for some $i\in I$. Therefore, since $x^{\prime}\in xg^{-1}$ and
$xg^{-1}\in\pi(f)$, we obtain $x(gh)=(xg)h=x^{\prime}f=xf$, which yields
$f=gh$. Hence $R_{f}\leq R_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. ∎

######  Theorem 5.6.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{R}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if
$\left(\chi^{(f)},\chi^{(g)}\right)\in\mathcal{R}$ in $\mathbb{S}(I)$ and
$\pi(f)=\pi(g)$.

###### Proof.

It follows directly from Lemma 5.5. ∎

Recall that $T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$. Using Theorem 5.6 and
Theorem 5.1(2), we immediately have the following characterization of the
relation $\mathcal{R}$ on $T(X,\mathcal{P})$, which was first appeared in [17,
Theorem 3.1].

######  Corollary 5.7.

Let $f,g\in T(X,\mathcal{P})$. Then $(f,g)\in\mathcal{R}$ in
$T(X,\mathcal{P})$ if and only if
$\pi\left(\chi^{(f)}\right)=\pi\left(\chi^{(g)}\right)$ and $\pi(f)=\pi(g)$.

We require the following result that is useful in establishing the relation
$\mathcal{D}$ on $T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Lemma 5.8.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$.

  1. (i)

If $L_{f}\leq L_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, then
$L_{\chi^{(f)}}\leq L_{\chi^{(g)}}$ in $\mathbb{S}(I)$. Subsequently, if
$(f,g)\in\mathcal{L}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, then
$(\chi^{(f)},\chi^{(g)})\in\mathcal{L}$ in $\mathbb{S}(I)$.

  2. (ii)

If $R_{f}\leq R_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, then
$R_{\chi^{(f)}}\leq R_{\chi^{(g)}}$ in $\mathbb{S}(I)$. Subsequently, if
$(f,g)\in\mathcal{R}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, then
$(\chi^{(f)},\chi^{(g)})\in\mathcal{R}$ in $\mathbb{S}(I)$.

###### Proof.

  1. (i)

If $L_{f}\leq L_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, then $f=hg$ for
some $h\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. This gives
$\chi^{(f)}=\chi^{(hg)}=\chi^{(h)}\chi^{(g)}$ by Lemma 3.1. Since
$\chi^{(f)},\chi^{(g)},\chi^{(h)}\in\mathbb{S}(I)$, it follows that
$L_{\chi^{(f)}}\leq L_{\chi^{(g)}}$ in $\mathbb{S}(I)$. The rest of the proof
is immediate from the previous statements.

  2. (ii)

It is the dual of the proof above.

∎

For every $A\subseteq X$ and every $f\in T(X)$, let
$\pi_{A}(f)=\\{M\in\pi(f)\colon M\cap A\neq\varnothing\\}$. If $A=X$, then we
write $\pi(f)$ instead of $\pi_{X}(f)$.

The following result characterizes the relation $\mathcal{D}$ on
$T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Theorem 5.9.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{D}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if there exist
$\alpha,\beta,\gamma\in\mathbb{S}(I)$ and a bijection
$\varphi\colon\pi(f)\to\pi(g)$ such that:

  1. (i)

$\chi^{(f)}=\alpha\gamma$, $\gamma=\beta\chi^{(f)}$, and
$\left(\gamma,\chi^{(g)}\right)\in\mathcal{R}$ in $\mathbb{S}(I)$;

  2. (ii)

$\pi_{X_{i}}(f)\varphi\subseteq\pi_{X_{i\alpha}}(g)$ for all $i\in I$;

  3. (iii)

$\pi_{X_{i}}(g)\varphi^{-1}\subseteq\pi_{X_{i\beta}}(f)$ for all $i\in I$.

