# Green’s relations and unit-regularity for semigroup of transformations whose
characters are bijective

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

Let $X$ be a nonempty set and $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a
partition of $X$. Denote by $T(X,\mathcal{P})$ the semigroup of all
transformations of $X$ that preserve $\mathcal{P}$. In this paper, we study
the semigroup $\mathcal{B}(X,\mathcal{P})$ of all transformations $f\in
T(X,\mathcal{P})$ such that $\chi^{(f)}\in\operatorname{Sym}(I)$, where
$\operatorname{Sym}(I)$ is the symmetric group on $I$ and $\chi^{(f)}\colon
I\to I$ is the character (map) of $f$ defined by $i\chi^{(f)}=j$ whenever
$X_{i}f\subseteq X_{j}$. We describe unit-regular elements in
$\mathcal{B}(X,\mathcal{P})$, and determine when $\mathcal{B}(X,\mathcal{P})$
is a unit-regular semigroup. We alternatively prove that
$\mathcal{B}(X,\mathcal{P})$ is a regular semigroup. We describe Green’s
relations on $\mathcal{B}(X,\mathcal{P})$, and prove that
$\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$ when $\mathcal{P}$
is finite. We also give a necessary and sufficient condition for
$\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$. We end the paper
with a conjecture.

###### Key words and phrases:

Transformation semigroups, Symmetric groups, Regular semigroups, Unit-regular
semigroups, Green’s relations, Set partitions.

###### 2010 Mathematics Subject Classification:

20M17, 20M20, 20B30.

##  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 $X$ corresponding to the partition $\mathcal{P}$.
Denote by $T(X)$ (resp. $\operatorname{Sym}(X)$) the full transformation
semigroup (resp. symmetric group) on $X$. The semigroup $T(X)$ and its
subsemigroups play a vital role in semigroup theory, since every semigroup can
be embedded in some $T(Z)$ (cf. [13, Theorem 1.1.2]). This famous result is
analogous to Cayley’s theorem for groups, which states that every group can be
embedded in some $\operatorname{Sym}(X)$.

We say that any transformation $f\colon X\to X$ _preserves_ the partition
$\mathcal{P}$ if for every $X_{i}\in\mathcal{P}$, there exists
$X_{j}\in\mathcal{P}$ such that $X_{i}f\subseteq X_{j}$. In 1994, Pei [14]
first studied the subsemigroup $T(X,\mathcal{P})$ of $T(X)$ consisting of all
transformations that preserve the partition $\mathcal{P}$. Using symbols,

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

In that paper, Pei [14, Theorem 2.8] proved that $T(X,\mathcal{P})$ is exactly
the semigroup of all continuous selfmaps on $X$ with respect to the topology
having $\mathcal{P}$ as a basis. The semigroup $T(X,\mathcal{P})$ and its
subsemigroups have been extensively studied by a number of authors (see [1, 2,
10, 11, 15, 16, 17, 18, 21, 23] for some references).

In what follows, the letter $I$ denotes the index set of the partition
$\mathcal{P}$. Let $f\in T(X,\mathcal{P})$. The _character (map)_ of $f$ is
the selfmap $\chi^{(f)}\colon I\to I$ defined by $i\chi^{(f)}=j$ whenever
$X_{i}f\subseteq X_{j}$. For finite $X$, the character $\chi^{(f)}$ with the
notation $\bar{f}$ has been studied by Araújo et al. [1], Dolinka and East
[8], and Dolinka et al. [9]. For arbitrary $X$, the character $\chi^{(f)}$ was
first considered by Purisang and Rakbud [19, p. 220]. The character
$\chi^{(f)}$ of $f$ has also been received attention (see [20, 21, 22, 23]).

Using the notion of character $\chi^{(f)}$, Purisang and Rakbud [19]
introduced the following subsemigroup of $T(X,\mathcal{P})$:

$\mathcal{B}(X,\mathcal{P})=\\{f\in
T(X,\mathcal{P})\colon\chi^{(f)}\in\operatorname{Sym}(I)\\}.$

In that paper, the authors [19, Theorem 3.5(1)] proved that
$\mathcal{B}(X,\mathcal{P})$ is a regular semigroup. The semigroup
$\mathcal{B}(X,\mathcal{P})$ generalizes both $T(X)$ and
$\operatorname{Sym}(X)$ in the sense that $\mathcal{B}(X,\mathcal{P})=T(X)$ if
$|\mathcal{P}|=1$, and $\mathcal{B}(X,\mathcal{P})=\operatorname{Sym}(X)$ if
$\mathcal{P}$ consists of singletons. When $\mathcal{P}$ is finite, the
authors [21, Corollary 3.5] proved that
$\mathcal{B}(X,\mathcal{P})=\Sigma(X,\mathcal{P})=T_{E^{*}}(X)$, where

$\displaystyle\Sigma(X,\mathcal{P})$ $\displaystyle=\\{f\in
T(X,\mathcal{P})\colon Xf\cap X_{i}\neq\varnothing\;\forall
X_{i}\in\mathcal{P}\\},$ $\displaystyle T_{E^{*}}(X)$ $\displaystyle=\\{f\in
T(X)\colon\forall x,y\in X,\;(x,y)\in E\iff(xf,yf)\in E\\}.$

For arbitrary $X$, both the semigroups $\Sigma(X,\mathcal{P})$ and
$T_{E^{*}}(X)$ have been studied (see [1, 2, 21, 22, 23] and [5, 6, 7, 24],
respectively). In particular, the authors [1, 2] considered the semigroup
$\Sigma(X,\mathcal{P})$ for finite $X$ to calculate the rank of the finite
semigroup $T(X,\mathcal{P})$.

This paper is motivated by various results on $T(X,\mathcal{P})$ and its two
subsemigroups $\Sigma(X,\mathcal{P})$ and $T_{E^{*}}(X)$. The rest of the
paper is organized as follows. In the next section, we define concepts,
introduce notation, and recall some results needed in this paper. In Section
3, we give an alternative proof of Theorem 3.5(1) in [19], which ascertain
that the semigroup $\mathcal{B}(X,\mathcal{P})$ is regular. We next describe
unit-regular elements in $\mathcal{B}(X,\mathcal{P})$, and determine when
$\mathcal{B}(X,\mathcal{P})$ is a unit-regular semigroup. In Section 4, we
describe Green’s relations on $\mathcal{B}(X,\mathcal{P})$. We also give a
necessary and sufficient condition for $\mathcal{D}=\mathcal{J}$ on
$\mathcal{B}(X,\mathcal{P})$.

##  2\. Preliminaries and Notation

Let $X$ be a nonempty set. The cardinality of $X$ is denoted by $|X|$, and the
identity map on $X$ is denoted by $\operatorname{id}_{X}$. For any sets $A$
and $B$, let $A\setminus B$ denote the set $\\{x\in A\colon x\notin B\\}$. A
_partition_ of $X$ is a collection of pairwise disjoint nonempty subsets,
called _blocks_ , whose union is $X$. A _trivial_ partition is a partition
that has only singleton blocks or a single block. A partition is _uniform_ if
all its blocks have the same cardinality. A _transversal_ of an equivalence
relation $\rho$ on $X$ is a subset of $X$ that contains exactly one element
from each $\rho$-class. The set of all positive integers is denoted by
$\mathbb{N}$. For $n\in\mathbb{N}$, let $[n]$ denote the set
$\\{1,\ldots,n\\}$.

We compose mappings from left to right and denote their composition by
juxtaposition. Let $f\colon X\to Y$ be a mapping. We denote by $xf$ the image
of an element $x\in X$ under $f$. For any $A\subseteq X$ (resp. $B\subseteq
Y$), we denote by $Af$ (resp. $Bf^{-1}$) the set $\\{af\colon a\in A\\}$
(resp. $\\{x\in X\colon xf\in B\\}$). Furthermore, if $B=\\{b\\}$, then we
write $bf^{-1}$ instead of $\\{b\\}f^{-1}$. Let $D(f)$ denote the set
$Y\setminus Xf$ and $\textnormal{d}(f)=|D(f)|$. The _kernel_ of $f$, denoted
by $\ker(f)$, is an equivalence 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 the symbol $T_{f}$ denote any transversal of $\ker(f)$. Note
that $|X\setminus T_{f}|$ is independent of the choice of transversal of
$\ker(f)$ (cf. [12, p. 1356]). Set $\textnormal{c}(f)=|X\setminus T_{f}|$. Let
$g\colon X\to X$ be a mapping. For any nonempty subset $A$ of the domain of
$g$, the _restriction_ of $g$ to $A$ is the mapping $g_{\upharpoonright
A}\colon A\to X$ defined by $x(g_{\upharpoonright A})=xg$ for all $x\in A$.
Moreover if $B$ is a subset of the codomain of $g$ such that $Ag\subseteq B$,
then we reserve the same notation $g_{\upharpoonright A}$ for the mapping from
$A$ to $B$ that assigns $xg$ to each $x\in A$.

Let $S$ be a semigroup and $a\in S$. We say that $a$ is regular in $S$ if
there exists $b\in S$ such that $aba=a$. The set of all regular elements in
$S$ is denoted by $\operatorname{reg}(S)$. If $\operatorname{reg}(S)=S$, then
$S$ is a _regular semigroup_. In addition, let $S$ contains the identity. Then
the set of all unit elements of $S$ is denoted by $U(S)$. We say that $a$ is
unit-regular in $S$ if there exists $u\in U(S)$ 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 $S$ is a _unit-regular semigroup_. Note that
$U(T(X))=\operatorname{Sym}(X)$, and denote by $S(X,\mathcal{P})$ the set of
all unit elements of $T(X,\mathcal{P})$. It is evident that
$S(X,\mathcal{P})=T(X,\mathcal{P})\cap\operatorname{Sym}(X)$ and
$U(\mathcal{B}(X,\mathcal{P}))=S(X,\mathcal{P})$.

