Green’s Relations on a Semigroup of Transformations with Restricted Range that Preserves an Equivalence Relation and a Cross-Section

Abstract

Let $T(X,Y)$ be the semigroup consisting of all total transformations from X into a fixed nonempty subset Y of X. For an equivalence relation $\rho$ on X, let $\widehat{\rho }$ be the restriction of $\rho$ on Y, R a cross-section of $Y/\widehat{\rho }$ and define $T(X,Y,\rho ,R)$ to be the set of all total transformations $\alpha$ from X into Y such that $\alpha$ preserves both $\rho$ (if $(a,b)\in \rho$, then $(a\alpha ,b\alpha )\in \rho$) and R (if $r\in R$, then $r\alpha \in R$). $T(X,Y,\rho ,R)$ is then a subsemigroup of $T(X,Y)$. In this paper, we give descriptions of Green’s relations on $T(X,Y,\rho ,R)$, and these results extend the results on $T(X,Y)$ and $T(X,\rho ,R)$ when taking $\rho$ to be the identity relation and $Y=X$, respectively.

1. Introduction

Let X be a nonempty set and $T\left(X\right)$ denote the semigroup containing all full transformations from X into itself with the composition. It is well-known that $T\left(X\right)$ is a regular semigroup, as shown in Reference [1]. Various subsemigroups of $T\left(X\right)$ have been investigated in different years. One of the subsemigroups of $T\left(X\right)$ is related to an equivalence relation $\rho$ on X and a cross-section R of the partition $X/\rho$ (i.e., each $\rho$-class contains exactly one element of R), namely $T(X,\rho ,R)$, which was first considered by Araújo and Konieczny in 2003 [2], and is defined by

$$T(X,\rho ,R)=\{\alpha \in T(X):R\alpha \subseteq R\mathrm{and}(a,b)\in \rho \Rightarrow (a\alpha ,b\alpha )\in \rho \},$$

where $Z\alpha =\{z\alpha :z\in Z\}.$ They studied automorphism groups of centralizers of idempotents. Moreover, they also determined Green’s relations and described the regular elements of $T(X,\rho ,R)$ in 2004 [3].

Let Y be a nonempty subset of the set X. Consider another subsemigroup of $T\left(X\right)$, which was first introduced by Symons [4] in 1975, called $T(X,Y)$, defined by

$$T(X,Y)=\{\alpha \in T(X):X\alpha \subseteq Y\},$$

when $X\alpha$ denotes the image of $\alpha$. He described all the automorphisms of $T(X,Y)$ and also determined when $T(X_1,Y_1)$ is isomorphic to $T(X_2,Y_2)$. In 2009, Sanwong, Singha and Sullivan [5] described all the maximal and minimal congruences on $T(X,Y)$. Later, in Reference [6], Sanwong and Sommanee studied other algebraic properties of $T(X,Y)$. They gave necessary and sufficient conditions for $T(X,Y)$ to be regular and also determined Green’s relations on $T(X,Y)$. Furthermore, they obtained a class of maximal inverse subsemigroups of $T(X,Y)$ and proved that the set

$$F(X,Y)=\{\alpha \in T(X,Y):X\alpha \subseteq Y\alpha \}$$

contains all regular elements in $T(X,Y)$, and is the largest regular subsemigroup of $T(X,Y)$.

From now on, we study the subsemigroup $T(X,Y,\rho ,R)$ of $T(X,Y)$ defined by

$$T(X,Y,\rho ,R)=\{\alpha \in T(X,Y):R\alpha \subseteq R\mathrm{and}(a,b)\in \rho \Rightarrow (a\alpha ,b\alpha )\in \rho \},$$

where $\rho$ is an equivalence relation on X and R is a cross-section of the partition $Y/\widehat{\rho }$ in which $\widehat{\rho }=\rho$ ∩ $(Y\times Y)$. If $Y=X$, then $T(X,Y,\rho ,R)=T(X,\rho ,R)$; and if $\rho$ is the identity relation, then $T(X,Y,\rho ,R)=T(X,Y)$, so we may regard $T(X,Y,\rho ,R)$ as a generalization of $T(X,\rho ,R)$ and $T(X,Y)$.

Green’s relations play a role in semigroup theory, and the aim of this paper is to characterize Green’s relations on $T(X,Y,\rho ,R)$. As consequences, we obtain Green’s relations on $T(X,\rho ,R)$ and $T(X,Y)$ as corollaries.