###### Proof.

Assume that $(f,g)\in\mathcal{D}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then,
since $\mathcal{D}=\mathcal{L}\circ\mathcal{R}$, there exists $h\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $(f,h)\in\mathcal{L}$ and
$(h,g)\in\mathcal{R}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. Write
$\chi^{(h)}=\gamma$. Clearly $\gamma\in\mathbb{S}(I)$. Since
$(f,h)\in\mathcal{L}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, by Theorem 5.3
there exist $\alpha,\beta\in\mathbb{S}(I)$ such that $\chi^{(f)}=\alpha\gamma$
and $\gamma=\beta\chi^{(f)}$; $X_{i}f\subseteq X_{i\alpha}h$ and
$X_{i}h\subseteq X_{i\beta}f$ for all $i\in I$. Also, since
$(h,g)\in\mathcal{R}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we get
$(\gamma,\chi^{(g)})\in\mathcal{R}$ in $\mathbb{S}(I)$ by Lemma 5.8(ii). Thus
$\rm(i)$ holds.

To prove (ii) and (iii), we first define a mapping
$\varphi\colon\pi(f)\to\pi(g)$ by $(af^{-1})\varphi=ah^{-1}$ whenever $a\in
Xf$. Since $(f,h)\in\mathcal{L}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we
have $(f,h)\in\mathcal{L}$ in $T(X)$, and so $Xf=Xh$ by Theorem 5.1(1). Since
$a\in Xf$, it follows that $a\in Xh$, and so $ah^{-1}\in\pi(h)$. Also, since
$(h,g)\in\mathcal{R}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we have
$(h,g)\in\mathcal{R}$ in $T(X)$, and so $\pi(h)=\pi(g)$ by Theorem 5.1(2).
Therefore, since $ah^{-1}\in\pi(h)$, we obtain $ah^{-1}\in\pi(g)$. Thus
$\varphi$ is well-defined. Moreover, since $Xf=Xh$ and $\pi(h)=\pi(g)$, it is
a routine matter to verify that $\varphi$ is bijective.

Now to prove $\rm(ii)$, let $i\in I$ and let $A\in\pi_{X_{i}}(f)$. Then $A\cap
X_{i}\neq\varnothing$. Write $Af=\\{a\\}$. Since $A\cap X_{i}\neq\varnothing$
and $f$ stabilizes $\mathcal{P}$, it is obvious that $a\in X_{i}f$. Note that
$X_{i}f\subseteq X_{i\alpha}h$, and so $a\in X_{i\alpha}h$. This gives
$ah^{-1}\cap X_{i\alpha}\neq\varnothing$. Therefore $A\varphi\cap
X_{i\alpha}=(af^{-1})\varphi\cap X_{i\alpha}=ah^{-1}\cap
X_{i\alpha}\neq\varnothing$, which yields $A\varphi\in\pi_{X_{i\alpha}}(g)$.
Hence $\pi_{X_{i}}(f)\varphi\subseteq\pi_{X_{i\alpha}}(g)$, and thus $\rm(ii)$
holds.

Next to prove $\rm(iii)$, let $i\in I$ and let $B\in\pi_{X_{i}}(g)$. Then
$B\cap X_{i}\neq\varnothing$. Since $\pi(h)=\pi(g)$, we get
$\pi_{X_{i}}(h)=\pi_{X_{i}}(g)$. It follows that $B\in\pi_{X_{i}}(h)$. Write
$Bh=\\{b\\}$. Clearly $b\in X_{i}h$. Note that $X_{i}h\subseteq X_{i\beta}f$,
and so $b\in X_{i\beta}f$. This gives $bf^{-1}\cap X_{i\beta}\neq\varnothing$.
Therefore $B\varphi^{-1}\cap X_{i\beta}=(bh^{-1})\varphi^{-1}\cap
X_{i\beta}=bf^{-1}\cap X_{i\beta}\neq\varnothing$, which yields
$B\varphi^{-1}\in\pi_{X_{i\beta}}(f)$. Hence
$\pi_{X_{i}}(g)\varphi^{-1}\subseteq\pi_{X_{i\beta}}(f)$, and thus $\rm(iii)$
holds.