Let $S$ be a semigroup and $a,b\in S$. The Green’s relations
$\mathcal{L},\mathcal{R},\mathcal{H},\mathcal{D}$, and $\mathcal{J}$ on $S$
defined as follows: $(a,b)\in\mathcal{L}$ if $S^{1}a=S^{1}b$,
$(a,b)\in\mathcal{R}$ if $aS^{1}=bS^{1}$,
$\mathcal{H}=\mathcal{L}\cap\mathcal{R}$,
$\mathcal{D}=\mathcal{L}\circ\mathcal{R}$, and $(a,b)\in\mathcal{J}$ if
$S^{1}aS^{1}=S^{1}bS^{1}$, where $S^{1}$ is the semigroup $S$ with an identity
adjoined (if necessary). If $\mathcal{K}$ is any Green’s relation on $S$, then
the equivalence class of $a$ with respect to $\mathcal{K}$ is denoted by
$K_{a}$. Since $\mathcal{L},\mathcal{R}$, and $\mathcal{J}$ are defined in
terms of ideals, which are partially ordered by inclusion, we have induced
partial orders on the sets of the equivalence classes of
$\mathcal{L},\mathcal{R}$, and $\mathcal{J}$ [13, p. 47, (2.1.3)]: $L_{a}\leq
L_{b}$ if $S^{1}a\subseteq S^{1}b$, $R_{a}\leq R_{b}$ if $aS^{1}\subseteq
bS^{1}$, and $J_{a}\leq J_{b}$ if $S^{1}aS^{1}\subseteq S^{1}bS^{1}$.

We refer the reader to [13] for any undefined concepts, notation, and results
of semigroup theory. We end this section by stating a list of preliminary
results.

######  Lemma 2.1.

[23, 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 $\ker(f)$.

######  Lemma 2.2.

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

######  Lemma 2.3.

[21, Lemma 5.2] Let $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a partition of
$X$ and $f\in T(X)$. Then $f\in T(X,\mathcal{P})$ if and only if there exists
a unique family $B(f,I):=\\{f_{\upharpoonright_{X_{i}}}\colon i\in I\\}$ such
that the codomain of each $f_{\upharpoonright_{X_{i}}}$ is a block of
$\mathcal{P}$.

######  Theorem 2.4.

[21, Theorem 5.8] Let $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a partition of
$X$ and $f\in T(X,\mathcal{P})$. Then $f\in S(X,\mathcal{P})$ if and only if

  1. (i)

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

  2. (ii)

$\chi^{(f)}\in\operatorname{Sym}(I)$.

##  3\. Unit-regularity for $\mathcal{B}(X,\mathcal{P})$

In this section, we describe unit-regular elements in
$\mathcal{B}(X,\mathcal{P})$ and then give a necessary and sufficient
condition for $\mathcal{B}(X,\mathcal{P})$ to be unit-regular. We begin by
giving an alternative proof of the following proposition which was first
appeared in [19, Theorem 3.5(1)].

######  Proposition 3.1.

The semigroup $\mathcal{B}(X,\mathcal{P})$ is regular.

###### Proof.

Let $f\in\mathcal{B}(X,\mathcal{P})$, and write $(\chi^{(f)})^{-1}=\alpha$.
Fix $x^{\prime}\in xf^{-1}$ for each $x\in Xf$, and also fix $x_{i}\in X_{i}$
for each $i\in I$. Define $g\in T(X)$ as follows: Given $x\in X$, there exists
$i\in I$ such that $x\in X_{i}$; we let

$\displaystyle xg=\begin{cases}x^{\prime}&\text{if $x\in X_{i}\cap Xf$;}\\\
x_{i\alpha}&\text{if $x\in X_{i}\setminus Xf$}.\end{cases}$

It is easy to see that $g\in T(X,\mathcal{P})$ and $\chi^{(g)}=\alpha$, so
$g\in\mathcal{B}(X,\mathcal{P})$. To prove $fgf=f$, let $x\in X$ and write
$xf=y$. Then $x\in X_{i}$ for some $i\in I$. Therefore, since $y^{\prime}\in
yf^{-1}$, we obtain $x(fgf)=(xf)gf=(yg)f=y^{\prime}f=y=xf$, which yields
$fgf=f$. Hence $f\in\operatorname{reg}(\mathcal{B}(X,\mathcal{P}))$ as
required.

∎

The following theorem describes unit-regular elements in
$\mathcal{B}(X,\mathcal{P})$.

######  Theorem 3.2.

Let $f\in\mathcal{B}(X,\mathcal{P})$. Then
$f\in\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$ if and only if
$\textnormal{c}(f_{\upharpoonright_{X_{i}}})=\textnormal{d}(f_{\upharpoonright_{X_{i}}})$
for all $i\in I$.

###### Proof.

Suppose that $f\in\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$. Then there
exists $g\in U(\mathcal{B}(X,\mathcal{P}))$ such that $fgf=f$. This gives
$\chi^{(f)}\chi^{(g)}\chi^{(f)}=\chi^{(f)}$ by Lemma 2.2. Therefore
$\chi^{(g)}=(\chi^{(f)})^{-1}$, since
$\chi^{(f)},\chi^{(g)}\in\operatorname{Sym}(I)$. Write $\chi^{(f)}=\alpha$.

To prove the desired result, let $i\in I$ and write $i\alpha=j$. Then
$j\chi^{(g)}=i$. Therefore, since $fgf=f$, it is easy to see that
$f_{\upharpoonright_{X_{i}}}g_{\upharpoonright_{X_{j}}}f_{\upharpoonright_{X_{i}}}=f_{\upharpoonright_{X_{i}}}$,
where $f_{\upharpoonright_{X_{i}}}\colon X_{i}\to X_{j}$ (resp.
$g_{\upharpoonright_{X_{j}}}\colon X_{j}\to X_{i}$) is a map of $B(f,I)$
(resp. $B(g,I)$). It follows from Lemma 2.1 that
$T_{f_{\upharpoonright_{X_{i}}}}:=X_{i}(f_{\upharpoonright_{X_{i}}}g_{\upharpoonright_{X_{j}}})$
is a transversal of $\ker(f_{\upharpoonright_{X_{i}}})$. Since $g\in
S(X,\mathcal{P})$, it follows from Theorem 2.4 that every map of $B(g,I)$ is
bijective. Therefore

$(X_{j}\setminus
X_{i}f_{\upharpoonright_{X_{i}}})g_{\upharpoonright_{X_{j}}}=X_{j}g_{\upharpoonright_{X_{j}}}\setminus
X_{i}(f_{\upharpoonright_{X_{i}}}g_{\upharpoonright_{X_{j}}})=X_{i}\setminus
T_{f_{\upharpoonright_{X_{i}}}},$

which yields $|X_{j}\setminus
X_{i}f_{\upharpoonright_{X_{i}}}|=|X_{i}\setminus
T_{f_{\upharpoonright_{X_{i}}}}|$. Hence
$\textnormal{c}(f_{\upharpoonright_{X_{i}}})=|X_{i}\setminus
T_{f_{\upharpoonright_{X_{i}}}}|=|X_{j}\setminus
X_{i}f_{\upharpoonright_{X_{i}}}|=\textnormal{d}(f_{\upharpoonright_{X_{i}}})$.

Conversely, suppose that
$\textnormal{c}(f_{\upharpoonright_{X_{i}}})=\textnormal{d}(f_{\upharpoonright_{X_{i}}})$
for all $i\in I$. To prove
$f\in\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$, we shall construct
$g\in U(\mathcal{B}(X,\mathcal{P}))$ such that $fgf=f$. For this, let $i\in I$
and write $i(\chi^{(f)})^{-1}=j$. Let $T_{f_{\upharpoonright_{X_{j}}}}$ be a
transversal of $\ker(f_{\upharpoonright_{X_{j}}})$, where
$f_{\upharpoonright_{X_{j}}}\colon X_{j}\to X_{i}$ is a map of $B(f,I)$. It is
easy to see that the map $\varphi_{i}\colon
X_{j}f_{\upharpoonright_{X_{j}}}\to T_{f_{\upharpoonright_{X_{j}}}}$ defined
by $x\varphi_{i}=x^{\prime}$ whenever $x(f_{\upharpoonright_{X_{j}}})^{-1}\cap
T_{f_{\upharpoonright_{X_{j}}}}=\\{x^{\prime}\\}$ is bijective. Since
$|X_{i}\setminus X_{j}f_{\upharpoonright_{X_{j}}}|=|X_{j}\setminus
T_{f_{\upharpoonright_{X_{j}}}}|$ by hypothesis, there exists a bijection
$\psi_{i}\colon X_{i}\setminus X_{j}f_{\upharpoonright_{X_{j}}}\to
X_{j}\setminus T_{f_{\upharpoonright_{X_{j}}}}$. Define $h_{i}\colon X_{i}\to
X_{j}$ by

$\displaystyle xh_{i}=\begin{cases}x\varphi_{i}&\text{if $x\in
X_{j}f_{\upharpoonright_{X_{j}}}$;}\\\ x\psi_{i}&\text{if $x\in X_{i}\setminus
X_{j}f_{\upharpoonright_{X_{j}}}$}.\end{cases}$

Clearly $h_{i}$ is bijective. We now prove that the map $g\in T(X)$ defined by
$xg=xh_{i}$ whenever $x\in X_{i}$ for some $i\in I$, is a unit element of
$\mathcal{B}(X,\mathcal{P})$. For this, let $i\in I$ and write
$i(\chi^{(f)})^{-1}=j$. Then, since $h_{i}\colon X_{i}\to X_{j}$ is bijective,
we get $X_{i}g=X_{i}h_{i}=X_{j}$. Therefore $g\in T(X,\mathcal{P})$ and
$\chi^{(g)}=(\chi^{(f)})^{-1}$. Since $\chi^{(g)}\in\operatorname{Sym}(I)$ and
every map $g_{\upharpoonright_{X_{i}}}=h_{i}$ of $B(g,I)$ is bijective, it
follows from Theorem 2.4 that $g\in U(\mathcal{B}(X,\mathcal{P}))$. Finally,
we prove that $fgf=f$. For this, let $x\in X$. Then $x\in X_{i}$ for some
$i\in I$. Write $xf_{\upharpoonright_{X_{i}}}=y$ and $i\chi^{(f)}=j$.
Therefore, since $y^{\prime}\in y(f_{\upharpoonright_{X_{i}}})^{-1}\cap
T_{f_{\upharpoonright_{X_{i}}}}$, we obtain
$x(fgf)=(xf_{\upharpoonright_{X_{i}}})g_{\upharpoonright_{X_{j}}}f_{\upharpoonright_{X_{i}}}=(yg_{\upharpoonright_{X_{j}}})f_{\upharpoonright_{X_{i}}}=y^{\prime}f_{\upharpoonright_{X_{i}}}=y=xf_{\upharpoonright_{X_{i}}}=xf$,
which yields $fgf=f$. Hence
$f\in\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$ as required. ∎

It is well-known that $T(X)$ is unit-regular if and only if $X$ is finite (cf.
[4, Proposition 5]). In the following theorem, we describe the unit-regularity
for $\mathcal{B}(X,\mathcal{P})$.