2. Preliminaries and Notations

For any semigroup S, let $S^1$ be a semigroup obtained from S by adjoining an identity if S has no identity and letting $S^1=S$ if it already contains an identity. Green’s relations of S are equivalence relations on the set S which were first defined by Green. According to such definitions, we define the $\mathcal{L}$-relation as follows. For any $a,b\in S$,

$$a\mathcal{L}b\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}{S}^{1}a=S^{1}b,$$

or equivalently, $a\mathcal{L}b$ if and only if $a=xb$ and $b=ya$ for some $x,y\in S^1$.

Furthermore, we dually define the $\mathcal{R}$-relation as follows.

$$a\mathcal{R}b\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}a{S}^1=b{S}^1,$$

or equivalently, $a\mathcal{R}b$ if and only if $a=bx$ and $b=ay$ for some $x,y\in S^1$.

Moreover, we define the $\mathcal{J}$-relation as follows.

$$a\mathcal{J}b\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}{S}^{1}a{S}^1=S^{1}b{S}^1,$$

or equivalently, $a\mathcal{J}b$ if and only if $a=xby$ and $b=uav$ for some $x,y,u,v\in S^1$.

Finally, we define $\mathcal{H}=\mathcal{L}\cap \mathcal{R}$ and $\mathcal{D}=\mathcal{L}\circ \mathcal{R}$, where ∘ is the composition of relations. Since the relations $\mathcal{L}$ and $\mathcal{R}$ commute, it follows that $\mathcal{L}\circ \mathcal{R}=\mathcal{R}\circ \mathcal{L}$.

In this paper, we write functions on the right; in particular, this means that for a composition $\alpha \beta$, $\alpha$ is applied first. Furthermore, the cardinality of a set A is denoted by $\left|A\right|$.

For each $\alpha \in T\left(X\right)$, we denote by $ker\left(\alpha \right)$ the kernel of $\alpha$, the set of ordered pairs in $X\times X$ having the same image under $\alpha$, that is,

$$ker\left(\alpha \right)=\left\{\right(a,b)\in X\times X:a\alpha =b\alpha \}.$$

Moreover, the symbol $\pi \left(\alpha \right)$ denotes the partition of X induced by the map $\alpha$, namely

$$\pi \left(\alpha \right)=\{x{\alpha }^{-1}:x\in X\alpha \}.$$

We observe that $ker\left(\alpha \right)$ is an equivalence relation on X in which the partition $X/ker\left(\alpha \right)$ and $\pi \left(\alpha \right)$ coincide. Moreover, for all $\alpha ,\beta \in T\left(X\right)$, we have $ker\left(\alpha \right)=ker\left(\beta \right)$ if and only if $\pi \left(\alpha \right)=\pi \left(\beta \right)$.

In addition, if $\rho$ is an equivalence relation on the set X and $a,b\in X$, we sometimes write $a\rho b$ instead of $(a,b)\in \rho$, and define $a\rho$ to be the equivalence class that contains a, that is, $a\rho =\{b\in X:b\rho a\}$.

For the subsemigroup $T(X,Y,\rho ,R)$ of $T\left(X\right)$ where $\rho$ is an equivalence relation on X, Y is a nonempty subset of X and R is a cross-section of $Y/\widehat{\rho }$ in which $\widehat{\rho }=\rho \cap (Y\times Y)$, we see that if $a\in X$ and $a\rho \cap Y\ne \varnothing$, then there exists a unique $r\in R$ such that $a\rho r$, and we denote this element by $r_a$. Furthermore, we observe that $F(X,Y)\cap T(X,Y,\rho ,R)$ contains all constant maps whose images belong to R. This implies that $F(X,Y)\cap T(X,Y,\rho ,R)$ is a subsemigroup of $T(X,Y,\rho ,R)$, which will be denoted by F.

An element a in a semigroup S is said to be regular if there exists $x\in S$ such that $a=axa$; and S is a regular semigroup if every element of S is regular.

In general, $T(X,Y,\rho ,R)$ is not a regular semigroup, so we cannot apply Hall’s Theorem to find the $\mathcal{L}$-relation and the $\mathcal{R}$-relation on $T(X,Y,\rho ,R)$.