Conversely, assume that the given conditions hold. To prove
$(f,g)\in\mathcal{D}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we will construct
$h\in T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $(f,h)\in\mathcal{L}$ and
$(h,g)\in\mathcal{R}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. Define $h\in
T(X)$ by $xh=y$ whenever $\big{(}(xgg^{-1})\varphi^{-1}\big{)}f=\\{y\\}$.
First, we show that $\chi^{(h)}=\gamma$. For this, let $i\in I$ and let $x\in
X_{i}$. Clearly $xgg^{-1}\cap X_{i}\neq\varnothing$, and so
$xgg^{-1}\in\pi_{X_{i}}(g)$. This gives
$(xgg^{-1})\varphi^{-1}\in\pi_{X_{i\beta}}(f)$ by $\rm(iii)$, and so
$(xgg^{-1})\varphi^{-1}\cap X_{i\beta}\neq\varnothing$. By (i), since
$\gamma=\beta\chi^{(f)}$, we have $(i\beta)\chi^{(f)}=i\gamma$, and so
$\big{(}(xgg^{-1})\varphi^{-1}\big{)}f=\\{y\\}\subseteq X_{i\gamma}$. By the
definition of $h$, we therefore have $xh=y\in X_{i\gamma}$. This gives
$X_{i}h\subseteq X_{i\gamma}$, and so $i\chi^{(h)}=i\gamma$. Hence
$\chi^{(h)}=\gamma$, and thus $h\in T_{\mathbb{S}(I)}(X,\mathcal{P})$.

It is a routine matter to verify that $\pi(h)=\pi(g)$. By (i), we also get
$\left(\chi^{(h)},\chi^{(g)}\right)\in\mathcal{R}$ in $\mathbb{S}(I)$.
Therefore $(h,g)\in\mathcal{R}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$ by
Theorem 5.6.

We only require to show that $(f,h)\in\mathcal{L}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$. Note that $\chi^{(h)}=\gamma$. By
$\rm(i)$, we have $\chi^{(f)}=\alpha\chi^{(h)}$ and
$\chi^{(h)}=\beta\chi^{(f)}$. Therefore by Theorem 5.3, it remains to prove
that $X_{i}f\subseteq X_{i\alpha}h$ and $X_{i}h\subseteq X_{i\beta}f$ for all
$i\in I$. For this, let $i\in I$. To prove $X_{i}f\subseteq X_{i\alpha}h$, let
$y\in X_{i}f$. Then $yf^{-1}\cap X_{i}\neq\varnothing$, and so
$yf^{-1}\in\pi_{X_{i}}(f)$. Therefore by $\rm(ii)$, we have
$(yf^{-1})\varphi\in\pi_{X_{i\alpha}}(g)$. This yields $(yf^{-1})\varphi\cap
X_{i\alpha}\neq\varnothing$. Choose $z\in(yf^{-1})\varphi\cap X_{i\alpha}$.
Then $zgg^{-1}\cap X_{i\alpha}\neq\varnothing$, and so
$zgg^{-1}\in\pi_{X_{i\alpha}}(g)$. Therefore, since $z\in zgg^{-1}$ and
$z\in(yf^{-1})\varphi$, we obtain $(yf^{-1})\varphi=zgg^{-1}$. This gives
$\big{(}(zgg^{-1})\varphi^{-1}\big{)}f=\\{y\\}$. By the definition of $h$, we
get $zh=y$. Hence $y\in X_{i\alpha}h$, and thus $X_{i}f\subseteq
X_{i\alpha}h$.