######  Theorem 3.3.

The semigroup $\mathcal{B}(X,\mathcal{P})$ is unit-regular if and only if

  1. (i)

$\mathcal{P}$ is a uniform partition;

  2. (ii)

the cardinality of every block of $\mathcal{P}$ is finite.

###### Proof.

Suppose that $\mathcal{B}(X,\mathcal{P})$ is a unit-regular semigroup.

  1. (i)

Suppose to the contrary that there are distinct $j,k\in I$ such that
$|X_{j}|\neq|X_{k}|$. Consider the following two possible cases:

Case 1: Suppose $|X_{j}|<|X_{k}|$. Then there exists a mapping $\varphi\colon
X_{j}\to X_{k}$ that is injection, but not surjection. Therefore
$\textnormal{c}(\varphi)=0$ and $\textnormal{d}(\varphi)\geq 1$.

Case 2: Suppose $|X_{j}|>|X_{k}|$. Then there exists a mapping $\varphi\colon
X_{j}\to X_{k}$ that is surjection, but not injection. Therefore
$\textnormal{c}(\varphi)\geq 1$ and $\textnormal{d}(\varphi)=0$.

In either case, there is a mapping $\varphi\colon X_{j}\to X_{k}$ such that
$\textnormal{c}(\varphi)\neq\textnormal{d}(\varphi)$. We now choose $x_{i}\in
X_{i}$ for each $i\in I\setminus\\{k\\}$, and consider
$\alpha\in\operatorname{Sym}(I)$ such that $j\alpha=k$. Define $f\in T(X)$ by

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

Clearly $f\in T(X,\mathcal{P})$ and $\chi^{(f)}=\alpha$, so
$f\in\mathcal{B}(X,\mathcal{P})$. However, since
$\textnormal{c}(\varphi)\neq\textnormal{d}(\varphi)$, we see that
$\textnormal{c}(f_{\upharpoonright_{X_{j}}})=\textnormal{c}(\varphi)\neq\textnormal{d}(\varphi)=\textnormal{d}(f_{\upharpoonright_{X_{j}}})$.
Therefore $f\notin\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$ by Theorem
3.2, a contradiction. Hence $\mathcal{P}$ is uniform.

  2. (ii)

Suppose to the contrary that $X_{i}$ is infinite for some $i\in I$. Then there
exists a map $\psi\colon X_{i}\to X_{i}$ that is surjection, but not
injection. Therefore $\textnormal{c}(\psi)\geq 1$ and
$\textnormal{d}(\psi)=0$, so $\textnormal{c}(\psi)\neq\textnormal{d}(\psi)$.
Define $g\in T(X)$ by

$\displaystyle xg=\begin{cases}x\psi&\text{if $x\in X_{i}$;}\\\
x&\text{otherwise.}\end{cases}$

Clearly $g\in T(X,\mathcal{P})$ and $\chi^{(g)}=\operatorname{id}_{I}$, so
$g\in\mathcal{B}(X,\mathcal{P})$. However, since
$\textnormal{c}(\psi)\neq\textnormal{d}(\psi)$, we see that
$\textnormal{c}(f_{\upharpoonright_{X_{i}}})=\textnormal{c}(\psi)\neq\textnormal{d}(\psi)=\textnormal{d}(f_{\upharpoonright_{X_{i}}})$.
Therefore $f\notin\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$ by Theorem
3.2, a contradiction. Hence the cardinality of every block of $\mathcal{P}$ is
finite.

Conversely, suppose that the given conditions hold. Let
$f\in\mathcal{B}(X,\mathcal{P})$ and let $i\in I$. Consider the map
$f_{\upharpoonright_{X_{i}}}\colon X_{i}\to X_{i\chi^{(f)}}$ of $B(f,I)$. By
condition (i), we have $|X_{i}|=|X_{i\chi^{(f)}}|$. Note from condition (ii)
that $X_{i}$ is finite, so
$\textnormal{c}(f_{\upharpoonright_{X_{i}}})=\textnormal{d}(f_{\upharpoonright_{X_{i}}})$.
Thus
$\textnormal{c}(f_{\upharpoonright_{X_{i}}})=\textnormal{d}(f_{\upharpoonright_{X_{i}}})$
for all $i\in I$, where $f_{\upharpoonright_{X_{i}}}\in B(f,I)$. Hence
$f\in\operatorname{ureg}(\mathcal{B}(X,\mathcal{P}))$ by Theorem 3.2 as
required. ∎

##  4\. Green’s Relations on $\mathcal{B}(X,\mathcal{P})$

In this section, we give a complete description of Green’s relations on
$\mathcal{B}(X,\mathcal{P})$. We begin with a lemma that is useful in
describing the relation $\mathcal{L}$ on $\mathcal{B}(X,\mathcal{P})$.

######  Lemma 4.1.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $L_{f}\leq L_{g}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if there exists
$\alpha\in\operatorname{Sym}(I)$ such that $X_{i}f\subseteq X_{i\alpha}g$ for
all $i\in I$.

###### Proof.

Suppose that $L_{f}\leq L_{g}$ in $\mathcal{B}(X,\mathcal{P})$. Then there
exists $h\in\mathcal{B}(X,\mathcal{P})$ such that $f=hg$. This gives
$\chi^{(f)}=\chi^{(h)}\chi^{(g)}$ by Lemma 2.2. Write $\chi^{(h)}=\alpha$.
Then, since $h\in\mathcal{B}(X,\mathcal{P})$, we have
$\alpha\in\operatorname{Sym}(I)$. To prove the desired result, let $i\in I$.
Note that $X_{i}h\subseteq X_{i\alpha}$, and therefore
$X_{i}f=X_{i}(hg)=(X_{i}h)g\subseteq X_{i\alpha}g$ as required.

Conversely, suppose that the given condition holds. To prove $L_{f}\leq L_{g}$
in $\mathcal{B}(X,\mathcal{P})$, we shall construct
$h\in\mathcal{B}(X,\mathcal{P})$ such that $f=hg$. For this, let $i\in I$. By
hypothesis, we have $X_{i}f\subseteq X_{i\alpha}g$. Therefore for each $x\in
X_{i}$, fix $x^{\prime}\in X_{i\alpha}$ such that $xf=x^{\prime}g$. Define
$h\in T(X)$ as follows: Given $x\in X$, there exists $i\in I$ such that $x\in
X_{i}$; we let $xh=x^{\prime}$. Clearly $X_{i}h\subseteq X_{i\alpha}$, since
$x^{\prime}\in X_{i\alpha}$. Therefore $f\in T(X,\mathcal{P})$ and
$\chi^{(h)}=\alpha$, so $h\in\mathcal{B}(X,\mathcal{P})$. Finally, we prove
that $f=hg$. For this, let $x\in X$. Then $x\in X_{i}$ for some $i\in I$.
Therefore, since $xf=x^{\prime}g$, we obtain $x(hg)=(xh)g=x^{\prime}g=xf$,
which yields $f=hg$. Thus $L_{f}\leq L_{g}$ in $\mathcal{B}(X,\mathcal{P})$. ∎

In the following theorem, we describe the relation $\mathcal{L}$ on
$\mathcal{B}(X,\mathcal{P})$.

######  Theorem 4.2.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{L}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if there exists
$\alpha\in\operatorname{Sym}(I)$ such that $X_{i}f=X_{i\alpha}g$ for all $i\in
I$.

###### Proof.

Suppose that $(f,g)\in\mathcal{L}$ in $\mathcal{B}(X,\mathcal{P})$. Then there
exist $h,h^{\prime}\in\mathcal{B}(X,\mathcal{P})$ such that $f=hg$ and
$g=h^{\prime}f$. It follows from Lemma 2.2 that
$\chi^{(f)}=\chi^{(h)}\chi^{(g)}$ and
$\chi^{(g)}=\chi^{(h^{\prime})}\chi^{(f)}$. Note that
$\chi^{(f)},\chi^{(g)},\chi^{(h)},\chi^{(h^{\prime})}\in\operatorname{Sym}(I)$,
write $\chi^{(h)}=\alpha$ and $\chi^{(h^{\prime})}=\beta$. Clearly
$\beta=\alpha^{-1}$. To prove the desired result, let $i\in I$. Note that
$X_{i}h\subseteq X_{i\alpha}$, and so $X_{i}f=X_{i}(hg)=(X_{i}h)g\subseteq
X_{i\alpha}g$. For the reverse inclusion, note that $\beta=\alpha^{-1}$ and
$X_{i\alpha}h^{\prime}\subseteq X_{(i\alpha)\beta}=X_{i}$. Therefore, since
$g=h^{\prime}f$, we obtain
$X_{i\alpha}g=X_{i\alpha}(h^{\prime}f)=(X_{i\alpha}h^{\prime})f\subseteq
X_{i}f$ as required.

Conversely, suppose that there exists $\alpha\in\operatorname{Sym}(I)$ such
that $X_{i}f=X_{i\alpha}g$ for all $i\in I$. Then $L_{f}\leq L_{g}$ in
$\mathcal{B}(X,\mathcal{P})$ by Lemma 4.1. Next, note that
$\alpha^{-1}\in\operatorname{Sym}(I)$ and write $\alpha^{-1}=\beta$. Then by
hypothesis, we get $X_{i}g=X_{(i\beta)\alpha}g=X_{i\beta}f$ for all $i\in I$.
Therefore $L_{g}\leq L_{f}$ in $\mathcal{B}(X,\mathcal{P})$ by Lemma 4.1. Thus
$(f,g)\in\mathcal{L}$ in $\mathcal{B}(X,\mathcal{P})$. ∎

To describe the relation $\mathcal{R}$ on $\mathcal{B}(X,\mathcal{P})$, we
need the following terminology and Lemma 4.3.