Now, we give an example of a non-regular element in $T(X,Y,\rho ,R)$. Let $X=\{1,2,3,4,5\}$, $Y=\{3,4,5\}$, $X/\rho =\left\{\right\{1,2\},\{3,4,5\left\}\right\}$, $Y/\widehat{\rho }=\left\{\{3,4,5\}\right\}$ and $R=\left\{3\right\}$. Define $\alpha \in T(X,Y,\rho ,R)$ by

$$\alpha =\left(\begin{array}{ccccc}1& 2& 3& 4& 5\\ 4& 3& 3& 5& 5\end{array}\right).$$

Suppose that $\alpha$ is regular. Then $\alpha =\alpha \beta \alpha$ for some $\beta \in T(X,Y,\rho ,R)$. We see that $4=1\alpha =1\left(\alpha \beta \alpha \right)=\left(4\beta \right)\alpha$, which implies that $1=4\beta \in Y$, a contradiction.

Throughout this paper, the set X we study can be a finite or an infinite set. For convenience, we will denote $T(X,Y,\rho ,R)$ by T.

3. Green’s Relations on $T(X,Y,\rho ,R)$

Unlike $T(X,\rho ,R)$, in general T has no identity, as shown in the following example.

Example 1.

Let $X=\{1,2,3,4,5,6\},Y=\{1,3\},$ $X/\rho =\left\{\right\{1,2\},\{3,4\},\{5,6\left\}\right\},$ $Y/\widehat{\rho }=\{\left\{1\right\},\left\{3\right\}\}$ and $R=\{1,3\}$. Suppose that ε is an identity element in T. Consider $\alpha \in T$ defined by

$$\alpha =\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 1& 1& 1& 3& 3\end{array}\right).$$

We see that $\left(5\epsilon \right)\alpha =5\left(\epsilon \alpha \right)=5\alpha =3$, which implies that $5\epsilon \in \{5,6\}$. This leads to a contradiction, since both 5 and 6 are not in Y.

Therefore, we use the semigroup T with identity adjoined, given by $T^1$, in studying its Green’s relations.

From now on, the notation $L_{\alpha }$ ($R_{\alpha },H_{\alpha },D_{\alpha }$) denote the set of all elements of T which are $\mathcal{L}$-related ($\mathcal{R}$-related, $\mathcal{H}$-related, $\mathcal{D}$-related) to $\alpha$, where $\alpha \in T$.

Let $\mathcal{A}$ and $\mathcal{B}$ be families of sets. If for each set $A\in \mathcal{A}$ there is a set $B\in \mathcal{B}$ such that $A\subseteq B$, we say that $\mathcal{A}$ refines $\mathcal{B}$, denoted by $\mathcal{A}\hookrightarrow \mathcal{B}$.

In what follows, most of the notation used are taken from Reference [3]. For each $\alpha \in T$, we denote by $▾\alpha$ the family $\left\{\right(x\rho )\alpha :x\in X\}$ and ${▾}^Y\alpha$ the family $\{\left(r\widehat{\rho }\right)\alpha :r\in R\}$. Furthermore, we define $\overline{▾}\alpha =\{\left(x\rho \right){\alpha }^{-1}:x\in X$ and $\left(x\rho \right){\alpha }^{-1}\ne \varnothing \}$. In fact, we see that $\overline{▾}\alpha =\{\left(r\widehat{\rho }\right){\alpha }^{-1}:r\in R$ and $\left(r\widehat{\rho }\right){\alpha }^{-1}\ne \varnothing \}$.

The following example describes the above notation.

Example 2.

Let $X=\{1,2,3,4,5,6,7,8\},Y=\{1,2,3,4,5\},X/\rho =\left\{\right\{1,2,3\},\{4,5,6\},\{7,8\left\}\right\}$, $Y/\widehat{\rho }=\{\{1,2,3\},\{4,5\}\}$ and $R=\{1,4\}$. Define $\alpha \in T$ by

$$\alpha =\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 4& 5& 5& 1& 2& 3& 2& 3\end{array}\right).$$