To prove $X_{i}h\subseteq X_{i\beta}f$, let $y\in X_{i}h$. Then $yh^{-1}\cap
X_{i}\neq\varnothing$, and so $yh^{-1}\in\pi_{X_{i}}(h)$. Since
$\pi(h)=\pi(g)$, it follows easily that $\pi_{X_{i}}(h)=\pi_{X_{i}}(g)$.
Therefore $yh^{-1}\in\pi_{X_{i}}(g)$. By (iii), we then have
$(yh^{-1})\varphi^{-1}\in\pi_{X_{i\beta}}(f)$, and so
$(yh^{-1})\varphi^{-1}\cap X_{i\beta}\neq\varnothing$. Choose
$z\in(yh^{-1})\varphi^{-1}\cap X_{i\beta}$. Then $zff^{-1}\cap
X_{i\beta}\neq\varnothing$, and so $zff^{-1}\in\pi_{X_{i\beta}}(f)$.
Therefore, since $z\in zff^{-1}$ and $z\in(yh^{-1})\varphi^{-1}$, we obtain
$(yh^{-1})\varphi^{-1}=zff^{-1}$. This gives
$\big{(}(yh^{-1})\varphi^{-1}\big{)}f=\\{zf\\}$. By the definition of $h$, we
get $zf=y$. Hence $y\in X_{i\beta}f$, and thus $X_{i}h\subseteq X_{i\beta}f$.
Thus $(f,h)\in\mathcal{L}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$ by Theorem
5.3. Hence, since $\mathcal{L}\circ\mathcal{R}=\mathcal{D}$, we conclude that
$(f,g)\in\mathcal{D}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. ∎

In 2005, Pei [17, Theorem 3.4] determined the relation $\mathcal{D}$ on
$T(X,\mathcal{P})$. Sangkhanan and Sanwong [22, Corollary 3.10] also obtained
a different description of $\mathcal{D}$ on $T(X,\mathcal{P})$. Using Theorem
5.9, we obtain a new characterization of $\mathcal{D}$ on $T(X,\mathcal{P})$
as follows.

######  Proposition 5.10.

Let $f,g\in T(X,\mathcal{P})$. Then $(f,g)\in\mathcal{D}$ in
$T(X,\mathcal{P})$ if and only if there exist $\gamma\in T(I)$ and a bijection
$\varphi\colon\pi(f)\to\pi(g)$ such that:

  1. (i)

$\left(\chi^{(f)},\gamma\right)\in\mathcal{L}$ and
$\left(\gamma,\chi^{(g)}\right)\in\mathcal{R}$ in $T(I)$;

  2. (ii)

for each $i\in I$, there exist $j\in\left(i\chi^{(f)}\right)\gamma^{-1}$ and
$k\in(i\gamma)\left(\chi^{(f)}\right)^{-1}$ such that
$\pi_{X_{i}}(f)\varphi\subseteq\pi_{X_{j}}(g)$ and
$\pi_{X_{i}}(g)\varphi^{-1}\subseteq\pi_{X_{k}}(f)$.

###### Proof.

Assume that $(f,g)\in\mathcal{D}$ in $T(X,\mathcal{P})$. By Theorem 5.9, since
$T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, there exist
$\alpha,\beta,\gamma\in T(I)$ and a bijection $\varphi\colon\pi(f)\to\pi(g)$
such that each condition of Theorem 5.9 holds. Thus (i) follows immediately
from Theorem 5.9(i).

To prove (ii), let $i\in I$. Write $i\alpha=j$ and $i\beta=k$. First, we show
that $j\in\left(i\chi^{(f)}\right)\gamma^{-1}$. By Theorem 5.9$\rm(i)$, we
have $\chi^{(f)}=\alpha\gamma$. Therefore
$i\chi^{(f)}=i(\alpha\gamma)=(i\alpha)\gamma=j\gamma$, which gives
$j\in\left(i\chi^{(f)}\right)\gamma^{-1}$. Moreover, by Theorem 5.9$\rm(ii)$
we deduce that $\pi_{X_{i}}(f)\varphi\subseteq\pi_{X_{j}}(g)$. Next by Theorem
5.9$\rm(i)$, we have $\gamma=\beta\chi^{(f)}$. A similar argument using
$\gamma=\beta\chi^{(f)}$ establishes that
$k\in(i\gamma)\left(\chi^{(f)}\right)^{-1}$. Then by Theorem 5.9(iii), we
deduce that $\pi_{X_{i}}(g)\varphi^{-1}\subseteq\pi_{X_{k}}(f)$. Thus (ii)
holds.