Let $f,g\in T(X)$. We say that $\pi(f)$ _refines_ $\pi(g)$, denoted by
$\pi(f)\preceq\pi(g)$, if $\ker(f)\subseteq\ker(g)$. Equivalently, we have
$\pi(f)\preceq\pi(g)$ if for every $P\in\pi(f)$, there exists $Q\in\pi(g)$
such that $P\subseteq Q$. We shall write $\pi(f)=\pi(g)$ if
$\pi(f)\preceq\pi(g)$ and $\pi(g)\preceq\pi(f)$.

######  Lemma 4.3.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $R_{f}\leq R_{g}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if $\pi(g)\preceq\pi(f)$.

###### Proof.

Suppose that $R_{f}\leq R_{g}$ in $\mathcal{B}(X,\mathcal{P})$. Then there
exists $h\in\mathcal{B}(X,\mathcal{P})$ such that $f=gh$. It follows from [3,
Lemma 2.6] that $\pi(g)\preceq\pi(f)$, since $f,g,h\in T(X)$.

Conversely, suppose that $\pi(g)\preceq\pi(f)$. To prove $R_{f}\leq R_{g}$ in
$\mathcal{B}(X,\mathcal{P})$, we shall construct
$h\in\mathcal{B}(X,\mathcal{P})$ such that $f=gh$. For this, fix $y_{i}\in
X_{i}$ for each $i\in I$; fix $x^{\prime}\in xg^{-1}$ for each $x\in Xg$. Note
that $\chi^{(f)},\chi^{(g)}\in\operatorname{Sym}(I)$, and write
$(\chi^{(g)})^{-1}\chi^{(f)}=\alpha$. Define $h\in T(X)$ as follows: Given
$x\in X$, there exists $i\in I$ such that $x\in X_{i}$; we let

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

Clearly $h$ is well-defined, since $\pi(g)\preceq\pi(f)$. We also observe that
$X_{i}h\subseteq X_{i\alpha}$, since $x^{\prime}\in X_{i(\chi^{(g)})^{-1}}$
whenever $x\in X_{i}\cap Xg$, and $(\chi^{(g)})^{-1}\chi^{(f)}=\alpha$.
Therefore $h\in T(X,\mathcal{P})$ and $\chi^{(h)}=\alpha$, so
$h\in\mathcal{B}(X,\mathcal{P})$. Finally, we prove that $f=gh$. For this, 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 $f=gh$. Thus $R_{f}\leq R_{g}$ in $\mathcal{B}(X,\mathcal{P})$. ∎

######  Theorem 4.4.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{R}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if $\pi(f)=\pi(g)$.

###### Proof.

This proof follows directly from Lemma 4.3. ∎

To describe the relation $\mathcal{D}$ on $\mathcal{B}(X,\mathcal{P})$, we
need the following terminology.

For any $A\subseteq X$ and any $f\in T(X)$, let
$\pi_{A}(f)=\\{M\in\pi(f)\colon M\cap A\neq\varnothing\\}$. We shall write
$\pi(f)$ instead of $\pi_{X}(f)$ if $A=X$.

######  Theorem 4.5.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{D}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if there exists
$\alpha\in\operatorname{Sym}(I)$ such that $|X_{i}f|=|X_{i\alpha}g|$ for all
$i\in I$.

###### Proof.

Suppose that $(f,g)\in\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$. Then there
exists $h\in\mathcal{B}(X,\mathcal{P})$ such that $(f,h)\in\mathcal{L}$ in
$\mathcal{B}(X,\mathcal{P})$ and $(h,g)\in\mathcal{R}$ in
$\mathcal{B}(X,\mathcal{P})$. Since $(f,h)\in\mathcal{L}$, it follows from
Theorem 4.2 that there exists $\alpha\in\operatorname{Sym}(I)$ such that
$X_{k}f=X_{k\alpha}h$ for all $k\in I$. To prove the desired result, let $i\in
I$ and write $i\alpha=j$. Then $X_{i}f=X_{j}h$, so $|X_{i}f|=|X_{j}h|$. Now
since $(h,g)\in\mathcal{R}$ in $\mathcal{B}(X,\mathcal{P})$, it follows from
Theorem 4.4 that $\pi(h)=\pi(g)$. This yields $\pi_{X_{j}}(h)=\pi_{X_{j}}(g)$,
so $|\pi_{X_{j}}(h)|=|\pi_{X_{j}}(g)|$. Notice that
$|\pi_{X_{j}}(h)|=|X_{j}h|$ and $|\pi_{X_{j}}(g)|=|X_{j}g|$, so
$|X_{j}h|=|X_{j}g|$. Thus $|X_{i}f|=|X_{j}g|$ as required.

Conversely, suppose that there exists $\alpha\in\operatorname{Sym}(I)$ such
that $|X_{i}f|=|X_{i\alpha}g|$ for all $i\in I$. To prove
$(f,g)\in\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$, we shall construct
$h\in\mathcal{B}(X,\mathcal{P})$ such that $(f,h)\in\mathcal{L}$ in
$\mathcal{B}(X,\mathcal{P})$ and $(h,g)\in\mathcal{R}$ in
$\mathcal{B}(X,\mathcal{P})$. For this, let $i\in I$ and write
$\beta=\alpha^{-1}$. Then $|X_{i}g|=|X_{(i\beta)\alpha}g|=|X_{i\beta}f|$ by
hypothesis, so there is a bijection $\varphi_{i}\colon X_{i}g\to X_{i\beta}f$.
Now define $h\in T(X)$ as follows: Given $x\in X$, there exists $i\in I$ such
that $x\in X_{i}$; we let $xh=x(g\varphi_{i})$. Clearly $h\in
T(X,\mathcal{P})$ and $\chi^{(h)}=\beta\chi^{(f)}$, since
$X_{i}h=X_{i}(g\varphi_{i})=X_{i\beta}f\subseteq X_{i(\beta\chi^{(f)})}$ for
all $i\in I$. Therefore $h\in\mathcal{B}(X,\mathcal{P})$. Recall that
$\beta\in\operatorname{Sym}(I)$ and $X_{i}h=X_{i\beta}f$ for all $i\in I$.
Thus $(f,h)\in\mathcal{L}$ in $\mathcal{B}(X,\mathcal{P})$ by Theorem 4.2 and
the fact that $\mathcal{L}$ is a symmetric relation. It is easy to see from
definition of $h$ that $\pi_{X_{i}}(h)=\pi_{X_{i}}(g)$ for all $i\in I$, which
yields $\pi(h)=\pi(g)$. Therefore $(h,g)\in\mathcal{R}$ in
$\mathcal{B}(X,\mathcal{P})$ by Theorem 4.4. Thus, since
$\mathcal{D}=\mathcal{L}\circ\mathcal{R}$, we conclude that
$(f,g)\in\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$. ∎

To describe the relation $\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$, we
need the following lemma.

######  Lemma 4.6.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $J_{f}\leq J_{g}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if there exists
$\alpha\in\operatorname{Sym}(I)$ such that $|X_{i}f|\leq|X_{i\alpha}g|$ for
all $i\in I$.

###### Proof.

Suppose that $J_{f}\leq J_{g}$ in $\mathcal{B}(X,\mathcal{P})$. Then there
exist $h,h_{1}\in\mathcal{B}(X,\mathcal{P})$ such that $f=hgh_{1}$. Set
$\chi^{(h)}=\alpha$. Clearly $\alpha\in\operatorname{Sym}(I)$, since
$h\in\mathcal{B}(X,\mathcal{P})$. To prove the desired result, let $i\in I$.
Notice that $X_{i}h\subseteq X_{i\alpha}$ and therefore
$|X_{i}f|=|(X_{i}h)gh_{1}|\leq|(X_{i\alpha}g)h_{1}|\leq|X_{i\alpha}g|$ as
required.

Conversely, suppose that there exists $\alpha\in\operatorname{Sym}(I)$ such
that $|X_{i}f|\leq|X_{i\alpha}g|$ for all $i\in I$. Then for every $i\in I$,
there exists an injection $\varphi_{i}\colon X_{i}f\to X_{i\alpha}g$. Now
choose $y^{\prime}\in(y\varphi_{i})g^{-1}$ for each $y\in X_{i}f$. To prove
the desired result, we shall construct $h,h_{1}\in\mathcal{B}(X,\mathcal{P})$
such that $f=hgh_{1}$.

We first define $h\in T(X)$ as follows: Given $x\in X$, there exists $i\in I$
such that $x\in X_{i}$; write $xf=y$, and we let $xh=y^{\prime}$. To prove
$h\in\mathcal{B}(X,\mathcal{P})$, let $i\in I$ and $x\in X_{i}$. Set $xf=z$.
Then $xh=z^{\prime}$ by definition of $h$, where
$z^{\prime}\in(z\varphi_{i})g^{-1}$. Since $z\in X_{i}f$, it follows that
$z\varphi_{i}\in X_{i\alpha}g$ by definition of $\varphi_{i}$. This gives
$(z\varphi_{i})g^{-1}\subseteq X_{i\alpha}$, since
$\chi^{(g)}\in\operatorname{Sym}(I)$. Therefore, since
$z^{\prime}\in(z\varphi_{i})g^{-1}$, we get $xh=z^{\prime}\in X_{i\alpha}$.
Thus $h\in T(X,\mathcal{P})$ and $\chi^{(h)}=\alpha$, so
$h\in\mathcal{B}(X,\mathcal{P})$.