It follows that

$$\begin{array}{cccc}\hfill ▾\alpha & =\left\{\right(x\rho )\alpha :x\in X\},\hfill & \hfill {▾}^Y\alpha & =\{\left(r\widehat{\rho }\right)\alpha :r\in R\},\hfill \\ \hfill & =\left\{\right\{1,2,3\}\alpha ,\{4,5,6\}\alpha ,\{7,8\left\}\alpha \right\},\hfill & \hfill & =\left\{\right\{1,2,3\}\alpha ,\{4,5\left\}\alpha \right\},\hfill \\ \hfill & =\left\{\right\{4,5\},\{1,2,3\},\{2,3\left\}\right\},\hfill & \hfill & =\left\{\right\{4,5\},\{1,2\left\}\right\},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \overline{▾}\alpha & =\{\left(r\widehat{\rho }\right){\alpha }^{-1}:r\in Rand\left(r\widehat{\rho }\right){\alpha }^{-1}\ne \varnothing \},\hfill \\ \hfill & =\{\{1,2,3\}{\alpha }^{-1},\{4,5\}{\alpha }^{-1}\},\hfill \\ \hfill & =\left\{\right\{4,5,6,7,8\},\{1,2,3\left\}\right\}.\hfill \end{array}$$

3.1. $\mathcal{L}$-Relation and $\mathcal{R}$-Relation

We begin with characterizing the Green’s $\mathcal{L}$-relation on T by using the idea of the proof for the $\mathcal{L}$-relation on $T(X,\rho ,R)$ (see Reference [3] [Lemma 2.4]) with the idea of restricted range concerned.

The following example shows why the restricted range is involved.

Example 3.

Let $X=\{1,2,3,4,5,6,7,8,9,10\},$ $Y=\{1,2,3,4,5,6,7\},$ $X/\rho =\left\{\right\{1,2,3\},\{4,5\},\{6,7,8\},$ $\{9,10\}\},$ $Y/\widehat{\rho }=\{\{1,2,3\},\{4,5\},\{6,7\}\}$ and $R=\{1,4,6\}$. Define $\alpha ,{\beta }_1,{\beta }_2,\gamma \in T$ as follows.

$$\alpha =\left(\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 1& 2& 2& 6& 7& 1& 2& 2& 7& 7\end{array}\right), {\beta }_1=\left(\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 1& 2& 3& 4& 5& 1& 3& 2& 6& 7\end{array}\right),$$

$${\beta }_2=\left(\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 1& 2& 2& 6& 7& 4& 5& 5& 3& 3\end{array}\right), \gamma =\left(\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 1& 2& 2& 4& 5& 1& 2& 2& 5& 5\end{array}\right).$$

We see that $\alpha \ne {\gamma }^{\prime }{\beta }_1$ for all ${\gamma }^{\prime }\in T$, for if $\alpha ={\gamma }^{\prime }{\beta }_1$ for some ${\gamma }^{\prime }\in T$, then $7=9\alpha =9{\gamma }^{\prime }{\beta }_1$, and thus $10=9{\gamma }^{\prime }\in Y$, a contradiction. However, $▾\alpha =\{\{1,2\},\{6,7\},\left\{7\right\}\}\hookrightarrow \{\{1,2,3\},\{4,5\},\{6,7\}\}=▾{\beta }_1$. But $\alpha =\gamma {\beta }_2$ since $▾\alpha \hookrightarrow {▾}^Y{\beta }_2=\{\{1,2\},\{6,7\},\{4,5\}\}$.

Theorem 1.

Let $\alpha ,\beta \in T$. Then $\alpha =\gamma \beta$ for some $\gamma \in T^1$ if and only if $\alpha =\beta$ or $▾\alpha \hookrightarrow {▾}^Y\beta$. Consequently, $\alpha \mathcal{L}\beta$ if and only if $\alpha =\beta$; or $▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha$.

Proof. 

Assume that $\alpha =\gamma \beta$ for some $\gamma \in T^1$. Suppose that $\alpha \ne \beta$. Thus $\gamma \ne 1$. We prove that $▾\alpha \hookrightarrow {▾}^Y\beta$. Let $A\in ▾\alpha$. Then $A=\left(x\rho \right)\alpha =\left(\right(x\rho \left)\gamma \right)\beta$ for some $x\in X$ and $\left(x\rho \right)\gamma \subseteq r\widehat{\rho }$ for some $r\in R$. Therefore $A\subseteq \left(r\widehat{\rho }\right)\beta \in {▾}^Y\beta ,$ and thus, $▾\alpha \hookrightarrow {▾}^Y\beta$.