Conversely, assume that the given conditions hold. Note from (i) that
$\left(\chi^{(f)},\gamma\right)\in\mathcal{L}$. For each $i\in I$, by (ii) we
fix $j_{i}\in\left(i\chi^{(f)}\right)\gamma^{-1}$ and
$k_{i}\in(i\gamma)\left(\chi^{(f)}\right)^{-1}$ such that
$\pi_{X_{i}}(f)\varphi\subseteq\pi_{X_{j_{i}}}(g)$ and
$\pi_{X_{i}}(g)\varphi^{-1}\subseteq\pi_{X_{k_{i}}}(f)$. Define
$\alpha,\beta\in T(I)$ given by $i\alpha=j_{i}$ and $i\beta=k_{i}$. Now we
show that $\chi^{(f)}=\alpha\gamma$. For this, let $i\in I$. Since
$j_{i}\in\left(i\chi^{(f)}\right)\gamma^{-1}$, we obtain
$i\chi^{(f)}=j_{i}\gamma=(i\alpha)\gamma=i(\alpha\gamma)$, which yields
$\chi^{(f)}=\alpha\gamma$. We can similarly prove that
$\gamma=\beta\chi^{(f)}$, since
$k_{i}\in(i\gamma)\left(\chi^{(f)}\right)^{-1}$. By (i), we have
$(\gamma,\chi^{(g)})\in\mathcal{R}$ in $T(I)$. Thus Theorem 5.9(i) holds. By
the definition of $\alpha$ and $\beta$, we also deduce that
$\pi_{X_{i}}(f)\varphi\subseteq\pi_{X_{i\alpha}}(g)$ for all $i\in I$, and
$\pi_{X_{i}}(g)\varphi^{-1}\subseteq\pi_{X_{i\beta}}(f)$ for all $i\in I$.
Thus each condition of Theorem 5.9 holds. Hence, since
$T_{T(I)}(X,\mathcal{P})=T(X,\mathcal{P})$, we conclude from Theorem 5.9 that
$(f,g)\in\mathcal{D}$ in $T(X,\mathcal{P})$. ∎

For every $x\in X$, denote by $\bar{x}$ the block of $\mathcal{P}$ that
contains $x$. Thus $\bar{x}=X_{i}$ for some $i\in I$.

We require the following result to characterize the relation $\mathcal{J}$ on
$T_{\mathbb{S}(I)}(X,\mathcal{P})$.

######  Lemma 5.11.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $J_{f}\leq J_{g}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if there exist
$\alpha,\beta\in\mathbb{S}(I)$ and a mapping $\varphi\colon Xg\to X$ such
that:

  1. (i)

$X_{i}f\subseteq(X_{i\alpha}g)\varphi$ for all $i\in I$;

  2. (ii)

$(X_{i}\cap Xg)\varphi\subseteq X_{i\beta}$ for all $i\in I\chi^{(g)}$.

###### Proof.

Assume that $J_{f}\leq J_{g}$ in $T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then
$f=h_{1}gh_{2}$ for some $h_{1},h_{2}\in T_{\mathbb{S}(I)}(X,\mathcal{P})$.
This gives
$\chi^{(f)}=\chi^{(h_{1}gh_{2})}=\chi^{(h_{1})}\chi^{(g)}\chi^{(h_{2})}$ by
Lemma 3.1. Write $\chi^{(h_{1})}=\alpha$ and $\chi^{(h_{2})}=\beta$.