To define $h_{1}\in\mathcal{B}(X,\mathcal{P})$, we observe for every $i\in I$
that $X_{i}(f\varphi_{i})=X_{i}(hg)$, since
$(xf)^{\prime}\in((xf)\varphi_{i})g^{-1}$ and
$x(hg)=(xh)g=(xf)^{\prime}g=(xf)\varphi_{i}$ for all $x\in X_{i}$. Also, fix
$x_{i}\in X_{i}$ for each $i\in I$. Let
$\beta=(\alpha\chi^{(g)})^{-1}\chi^{(f)}$. Clearly
$\beta\in\operatorname{Sym}(I)$ and $\chi^{(f)}=\alpha\chi^{(g)}\beta$, since
$\chi^{(f)},\chi^{(g)},\alpha\in\operatorname{Sym}(I)$. Define $h_{1}\in T(X)$
as follows: Given $x\in X$, there exists $i\in I$ such that $x\in X_{i}$;
write $i(\alpha\chi^{(g)})^{-1}=j$, and we let

$\displaystyle xh_{1}=\begin{cases}y&\text{if $x\in X_{i}\cap X(hg)$ and
$x\varphi_{j}^{-1}=\\{y\\}$;}\\\ x_{i\beta}&\text{if $x\in X_{i}\setminus
X(hg)$}.\end{cases}$

To prove $h_{1}\in\mathcal{B}(X,\mathcal{P})$, let $i\in I$ and $x\in X_{i}$.
Consider the following two possible cases:

Case 1: Suppose $x\in X_{i}\cap X(hg)$. Write $i(\alpha\chi^{(g)})^{-1}=k$.
Then, since $\chi^{(h)}=\alpha$, we see that $X_{k}(hg)\subseteq
X_{k\alpha}g\subseteq X_{k(\alpha\chi^{(g)})}=X_{i}$. As
$\alpha,\chi^{(g)}\in\operatorname{Sym}(I)$, it follows that
$X_{k}(hg)=X_{i}\cap X(hg)$. Therefore $x\in X_{k}(hg)\subseteq X_{k\alpha}g$.
Note that $\varphi_{k}\colon X_{k}f\to X_{k\alpha}g$ is injective and
$X_{k}(f\varphi_{k})=X_{k}(hg)$. It follows that $x\varphi_{k}^{-1}=\\{z\\}$
for some $z\in X_{k}f$. Therefore, since $i(\alpha\chi^{(g)})^{-1}=k$ and
$\beta=(\alpha\chi^{(g)})^{-1}\chi^{(f)}$, we obtain
$k\chi^{(f)}=(i(\alpha\chi^{(g)})^{-1})\chi^{(f)}=i\beta$, which yields
$X_{k}f\subseteq X_{i\beta}$. Thus, since $z\in X_{k}f$, we get $xh_{1}=z\in
X_{i\beta}$.

Case 2: Suppose $x\in X_{i}\setminus X(hg)$. Then $xh_{1}=x_{i\beta}$ by
definition of $h_{1}$, so $xh_{1}\in X_{i\beta}$.

In either case, we have $xh_{1}\in X_{i\beta}$. Hence $X_{i}h_{1}\subseteq
X_{i\beta}$. This gives $h_{1}\in T(X,\mathcal{P})$ and
$\chi^{(h_{1})}=\beta$, so $h_{1}\in\mathcal{B}(X,\mathcal{P})$.

Finally, we prove that $f=hgh_{1}$. For this, let $x\in X$. Then $x\in X_{i}$
for some $i\in I$. Recall that $\varphi_{i}\colon X_{i}f\to X_{i\alpha}g$ is
an injection. Therefore, since
$xh=(xf)^{\prime}\in((xf)\varphi_{i})g^{-1}\subseteq X_{i\alpha}$, we obtain
$x(hgh_{1})=(xh)(gh_{1})=(xf)^{\prime}(gh_{1})=(x(f\varphi_{i}))h_{1}=xf$,
which yields $f=hgh_{1}$ as required. ∎

######  Theorem 4.7.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{J}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if there exist
$\alpha,\beta\in\operatorname{Sym}(I)$ such that $|X_{i}f|\leq|X_{i\alpha}g|$
and $|X_{i}g|\leq|X_{i\beta}f|$ for all $i\in I$.

###### Proof.

This proof follows directly from Lemma 4.6. ∎

In the following proposition, we prove that the relations $\mathcal{D}$ and
$\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$ coincide when $\mathcal{P}$ is
finite.

######  Proposition 4.8.

If $\mathcal{P}$ is a finite partition of $X$, then $\mathcal{D}=\mathcal{J}$
on $\mathcal{B}(X,\mathcal{P})$.

###### Proof.

Let $\mathcal{P}=\\{X_{i}\colon i\in[m]\\}$, where $m\in\mathbb{N}$. In
general, we have $\mathcal{D}\subseteq\mathcal{J}$. For the reverse inclusion,
let $(f,g)\in\mathcal{J}$ in $\mathcal{B}(X,\mathcal{P})$. Then by Theorem
4.7, there exist $\alpha,\beta\in\operatorname{Sym}([m])$ such that
$|X_{i}f|\leq|X_{i\alpha}g|$ and $|X_{i}g|\leq|X_{i\beta}f|$ for all
$i\in[m]$. Therefore for every $i\in[m]$, we obtain

$\displaystyle\qquad\qquad|X_{i}f|\leq|X_{i\alpha}g|\leq|X_{(i\alpha)\beta}f|=|X_{i(\alpha\beta)}f|\qquad\qquad\cdots\text{(1).}$

Note that $m$ is finite and $\alpha\beta\in\operatorname{Sym}([m])$. Therefore
there exists $k\in\mathbb{N}$ such that $j(\alpha\beta)^{k}=j$ for all
$j\in[m]$.

In view of Theorem 4.5, in order to show that $(f,g)\in\mathcal{D}$ in
$\mathcal{B}(X,\mathcal{P})$, it suffices to prove that
$|X_{i}f|=|X_{i\alpha}g|$ for all $i\in[m]$. For this, let $i\in[m]$. Then
from inequality (1), we have $|X_{i}f|\leq|X_{i(\alpha\beta)}f|$. Therefore,
since $(\alpha\beta)^{k}$ is identity map on $[m]$, we obtain by using
inequality (1) that

$|X_{i}f|\leq|X_{i(\alpha\beta)}f|\leq|X_{i(\alpha\beta)^{2}}f|\leq\cdots\leq|X_{i(\alpha\beta)^{k}}f|=|X_{i}f|$

whence $|X_{i}f|=|X_{i(\alpha\beta)}f|$. Note from inequality (1) that
$|X_{i}f|\leq|X_{i\alpha}g|\leq|X_{i(\alpha\beta)}f|$, so
$|X_{i}f|=|X_{i\alpha}g|$. Hence, since $\alpha\in\operatorname{Sym}([m])$, we
conclude from Theorem 4.5 that $(f,g)\in\mathcal{D}$ in
$\mathcal{B}(X,\mathcal{P})$ as required. ∎

We now introduce the following terminology.

For any $f\in\mathcal{B}(X,\mathcal{P})$, let $\mathcal{P}_{f}=\\{X_{i}f\colon
i\in I\\}$. It is clear that $\mathcal{P}_{f}$ is a partition of the range set
$Xf$ of $f$, since $\chi^{(f)}\in\operatorname{Sym}(I)$.

Note that if $\mathcal{P}$ is an infinite partition consisting of only
singleton blocks, then $\mathcal{B}(X,\mathcal{P})=\operatorname{Sym}(X)$, so
$\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$. However, for an
infinite partition $\mathcal{P}$, it is not always true that
$\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$, as shown in the
following example.

######  Example 4.9.

Let $X=\mathbb{N}$. Consider the partition $\mathcal{P}=\\{X_{i}\colon i\in
I\\}$ of $X$, where $I=\\{1,3,5k,5k+1\colon k\in\mathbb{N}\\}$;
$X_{1}=\\{1,2\\}$, $X_{3}=\\{3,4\\}$, and for each $k\in\mathbb{N}$,
$X_{5k}=\\{5k\\}$, $X_{5k+1}=\\{5k+1,5k+2,5k+3,5k+4\\}$. Take mappings
$f,g\colon X\to X$ defined by $xg=x$ and

$\displaystyle xf=\begin{cases}1&\text{if $x\in X_{1}$;}\\\
x&\text{otherwise}.\end{cases}$

Clearly $f,g\in T(X,\mathcal{P})$ and
$\chi^{(f)}=\chi^{(g)}=\operatorname{id}_{I}$, so
$f,g\in\mathcal{B}(X,\mathcal{P})$. Notice that $\mathcal{P}_{f}$ has exactly
one block of cardinality $2$, while $\mathcal{P}_{g}$ has two blocks of
cardinality $2$. Therefore there does not exist any
$\alpha\in\operatorname{Sym}(I)$ such that $|X_{i}f|=|X_{i\alpha}g|$ for all
$i\in I$. Hence $(f,g)\notin\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$ by
Theorem 4.5.

Finally, we prove that $(f,g)\in\mathcal{J}$ in $\mathcal{B}(X,\mathcal{P})$.
Consider $\alpha,\beta\in T(I)$ defined by $i\alpha=i$ and

$\displaystyle i\beta=\begin{cases}6&\text{if $i=1$;}\\\ 3&\text{if $i=3$;}\\\
1&\text{if $i=5$;}\\\ 5(k-1)&\text{if $i=5k$, where $k\geq 2$;}\\\
5(k+1)+1&\text{if $i=5k+1$, where $k\geq 1$}.\end{cases}$

Observe that $\alpha,\beta\in\operatorname{Sym}(I)$. For every $i\in I$, it is
also routine to verify that $|X_{i}f|\leq|X_{i\alpha}g|$ and
$|X_{i}g|\leq|X_{i\beta}f|$. Therefore $(f,g)\in\mathcal{J}$ in
$\mathcal{B}(X,\mathcal{P})$ by Theorem 4.7, and thus
$(f,g)\in\mathcal{J}\setminus\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$.

In the following proposition, we give a sufficient condition under which the
relations $\mathcal{D}$ and $\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$ do
not coincide.

######  Proposition 4.10.

Let $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a partition of $X$. If
$J=\\{i\in I\colon|X_{i}|\geq 3\\}$ is an infinite subset of $I$, then
$\mathcal{D}\neq\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$.