Conversely, assume that $\alpha =\beta$ or $▾\alpha \hookrightarrow {▾}^Y\beta$. If $\alpha =\beta$, then $\alpha =1\beta$, where $\gamma =1\in T^1$. For the case $▾\alpha \hookrightarrow {▾}^Y\beta$, we define γ on each ρ-class as follows. Let $x\rho \in X/\rho$. Then $\left(x\rho \right)\alpha \subseteq \left(r\widehat{\rho }\right)\beta \subseteq s\widehat{\rho }$ for some $r,s\in R$ (since $▾\alpha \hookrightarrow {▾}^Y\beta$). So, for each $a\in x\rho$, we choose $b_a\in r\widehat{\rho }$ such that $a\alpha =b_a\beta$ (if $a=t\in R$, we choose $b_a=r$ since $t\alpha =s=r\beta$) and define $a\gamma =b_a$. From $b_a\in r\widehat{\rho }$ for all $a\in x\rho$ we obtain $\left(x\rho \right)\gamma \subseteq r\widehat{\rho }$. By the definition of γ, $t\gamma =b_t=r$. Since $x\rho$ is arbitrary, we conclude that $\gamma \in T$. To see that $\alpha =\gamma \beta$, let $a\in X$. Then $a\gamma \beta =\left(a\gamma \right)\beta =b_a\beta =a\alpha$, and so $\alpha =\gamma \beta$. ☐

If we replace Y with X in Theorem 1, then $T=T(X,\rho ,R)$, and ${▾}^Y\alpha =▾\alpha$ for all $\alpha \in T$. Therefore, we have the $\mathcal{L}$-relation on $T(X,\rho ,R)$.

Corollary 1. [3] [Theorem 2.5]

Let $\alpha ,\beta \in T(X,\rho ,R)$. Then $\alpha \mathcal{L}\beta$ if and only if $▾\alpha \hookrightarrow ▾\beta$ and $▾\beta \hookrightarrow ▾\alpha$.

Similarly to $T(X,Y)$, there are two types of $\mathcal{L}$-classes on T. In order to describe these $\mathcal{L}$-classes, the following lemma is needed.

Lemma 1.

Let $\alpha ,\beta \in T$ be such that $▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha$. Then $\alpha ,\beta \in F$ and $X\alpha =X\beta$.

Proof. 

For each $a\alpha \in X\alpha$, we have $a\alpha \in \left(a\rho \right)\alpha \subseteq \left(r\widehat{\rho }\right)\beta$ for some $r\in R$, since $▾\alpha \hookrightarrow {▾}^Y\beta$. Thus, $X\alpha \subseteq Y\beta$. Similarly, $X\beta \subseteq Y\alpha$, since $▾\beta \hookrightarrow {▾}^Y\alpha$. It follows that $X\alpha \subseteq Y\beta \subseteq X\beta \subseteq Y\alpha \subseteq X\alpha$, and thus $\alpha ,\beta \in F$ and $X\alpha =X\beta$. ☐

Corollary 2.

For $\alpha \in T$, the following statements hold.

(i)

If $\alpha \in F$, then $L_{\alpha }=\left\{\alpha \right\}\cup \{\beta \in F:▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha \}$.

(ii)

If $\alpha \in T\backslash F$, then $L_{\alpha }=\left\{\alpha \right\}$.

Proof. 

Let α be any element in T and let $\beta \in L_{\alpha }$. Then $\alpha \mathcal{L}\beta$, which implies that $\alpha =\beta$; or $▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha$ by Theorem 1.

(i) Assume that $\alpha \in F$. It is clear by Theorem 1 that $\left\{\alpha \right\}\cup \{\beta \in F:▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha \}\subseteq L_{\alpha }$. To prove the other containment, we consider when $\beta \ne \alpha$. Since $\beta \mathcal{L}\alpha$, we obtain $▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha$. By Lemma 1, we obtain $\beta \in F$. Thus, we have $L_{\alpha }\subseteq \left\{\alpha \right\}\cup \{\beta \in F:▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha \}$.

(ii) Assume that $L_{\alpha }\ne \left\{\alpha \right\}$. Then there is $\gamma \ne \alpha$ such that $\gamma \mathcal{L}\alpha$, so $▾\alpha \hookrightarrow {▾}^Y\gamma$ and $▾\gamma \hookrightarrow {▾}^Y\alpha$. By Lemma 1, we get $\alpha \in F$. ☐

As a direct consequence of Corollary 2, we obtain the $\mathcal{L}$-relation on $T(X,Y)$ as follows.

Corollary 3. [6] [Theorem 3.2]

For $\alpha \in T(X,Y)$, the following statements hold.

(i)

If $\alpha \in F(X,Y)$, then $L_{\alpha }=\{\beta \in F(X,Y):X\alpha =X\beta \}$.