If $Xg\setminus X(h_{1}g)\neq\varnothing$, then choose $y\in\bar{x}\cap
X(h_{1}g)$ for each $x\in Xg\setminus X(h_{1}g)$ with $\bar{x}\cap
X(h_{1}g)\neq\varnothing$; choose $z\in X_{i\beta}$ for each $x\in Xg\setminus
X(h_{1}g)$ with $\bar{x}\cap X(h_{1}g)=\varnothing$, where $\bar{x}=X_{i}$ for
some $i\in I$. Define a mapping $\varphi\colon Xg\to X$ by:

$\displaystyle x\varphi=\begin{cases}xh_{2}&\text{if $x\in X(h_{1}g)$};\\\
yh_{2}&\text{if $x\in Xg\setminus X(h_{1}g)$ and $\bar{x}\cap
X(h_{1}g)\neq\varnothing$};\\\ z&\text{if $x\in Xg\setminus X(h_{1}g)$ and
$\bar{x}\cap X(h_{1}g)=\varnothing$}.\end{cases}$

To prove (i), let $i\in I$. Note that $X_{i}(h_{1}g)\subseteq X(h_{1}g)$, and
so $x\varphi=xh_{2}$ for all $x\in X_{i}(h_{1}g)$ by the definition of
$\varphi$. Therefore, since $f=h_{1}gh_{2}$, we obtain

$X_{i}f=X_{i}(h_{1}gh_{2})=(X_{i}(h_{1}g))h_{2}=(X_{i}(h_{1}g))\varphi=((X_{i}h_{1})g)\varphi\subseteq(X_{i\alpha}g)\varphi.$

Thus $\rm(i)$ holds. To prove (ii), let $i\in I\chi^{(g)}$ and $x\in X_{i}\cap
Xg$. Then there are two cases:

Case 1: Suppose $x\in X_{i}\cap X(h_{1}g)$. Then $x\in X(h_{1}g)$. Therefore,
since $\chi^{(h_{2})}=\beta$, we obtain $x\varphi=xh_{2}\in X_{i\beta}$.

Case 2: Suppose $x\in Xg\setminus X(h_{1}g)$. Then $x\varphi=yh_{2}\in
X_{i\beta}$ or $x\varphi=z\in X_{i\beta}\in X_{i\beta}$.

Thus, in either case, we have $x\varphi\in X_{i\beta}$. Therefore $(X_{i}\cap
Xg)\varphi\subseteq X_{i\beta}$, and thus $\rm(ii)$ holds.

Conversely, assume that the given conditions hold. To prove $J_{f}\leq J_{g}$
in $T_{\mathbb{S}(I)}(X,\mathcal{P})$, we will construct $h_{1},h_{2}\in
T_{\mathbb{S}(I)}(X,\mathcal{P})$ such that $f=h_{1}gh_{2}$. To define
$h_{1}\in T(X)$, let $i\in I$. For each $x\in X_{i}$, by $\rm(i)$ we fix $y\in
X_{i\alpha}$ such that $xf=y(g\varphi)$. Define $h_{1}\in T(X)$ by $xh_{1}=y$
whenever $x\in X_{i}$. It is easy to verify that $\chi^{(h_{1})}=\alpha$, and
so $h_{1}\in T_{\mathbb{S}(I)}(X,\mathcal{P})$.

Next for each $i\in I$, we fix $x_{i}\in X_{i}$. Define $h_{2}\in T(X)$ by:

$\displaystyle xh_{2}=\begin{cases}x\varphi&\text{if $x\in X_{i}\cap Xg$,
where $i\in I\chi^{(g)}$};\\\ x_{i\beta}&\text{if $x\in X_{i}\setminus Xg$,
where $i\in I\chi^{(g)}$};\\\ x_{j\beta}&\text{if $x\in X_{j}$, where $j\in
I\setminus I\chi^{(g)}$}.\end{cases}$

By (ii) and the definition of $h_{2}$, it is easy to see that
$\chi^{(h_{2})}=\beta$, and so $h_{2}\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. To
prove $f=h_{1}gh_{2}$, let $x\in X$. Then $x\in X_{i}$ for some $i\in I$. Note
that $xh_{1}=y$ and $xf=y(g\varphi)$. Therefore
$x(h_{1}gh_{2})=(xh_{1})gh_{2}=(yg)h_{2}=(yg)\varphi=xf$, which yields
$f=h_{1}gh_{2}$. Hence $J_{f}\leq J_{g}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$. ∎