###### Proof.

Since $J$ is infinite, there exists a proper subset $K$ of $J$ such that
$|K|=|J\setminus K|=|J|$. Let $k\in K$, and fix distinct
$y_{k},y_{k}^{\prime}\in X_{k}$. Also, fix distinct $u_{i},v_{i},w_{i}\in
X_{i}$ for each $i\in K\setminus\\{k\\}$; fix $z_{i}\in X_{i}$ for each $i\in
I\setminus K$. Consider mappings $f,g\colon X\to X$ defined by

$\displaystyle xf=\begin{cases}y_{k}&\text{if $x=y_{k}$;}\\\
y_{k}^{\prime}&\text{if $x\in X_{k}\setminus\\{y_{k}\\}$;}\\\ u_{i}&\text{if
$x=u_{i}\in X_{i}$, where $i\in K\setminus\\{k\\}$;}\\\ v_{i}&\text{if
$x=v_{i}\in X_{i}$, where $i\in K\setminus\\{k\\}$;}\\\ w_{i}&\text{if $x\in
X_{i}\setminus\\{u_{i},v_{i}\\}$, where $i\in K\setminus\\{k\\}$;}\\\
z_{i}&\text{if $x\in X_{i}$, where $i\in I\setminus K$}\end{cases}$

and

$\displaystyle xg=\begin{cases}y_{k}&\text{if $x\in X_{k}$;}\\\ u_{i}&\text{if
$x=u_{i}\in X_{i}$, where $i\in K\setminus\\{k\\}$;}\\\ v_{i}&\text{if
$x=v_{i}\in X_{i}$, where $i\in K\setminus\\{k\\}$;}\\\ w_{i}&\text{if $x\in
X_{i}\setminus\\{u_{i},v_{i}\\}$, where $i\in K\setminus\\{k\\}$;}\\\
z_{i}&\text{if $x\in X_{i}$, where $i\in I\setminus K$}.\end{cases}$

For every $i\in I$, we observe that $X_{i}f\subseteq X_{i}$ and
$X_{i}g\subseteq X_{i}$. Therefore $f,g\in T(X,\mathcal{P})$ and
$\chi^{(f)}=\operatorname{id}_{I}=\chi^{(g)}$. It follows that
$f,g\in\mathcal{B}(X,\mathcal{P})$. Notice that $|X_{k}f|=2$, while
$|X_{i}g|\neq 2$ for all $i\in I$. Therefore there does not exist
$\alpha\in\operatorname{Sym}(I)$ such that $|X_{k}f|=|X_{k\alpha}g|$. Hence
$(f,g)\notin\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$ by Theorem 4.5.

Finally, we prove that $(f,g)\in\mathcal{J}$ in $\mathcal{B}(X,\mathcal{P})$.
For this, we first fix $k^{\prime}(\neq k)\in K$. Since $K$ is infinite, we
see that $|K\setminus\\{k\\}|=|K\setminus\\{k,k^{\prime}\\}|$. Therefore there
is a bijection $\varphi\colon K\setminus\\{k\\}\to
K\setminus\\{k,k^{\prime}\\}$. Recall that $J$ is an infinite subset of $I$.
Since $|J\setminus K|=|J|$ and $J\setminus K\subseteq I\setminus K$, it
follows that $I\setminus K$ is infinite. Therefore $|I\setminus
K|=|(I\setminus K)\cup\\{k\\}|$, so there is a bijection $\psi\colon
I\setminus K\to(I\setminus K)\cup\\{k\\}$. Consider mappings $\alpha,\beta\in
T(I)$ defined by $i\beta=i$ and

$\displaystyle i\alpha=\begin{cases}k^{\prime}&\text{if $i=k$;}\\\
i\varphi&\text{if $i\in K\setminus\\{k\\}$;}\\\ i\psi&\text{$i\in I\setminus
K$}.\end{cases}$

Notice that $\alpha,\beta\in\operatorname{Sym}(I)$. For every $i\in I$, it is
routine to verify that $|X_{i}f|\leq|X_{i\alpha}g|$ and
$|X_{i}g|\leq|X_{i\beta}f|$. Therefore $(f,g)\in\mathcal{J}$ in
$\mathcal{B}(X,\mathcal{P})$ by Theorem 4.7. Thus $\mathcal{D}\neq\mathcal{J}$
on $\mathcal{B}(X,\mathcal{P})$. ∎

Now we establish an alternative characterization of the relation $\mathcal{D}$
on $\mathcal{B}(X,\mathcal{P})$ in Proposition 4.11. Before that, we introduce
the following terminology.

Let $f\in\mathcal{B}(X,\mathcal{P})$. Recall that
$\mathcal{P}_{f}=\\{X_{i}f\colon i\in I\\}$ is a partition of the range set
$Xf$ of $f$. For any cardinal $\lambda$, let $I_{\lambda}^{f}=\\{i\in
I\colon|X_{i}f|=\lambda\\}$ and $n_{\lambda}^{f}=|I_{\lambda}^{f}|$. Then it
is clear that the collection
$\\{I_{\lambda}^{f}\colon\lambda\leq|X|~{}\text{such
that}~{}I_{\lambda}^{f}\neq\varnothing\\}$ is a partition of $I$.

######  Proposition 4.11.

Let $f,g\in\mathcal{B}(X,\mathcal{P})$. Then $(f,g)\in\mathcal{D}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if $n_{\lambda}^{f}=n_{\lambda}^{g}$
for all cardinals $\lambda$.

###### Proof.

Suppose that $(f,g)\in\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$. Then by
Theorem 4.5, there exists $\alpha\in\operatorname{Sym}(I)$ such that
$|X_{i}f|=|X_{i\alpha}g|$ for all $i\in I$. To prove the desired result, let
$\lambda$ be a cardinal. It is clear that $n_{\lambda}^{f}=0$ if and only if
$n_{\lambda}^{g}=0$, since $|X_{i}f|\neq\lambda$ for all $i\in I$ if and only
if $|X_{i}g|\neq\lambda$ for all $i\in I$. Assume that $n_{\lambda}^{f}\geq
1$, and let $i\in I_{\lambda}^{f}$. Then $|X_{i\alpha}g|=|X_{i}f|=\lambda$, so
$i\alpha\in I_{\lambda}^{g}$. Since $\alpha$ is injective, we thus get
$n_{\lambda}^{f}\leq n_{\lambda}^{g}$. Now, let $i\in I_{\lambda}^{g}$. Then
$|X_{i\alpha^{-1}}f|=|X_{(i\alpha^{-1})\alpha}g|=|X_{i}g|=\lambda$, so
$i\alpha^{-1}\in I_{\lambda}^{f}$. Since $\alpha^{-1}$ is injective, we thus
get $n_{\lambda}^{f}\geq n_{\lambda}^{g}$. Hence
$n_{\lambda}^{f}=n_{\lambda}^{g}$ as required.

Conversely, suppose that the given condition holds. In view of Theorem 4.5, in
order to prove $(f,g)\in\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$, it
suffices to construct $\alpha\in\operatorname{Sym}(I)$ such that
$|X_{i}f|=|X_{i\alpha}g|$ for all $i\in I$. For this, consider a cardinal
$\lambda$. Note by hypothesis that $I_{\lambda}^{f}=\varnothing$ if and only
if $I_{\lambda}^{g}=\varnothing$. If $I_{\lambda}^{f}\neq\varnothing$, then
$|I_{\lambda}^{f}|=|I_{\lambda}^{g}|$ by hypothesis. Therefore there exists a
bijection $\alpha_{\lambda}\colon I_{\lambda}^{f}\to I_{\lambda}^{g}$. Define
$\alpha\in T(I)$ by $i\alpha=i\alpha_{\lambda}$ whenever $i\in
I_{\lambda}^{f}$. Clearly $\alpha$ is well-defined, since $\lambda$ is
arbitrary. Also, since every $\alpha_{\lambda}$ is bijective and
$\displaystyle\bigcup_{\lambda}I_{\lambda}^{f}=I=\displaystyle\bigcup_{\lambda}I_{\lambda}^{g}$,
we see that $\alpha\in\operatorname{Sym}(I)$. It is also routine to verify
that $|X_{i}f|=|X_{i\alpha}g|$ for all $i\in I$. Hence $(f,g)\in\mathcal{D}$
by Theorem 4.5. ∎

In the following theorem, we give a sufficient condition for $D_{f}=J_{f}$ in
$\mathcal{B}(X,\mathcal{P})$.

######  Theorem 4.12.

Let $f\in\mathcal{B}(X,\mathcal{P})$. If there exists two consecutive
cardinals $\lambda_{1},\lambda_{2}$ such that $n_{\lambda}^{f}=0$ for all
cardinals $\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$, then $D_{f}=J_{f}$ in
$\mathcal{B}(X,\mathcal{P})$.

###### Proof.

Clearly $D_{f}\subseteq J_{f}$, since $\mathcal{D}\subseteq\mathcal{J}$. For
the reverse inclusion, let $g\in J_{f}$. Then by Theorem 4.2, there exist
$\alpha,\beta\in\operatorname{Sym}(I)$ such that for all $i\in I$,

$\displaystyle|X_{i}f|\leq|X_{i\alpha}g|\hskip
28.45274pt\cdots\text{(i)}\qquad\text{and}\qquad|X_{i}g|\leq|X_{i\beta}f|\hskip
28.45274pt\cdots\text{(ii)}$

First, we claim that $n_{\lambda}^{g}=0$ for all cardinals
$\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$. Suppose to the contrary that
there exists $\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$ such that
$n_{\lambda}^{g}\neq 0$. Then $|X_{i}g|=\lambda$ for some $i\in I$. Consider
the following two possible cases:

Case 1: Suppose $\lambda<\lambda_{1}$. By inequality (i), we obtain
$|X_{i\alpha^{-1}}f|\leq|X_{(i\alpha^{-1})\alpha}g|=|X_{i}g|=\lambda$. This
yields $|X_{i\alpha^{-1}}f|<\lambda_{1}$, a contradiction of our hypothesis.

Case 2: Suppose $\lambda>\lambda_{2}$. By inequality (ii), we obtain
$\lambda=|X_{i}g|\leq|X_{i\beta}f|$. This yields $|X_{i\beta}f|>\lambda_{2}$,
a contradiction of our hypothesis.

In either case, we get a contradiction. Hence $n_{\lambda}^{g}=0$ for all
cardinals $\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$, and thus
$n_{\lambda}^{f}=0=n_{\lambda}^{g}$ for all cardinals
$\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$.

Next, we prove that $n_{\lambda_{1}}^{f}=n_{\lambda_{1}}^{g}$. Consider the
following two possible cases:

Case 1: Suppose $I_{\lambda_{1}}^{f}=\varnothing$. Suppose to the contrary
that $I_{\lambda_{1}}^{g}\neq\varnothing$, and let $i\in I_{\lambda_{1}}^{g}$.
Then $|X_{i}g|=\lambda_{1}$, and so
$|X_{i\alpha^{-1}}f|\leq|X_{(i\alpha^{-1})\alpha}g|=|X_{i}g|=\lambda_{1}$ by
inequality (i). This yields $|X_{i\alpha^{-1}}f|=\lambda_{1}$ by hypothesis,
and so $i\alpha^{-1}\in I_{\lambda_{1}}^{f}$. This is a contradiction, because
$I_{\lambda_{1}}^{f}=\varnothing$. Hence $I_{\lambda_{1}}^{g}=\varnothing$,
and thus $n_{\lambda_{1}}^{f}=n_{\lambda_{1}}^{g}$.

Case 2: Suppose $I_{\lambda_{1}}^{f}\neq\varnothing$. Let $i\in
I_{\lambda_{1}}^{f}$. Then $|X_{i}f|=\lambda_{1}$, and so
$|X_{i\beta^{-1}}g|\leq|X_{(i\beta^{-1})\beta}f|=|X_{i}f|=\lambda_{1}$ by
inequality (ii). Since $n_{\lambda}^{g}=0$ for all cardinals
$\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$, it follows that
$|X_{i\beta^{-1}}g|=\lambda_{1}$ whence $i\beta^{-1}\in I_{\lambda_{1}}^{g}$.
Since $\beta^{-1}$ is injective, we thus get $n_{\lambda_{1}}^{f}\leq
n_{\lambda_{1}}^{g}$. Similarly, we can show that $i\alpha^{-1}\in
I_{\lambda_{1}}^{f}$ for every $i\in I_{\lambda_{1}}^{g}$, and further prove
that $n_{\lambda_{1}}^{g}\leq n_{\lambda_{1}}^{f}$ by using the fact that
$\alpha^{-1}$ is injective. Thus $n_{\lambda_{1}}^{f}=n_{\lambda_{1}}^{g}$.

In either case, we have $n_{\lambda_{1}}^{f}=n_{\lambda_{1}}^{g}$.

Next, we prove that $n_{\lambda_{2}}^{f}=n_{\lambda_{2}}^{g}$. Consider the
following two possible cases:

Case 1: Suppose $I_{\lambda_{2}}^{f}=\varnothing$. Suppose to the contrary
that $I_{\lambda_{2}}^{g}\neq\varnothing$, and let $i\in I_{\lambda_{2}}^{g}$.
Then $|X_{i}g|=\lambda_{2}$, and so $\lambda_{2}=|X_{i}g|\leq|X_{i\beta}f|$ by
inequality (ii). This yields $|X_{i\beta}f|=\lambda_{1}$ by hypothesis, and so
$i\beta\in I_{\lambda_{2}}^{f}$. This is a contradiction, because
$I_{\lambda_{2}}^{f}=\varnothing$. Hence $I_{\lambda_{2}}^{g}=\varnothing$,
and thus $n_{\lambda_{2}}^{f}=n_{\lambda_{2}}^{g}$.

Case 2: Suppose $I_{\lambda_{2}}^{f}\neq\varnothing$. Let $i\in
I_{\lambda_{2}}^{f}$. Then $|X_{i}f|=\lambda_{2}$, and so
$\lambda_{2}=|X_{i}f|\leq|X_{i\alpha}g|$ by inequality (i). Since
$n_{\lambda}^{g}=0$ for all cardinals
$\lambda\notin\\{\lambda_{1},\lambda_{2}\\}$, it follows that
$|X_{i\alpha}g|=\lambda_{2}$ whence $i\alpha\in I_{\lambda_{2}}^{g}$. Since
$\alpha$ is injective, we thus get $n_{\lambda_{2}}^{f}\leq
n_{\lambda_{2}}^{g}$. Similarly, we can show that $i\beta\in
I_{\lambda_{2}}^{f}$ for every $i\in I_{\lambda_{2}}^{g}$, and further prove
that $n_{\lambda_{2}}^{g}\leq n_{\lambda_{2}}^{f}$ by using the fact that
$\beta$ is injective. Thus $n_{\lambda_{2}}^{f}=n_{\lambda_{2}}^{g}$.

In either case, we have $n_{\lambda_{2}}^{f}=n_{\lambda_{2}}^{g}$. Thus, since
$n_{\lambda}^{f}=n_{\lambda}^{g}$ for all cardinals $\lambda$, we conclude
from Proposition 4.11 that $(f,g)\in\mathcal{D}$ in
$\mathcal{B}(X,\mathcal{P})$. Hence $g\in D_{f}$ as required. ∎

As an immediate consequence of Theorem 4.12, we get

######  Corollary 4.13.

Let $f\in\mathcal{B}(X,\mathcal{P})$. If $n_{\lambda}^{f}=0$ for all cardinals
$\lambda\geq 3$, then $D_{f}=J_{f}$ in $\mathcal{B}(X,\mathcal{P})$.

In the following theorem, we give a necessary and sufficient condition for
$\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$.

######  Theorem 4.14.

Let $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a partition of $X$. Then
$\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$ if and only if
$J=\\{i\in I\colon|X_{i}|\geq 3\\}$ is finite.

###### Proof.

Suppose that $\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$. If
$\mathcal{P}$ is finite, then there is nothing to prove. Assume that
$\mathcal{P}$ is infinite. Suppose to the contrary that $J=\\{i\in
I\colon|X_{i}|\geq 3\\}$ is infinite. Then $\mathcal{D}\neq\mathcal{J}$ on
$\mathcal{B}(X,\mathcal{P})$ by Proposition 4.10, a contradiction. Hence
$J=\\{i\in I\colon|X_{i}|\geq 3\\}$ is finite.

Conversely, suppose that $J=\\{i\in I\colon|X_{i}|\geq 3\\}$ is finite. If
$\mathcal{P}$ is finite, then $\mathcal{D}=\mathcal{J}$ on
$\mathcal{B}(X,\mathcal{P})$ by Proposition 4.8. Assume that $\mathcal{P}$ is
infinite. In general, we have $\mathcal{D}\subseteq\mathcal{J}$. For the
reverse inclusion, let $(f,g)\in\mathcal{J}$ in $\mathcal{B}(X,\mathcal{P})$.
Then by Theorem 4.7, there exist $\alpha,\beta\in\operatorname{Sym}(I)$ such
that for all $i\in I$,

$\displaystyle|X_{i}f|\leq|X_{i\alpha}g|\hskip
28.45274pt\cdots\text{(i)}\qquad\text{and}\qquad|X_{i}g|\leq|X_{i\beta}f|\hskip
28.45274pt\cdots\text{(ii)}$

In view of Proposition 4.11, in order to show that $(f,g)\in\mathcal{D}$ in
$\mathcal{B}(X,\mathcal{P})$, it suffices to prove that
$n_{\lambda}^{f}=n_{\lambda}^{g}$ for all cardinals $\lambda$. Consider the
following two possible cases:

Case 1: Suppose $n_{\lambda}^{f}=0$ for all $\lambda\geq 3$. Then
$(f,g)\in\mathcal{D}$ by Corollary 4.13.

Case 2: Suppose $n_{\lambda}^{f}\neq 0$ for some cardinal $\lambda\geq 3$. Let
$i\in I_{\lambda}^{f}$. Then $|X_{i}f|=\lambda$, and so $3\leq|X_{i\alpha}g|$
by inequality (i). Therefore $n_{\lambda}^{g}\neq 0$ for some cardinal
$\lambda\geq 3$.

Let $\lambda_{1},\ldots,\lambda_{k}$, where $\lambda_{1}>\cdots>\lambda_{k}$,
be the cardinalities of blocks in $\mathcal{P}_{f}$ of cardinalities at least
three; let $\mu_{1},\ldots,\mu_{t}$, where $\mu_{1}>\cdots>\mu_{t}$, be the
cardinalities of blocks in $\mathcal{P}_{g}$ of cardinalities at least three.
Note by hypothesis that both $k$ and $t$ are finite.

First, we prove that $\lambda_{1}=\mu_{1}$ and subsequently
$n_{\lambda_{1}}^{f}=n_{\mu_{1}}^{g}$. For this, let $i\in
I_{\lambda_{1}}^{f}$. Then by inequality (i), we have
$\lambda_{1}=|X_{i}f|\leq|X_{i\alpha}g|\leq\mu_{1}$. Now, let $i\in
I_{\mu_{1}}^{g}$. Then by inequality (ii), we have
$\mu_{1}=|X_{i}g|\leq|X_{i\beta}f|\leq\lambda_{1}$. Thus
$\lambda_{1}=\mu_{1}$. To prove $n_{\lambda_{1}}^{f}=n_{\mu_{1}}^{g}$, we
first observe for every $i\in I_{\lambda_{1}}^{f}$ that
$|X_{i\alpha}g|=\lambda_{1}$, whence $i\alpha\in I_{\lambda_{1}}^{g}$.
Therefore, since $\alpha$ is injective, we get $n_{\lambda_{1}}^{f}\leq
n_{\lambda_{1}}^{g}$. Similarly, by using the facts that $\beta$ is injective
and $i\beta\in I_{\lambda_{1}}^{f}$ for every $i\in I_{\lambda_{1}}^{g}$, we
get $n_{\lambda_{1}}^{g}\leq n_{\lambda_{1}}^{f}$. Thus
$n_{\lambda_{1}}^{f}=n_{\lambda_{1}}^{g}$. Moreover, recall by hypothesis that
both $I_{\lambda_{1}}^{f}$ and $I_{\lambda_{1}}^{g}$ are finite. Therefore,
since $\alpha,\beta$ are injective, both maps $\alpha_{\lambda_{1}}\colon
I_{\lambda_{1}}^{f}\to I_{\lambda_{1}}^{g}$ and $\beta_{\lambda_{1}}\colon
I_{\lambda_{1}}^{g}\to I_{\lambda_{1}}^{f}$ defined by
$i\alpha_{\lambda_{1}}=i\alpha$ and $i\beta_{\lambda_{1}}=i\beta$,
respectively, are bijective.