######  Theorem 5.12.

Let $f,g\in T_{\mathbb{S}(I)}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{J}$ in
$T_{\mathbb{S}(I)}(X,\mathcal{P})$ if and only if there exist
$\alpha,\beta,\gamma,\delta\in\mathbb{S}(I)$ and mappings $\varphi\colon Xg\to
X$, $\psi\colon Xf\to X$ such that:

  1. (i)

$X_{i}f\subseteq(X_{i\alpha}g)\varphi$ and $X_{i}g\subseteq(X_{i\gamma}f)\psi$
for all $i\in I$;

  2. (ii)

$(X_{i}\cap Xg)\varphi\subseteq X_{i\beta}$ for all $i\in I\chi^{(g)}$, and
$(X_{j}\cap Xf)\psi\subseteq X_{j\delta}$ for all $j\in I\chi^{(f)}$.

###### Proof.

It follows directly from Lemma 5.11. ∎

Let $Y$ and $Z$ be nonempty subsets of $X$, and let $\varphi\colon Y\to Z$ be
a mapping. We say that $\varphi$ is $E$-preserving if for all $x,y\in Y$,
$(x,y)\in E$ implies that $(x\varphi,y\varphi)\in E$ (cf. [17, p. 112]).
Equivalently, the mapping $\varphi$ is $E$-preserving if and only if for each
$X_{i}\in\mathcal{P}$, there exists $X_{j}\in\mathcal{P}$ such that
$(X_{i}\cap Y)\varphi\subseteq X_{j}\cap Z$.

In 2005, Pei [17, Theorem 3.8] determined the relation $\mathcal{J}$ on
$T(X,\mathcal{P})$ for a specific case. After that in 2007, Pei et al. [18,
Theorem 3.8] gave a complete description of the relation $\mathcal{J}$ on
$T(X,\mathcal{P})$. Sangkhanan and Sanwong [22, Corollary 3.16] also obtained
the same description of the relation $\mathcal{J}$ on $T(X,\mathcal{P})$.
Using Theorem 5.12, we find a more concrete characterization of the relation
$\mathcal{J}$ on $T(X,\mathcal{P})$ as follows.

######  Proposition 5.13.

Let $f,g\in T(X,\mathcal{P})$. Then $(f,g)\in\mathcal{J}$ in
$T(X,\mathcal{P})$ if and only if there exist $E$-preserving mappings
$\varphi\colon Xg\to X$ and $\psi\colon Xf\to X$ such that for each $i\in I$,
there exist $j,k\in I$ such that $X_{i}f\subseteq(X_{j}g)\varphi$ and
$X_{i}g\subseteq(X_{k}f)\psi$.

###### Proof.

Assume that $(f,g)\in\mathcal{J}$ in $T(X,\mathcal{P})$. By Theorem 5.12,
since $T(X,\mathcal{P})=T_{T(I)}(X,\mathcal{P})$, there exist
$\alpha,\beta,\gamma,\delta\in T(I)$ and mappings $\varphi\colon Xg\to X$,
$\psi\colon Xf\to X$ such that each condition of Theorem 5.12 hold. By Theorem
5.12(ii), it is clear that both mappings $\varphi$ and $\psi$ are
$E$-preserving. To prove the remaining parts, let $i\in I$. Write $i\alpha=j$
and $i\beta=k$. By Theorem 5.12(i), we then deduce that
$X_{i}f\subseteq(X_{j}g)\varphi$ and $X_{i}g\subseteq(X_{k}f)\psi$.

Conversely, assume that the given conditions hold. For each $i\in I$, by
hypothesis we choose $j_{i}\in I$ such that
$X_{i}f\subseteq(X_{j_{i}}g)\varphi$. Define $\alpha\in T(I)$ by
$i\alpha=j_{i}$. Then by hypothesis, we have
$X_{i}f\subseteq(X_{i\alpha}g)\varphi$ for all $i\in I$.