Next, we prove that $\lambda_{2}=\mu_{2}$ and subsequently
$n_{\lambda_{2}}^{f}=n_{\mu_{2}}^{g}$. For this, let $i\in
I_{\lambda_{2}}^{f}$. Recall that $\alpha_{\lambda_{1}}\colon
I_{\lambda_{1}}^{f}\to I_{\lambda_{1}}^{g}$ is bijective, and so
$i\alpha\notin I_{\lambda_{1}}^{g}$. Therefore $|X_{i\alpha}g|\leq\mu_{2}$,
which gives $\lambda_{2}=|X_{i}f|\leq|X_{i\alpha}g|\leq\mu_{2}$ by inequality
(i). Now, let $i\in I_{\mu_{2}}^{g}$. Recall that $\beta_{\lambda_{1}}\colon
I_{\lambda_{1}}^{g}\to I_{\lambda_{1}}^{f}$ is bijective, and so $i\beta\notin
I_{\lambda_{1}}^{f}$. Therefore $|X_{i\alpha}f|\leq\lambda_{2}$, which gives
$\mu_{2}=|X_{i}g|\leq|X_{i\beta}f|\leq\lambda_{2}$ by inequality (ii). Thus
$\lambda_{2}=\mu_{2}$. To prove $n_{\lambda_{2}}^{f}=n_{\mu_{2}}^{g}$, we
first observe for every $i\in I_{\lambda_{2}}^{f}$ that
$|X_{i\alpha}g|=\lambda_{2}$, whence $i\alpha\in I_{\lambda_{2}}^{g}$.
Therefore, since $\alpha$ is injective, we get $n_{\lambda_{2}}^{f}\leq
n_{\lambda_{2}}^{g}$. Similarly, by using the facts that $\beta$ is injective
and $i\beta\in I_{\lambda_{2}}^{f}$ for every $i\in I_{\lambda_{2}}^{g}$, we
get $n_{\lambda_{2}}^{g}\leq n_{\lambda_{2}}^{f}$. Thus
$n_{\lambda_{2}}^{f}=n_{\lambda_{2}}^{g}$. Moreover, recall by hypothesis that
both $I_{\lambda_{2}}^{f}$ and $I_{\lambda_{2}}^{g}$ are finite. Therefore,
since $\alpha,\beta$ are injective, both maps $\alpha_{\lambda_{2}}\colon
I_{\lambda_{2}}^{f}\to I_{\lambda_{2}}^{g}$ and $\beta_{\lambda_{2}}\colon
I_{\lambda_{2}}^{g}\to I_{\lambda_{2}}^{f}$ defined by
$i\alpha_{\lambda_{2}}=i\alpha$ and $i\beta_{\lambda_{2}}=i\beta$,
respectively, are bijective.

We can prove in similar way that $\lambda_{j}=\mu_{j}$ and
$n_{\lambda_{j}}^{f}=n_{\mu_{j}}^{g}$ for all $j=3,\ldots,\min\\{k,t\\}$.
Moreover, we observe for every $j=3,\ldots,\min\\{k,t\\}$ that both maps
$\alpha_{\lambda_{j}}\colon I_{\lambda_{j}}^{f}\to I_{\lambda_{j}}^{g}$ and
$\beta_{\lambda_{j}}\colon I_{\lambda_{j}}^{g}\to I_{\lambda_{j}}^{f}$ defined
by $i\alpha_{\lambda_{j}}=i\alpha$ and $i\beta_{\lambda_{j}}=i\beta$,
respectively, are bijective.

We now claim that $k=t$. Suppose to the contrary that $k\neq t$. Assume
without loss of generality that $k<t$. Then $\min\\{k,t\\}=k$, and so
$\lambda_{j}=\mu_{j}$ for all $j\in[k]$. Now, let $i\in I_{\mu_{t}}^{g}$. Then
we see that $i\beta\notin I_{\lambda_{j}}^{f}$ for all $j\in[k]$, since
$\beta_{\lambda_{j}}\colon I_{\lambda_{j}}^{g}\to I_{\lambda_{j}}^{f}$ is
bijective for all $j\in[k]$. Therefore $i\beta\in I_{1}^{f}\cup I_{2}^{f}$,
and so $|X_{i\beta}f|\leq 2$. Recall that $|X_{i}g|=\mu_{t}\geq 3$. By
inequality (ii), we therefore obtain $3\leq|X_{i}g|\leq|X_{i\beta}f|\leq 2$, a
contradiction. Hence $k=t$.

Next, we prove that $n_{2}^{f}=n_{2}^{g}$. Consider the following two possible
cases:

Case I: Suppose $I_{2}^{f}=\varnothing$. Suppose to the contrary that
$I_{2}^{g}\neq\varnothing$, and let $i\in I_{2}^{g}$. Then $|X_{i}g|=2$.
Notice that $i\beta\notin I_{\lambda_{j}}^{f}$ for all $j\in[k]$, since
$\beta_{\lambda_{j}}\colon I_{\lambda_{j}}^{g}\to I_{\lambda_{j}}^{f}$ is
bijective for all $j\in[k]$. Therefore $i\beta\in I_{1}^{f}$, and so
$|X_{i\beta}f|=1$. By inequality (ii), we obtain
$2=|X_{i}g|\leq|X_{i\beta}f|=1$, a contradiction. Hence
$I_{2}^{g}=\varnothing$, and thus $n_{2}^{f}=n_{2}^{g}$.

Case II: Suppose $I_{2}^{f}\neq\varnothing$. Let $i\in I_{2}^{f}$. Then
$|X_{i}f|=2$, and so $2\leq|X_{i\alpha}g|$ by inequality (i). Notice that
$i\alpha\notin I_{\lambda_{j}}^{g}$ for all $j\in[k]$, since
$\alpha_{\lambda_{j}}\colon I_{\lambda_{j}}^{f}\to I_{\lambda_{j}}^{g}$ is
bijective for all $j\in[k]$. Therefore $i\alpha\in I_{2}^{g}$. Since $\alpha$
is injective, we thus get $n_{2}^{f}\leq n_{2}^{g}$. Now, let $i\in
I_{2}^{g}$. Then $|X_{i}g|=2$, and so $2\leq|X_{i\beta}f|$ by inequality (ii).
Notice that $i\beta\notin I_{\lambda_{j}}^{f}$ for all $j\in[k]$, since
$\beta_{\lambda_{j}}\colon I_{\lambda_{j}}^{g}\to I_{\lambda_{j}}^{f}$ is
bijective for all $j\in[k]$. Therefore $i\beta\in I_{2}^{f}$. Since $\beta$ is
injective, we thus get $n_{2}^{f}\geq n_{2}^{g}$. Hence $n_{2}^{f}=n_{2}^{g}$.

In either case, we have $n_{2}^{f}=n_{2}^{g}$. Finally, we prove that
$n_{1}^{f}=n_{1}^{g}$. Consider the following two possible cases:

Case I: Suppose $I_{1}^{f}=\varnothing$. Suppose to the contrary that
$I_{1}^{g}\neq\varnothing$, and let $i\in I_{1}^{g}$. Then $|X_{i}g|=1$, and
so $|X_{i\alpha^{-1}}f|\leq|X_{(i\alpha^{-1})\alpha}g|=|X_{i}g|$ by inequality
(i). It follows that $|X_{i\alpha^{-1}}f|=1$, and so $i\alpha^{-1}\in
I_{1}^{f}$, a contradiction. Hence $I_{1}^{g}=\varnothing$, and thus
$n_{1}^{f}=n_{1}^{g}$.

Case II: Suppose $I_{1}^{f}\neq\varnothing$. Let $i\in I_{1}^{f}$. Then
$|X_{i}f|=1$, and so
$|X_{i\beta^{-1}}g|\leq|X_{(i\beta^{-1})\beta}f|=|X_{i}f|$ by inequality (ii).
It follows that $|X_{i\beta^{-1}}g|=1$, and so $i\beta^{-1}\in I_{1}^{g}$.
Since $\beta^{-1}$ is injective, we thus get $n_{1}^{f}\leq n_{1}^{g}$. Now,
let $i\in I_{1}^{g}$. Then $|X_{i}g|=1$, and so
$|X_{i\alpha^{-1}}f|\leq|X_{(i\alpha^{-1})\alpha}g|=|X_{i}g|$ by inequality
(i). It follows that $|X_{i\alpha^{-1}}f|=1$, and so $i\alpha^{-1}\in
I_{1}^{f}$. Since $\alpha^{-1}$ is injective, we thus get $n_{1}^{f}\geq
n_{1}^{g}$. Hence $n_{1}^{f}=n_{1}^{g}$.

In either case, we have $n_{1}^{f}=n_{1}^{g}$. In addition, we note for every
cardinal $\lambda\notin\\{1,2,\lambda_{k},\ldots,\lambda_{1}\\}$ that
$n_{\lambda}^{f}=0=n_{\lambda}^{g}$. Thus we conclude from Proposition 4.11
that $(f,g)\in\mathcal{D}$ in $\mathcal{B}(X,\mathcal{P})$ as required. ∎

Note that $\mathcal{D}=\mathcal{J}$ on $\mathcal{B}(X,\mathcal{P})$ if
$\mathcal{P}$ is a trivial partition of $X$. In connection with Proposition
4.8, Theorem 4.12, and Theorem 4.14, we end this section with the following
conjecture.

Conjecture. Let $\mathcal{P}=\\{X_{i}\colon i\in I\\}$ be a partition of $X$
and $f\in\mathcal{B}(X,\mathcal{P})$. Then $D_{f}=J_{f}$ in
$\mathcal{B}(X,\mathcal{P})$ if and only if there exist any two consecutive
cardinals $\lambda_{1},\lambda_{2}$ and a finite subset $K$ of $I$ such that
$|X_{i}f|\in\\{\lambda_{1},\lambda_{2}\\}$ for all $i\in I\setminus K$.

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