Note by hypothesis that the mapping $\varphi$ is $E$-preserving. Therefore for
each $i\in I\chi^{(g)}$, there exists unique $i^{\prime}\in I$ such that
$(X_{i}\cap Xg)\varphi\subseteq X_{i^{\prime}}$. Define $\beta\in T(I)$ by:

$\displaystyle i\beta=\begin{cases}i^{\prime}&\text{if $i\in I\chi^{(g)}$};\\\
i&\text{otherwise}.\end{cases}$

It is clear from the definition of $\beta$ that $(X_{i}\cap
Xg)\varphi\subseteq X_{i\beta}$ for all $i\in I\chi^{(g)}$. To prove
$\chi^{(f)}=\alpha\chi^{(g)}\beta$, let $i\in I$. Write $i\alpha=j$ and
$j\chi^{(g)}=k$. Since $k\in I\chi^{(g)}$, we get $k\beta=k^{\prime}$. It
follows that $(X_{k}\cap Xg)\varphi\subseteq X_{k^{\prime}}$ by the definition
of $\beta$. Since $j\chi^{(g)}=k$, we obtain

$X_{i}f\subseteq(X_{i\alpha}g)\varphi=(X_{j}g)\varphi\subseteq(X_{k}\cap
Xg)\varphi\subseteq X_{k^{\prime}},$

which gives $i\chi^{(f)}=k^{\prime}$. Then
$i(\alpha\chi^{(g)}\beta)=(i\alpha)\chi^{(g)}\beta=\left(j\chi^{(g)}\right)\beta=k\beta=k^{\prime}=i\chi^{(f)}$,
and therefore $\chi^{(f)}=\alpha\chi^{(g)}\beta$.

Again by hypothesis, for each $i\in I$ we choose $k_{i}\in I$ such that
$X_{i}g\subseteq(X_{k_{i}}f)\psi$. Define $\gamma\in T(I)$ by $i\gamma=k_{i}$.
Then by hypothesis, we have $X_{i}g\subseteq(X_{i\gamma}f)\psi$ for all $i\in
I$.

Finally by hypothesis, the mapping $\psi$ is $E$-preserving. Therefore for
each $j\in I\chi^{(f)}$, there exists unique $j^{\prime}\in I$ such that
$(X_{j}\cap Xf)\psi\subseteq X_{j^{\prime}}$. Define $\delta\in T(I)$ by:

$\displaystyle j\delta=\begin{cases}j^{\prime}&\text{if $j\in
I\chi^{(f)}$};\\\ j&\text{otherwise}.\end{cases}$

It is clear from the definition of $\delta$ that $(X_{i}\cap Xf)\psi\subseteq
X_{i\delta}$ for all $i\in I\chi^{(f)}$. To prove
$\chi^{(g)}=\gamma\chi^{(f)}\delta$, let $i\in I$. Write $i\gamma=j$ and
$j\chi^{(f)}=k$. Since $k\in I\chi^{(f)}$, we get $k\delta=k^{\prime}$. It
follows that $(X_{k}\cap Xf)\psi\subseteq X_{k^{\prime}}$ by the definition of
$\delta$. Since $j\chi^{(f)}=k$, we obtain

$X_{i}g\subseteq(X_{i\gamma}f)\psi=(X_{j}f)\psi\subseteq(X_{k}\cap
Xf)\psi\subseteq X_{k^{\prime}},$

which gives $i\chi^{(g)}=k^{\prime}$. Then
$i\left(\gamma\chi^{(f)}\delta\right)=(i\gamma)\chi^{(f)}\delta=\left(j\chi^{(f)}\right)\delta=k\delta=k^{\prime}=i\chi^{(g)}$,
and therefore $\chi^{(g)}=\gamma\chi^{(f)}\delta$.

Thus, since $T_{T(I)}(X,\mathcal{P})=T(X,\mathcal{P})$, we conclude from
Theorem 5.12 that $(f,g)\in\mathcal{J}$ in $T(X,\mathcal{P})$. ∎

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