(ii)

If $\alpha \in T(X,Y)\backslash F(X,Y)$, then $L_{\alpha }=\left\{\alpha \right\}$.

Proof. 

If we replace ρ with the identity relation in Corollary 2, then $T=T(X,Y)$, $F=F(X,Y)\cap T(X,Y)=F(X,Y)$ and $Y=R$. Therefore, (ii) holds. To see that (i) holds, it suffices to prove that for $\alpha ,\beta \in F$,

$$▾\alpha \hookrightarrow {▾}^Y\beta \mathrm{and}▾\beta \hookrightarrow {▾}^Y\alpha \mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}X\alpha =X\beta .$$

By Lemma 1, we have the “only if” part of the above statement. Now, if $\alpha ,\beta \in F$ and $X\alpha =X\beta$, then for each $x\alpha \in X\alpha$ there exist $y\in X$ and $r\in Y$ such that $x\alpha =y\beta =r\beta$, since $X\alpha \subseteq X\beta$ and $\beta \in F$. Hence, $▾\alpha =\{\left\{x\alpha \right\}:x\in X\}\hookrightarrow \{\left\{r\beta \right\}:r\in Y\}={▾}^Y\beta$. Similarly, by using $X\beta \subseteq X\alpha$ and $\alpha \in F$, we obtain $▾\beta \hookrightarrow {▾}^Y\alpha$. ☐

As we know, $\alpha \mathcal{R}\beta$ on $T(X,Y)$ (or $T(X,\rho ,R)$) if and only if $ker\left(\alpha \right)=ker\left(\beta \right)$. However, for the semigroup T, this is true only on F (see Corollary 5). For $\alpha ,\beta$ outside F, there are more terminologies involved.

The following example shows that there are $\alpha ,\beta \in T$ with $ker\left(\beta \right)\subseteq ker\left(\alpha \right)$ but $\alpha \ne \beta \gamma$ for all $\gamma \in T$.

Example 4.

Considering $\alpha ,{\beta }_1$ and ${\beta }_2$ defined in Example 3, we see that $ker\left({\beta }_1\right)\subseteq ker\left(\alpha \right)$ but $\alpha \ne {\beta }_1\gamma$ for all $\gamma \in T$, for if $\alpha ={\beta }_1\gamma$ for some $\gamma \in T$, then $7=9\alpha =9{\beta }_1\gamma =6\gamma \in R$, a contradiction. Moreover, we have $\overline{▾}{\beta }_1=\{\{1,2,3,6,7,8\},\{4,5\},$$\{9,10\}\}\hookrightarrow \{\{1,2,3,6,7,8\},\{4,5,9,10\}\}=\overline{▾}\alpha$ but $\left(R{\beta }_1^{-1}\right)\alpha =\{1,4,6,9\}\alpha ⊈R$. In the same way, $ker\left({\beta }_2\right)\subseteq ker\left(\alpha \right)$ but $\alpha \ne {\beta }_2{\gamma }^{\prime }$ for all ${\gamma }^{\prime }\in T$, for if $\alpha ={\beta }_2{\gamma }^{\prime }$ for some ${\gamma }^{\prime }\in T$, then $(1{\gamma }^{\prime },3{\gamma }^{\prime })=(1{\beta }_2{\gamma }^{\prime },9{\beta }_2{\gamma }^{\prime })=(1\alpha ,9\alpha )=(1,7)\notin \rho$, which is a contradiction. Furthermore, $\left(R{\beta }_2^{-1}\right)\alpha \subseteq R$ but $\overline{▾}{\beta }_2=\{\{1,2,3,9,10\},\{4,5\},\{6,7,8\}\}\neg \hookrightarrow \overline{▾}\alpha$.

The proof below is completely different from those for $T(X,Y)$ and $T(X,\rho ,R)$, especially when proving the existence of such $\gamma \in T$.

Theorem 2.

Let $\alpha ,\beta \in T$. Then $\alpha =\beta \gamma$ for some $\gamma \in T^1$ if and only if $ker\left(\beta \right)\subseteq ker\left(\alpha \right),\overline{▾}\beta \hookrightarrow \overline{▾}\alpha$ and $\left(R{\beta }^{-1}\right)\alpha \subseteq R$. Consequently, $\alpha \mathcal{R}\beta$ if and only if $ker\left(\alpha \right)=ker\left(\beta \right)$, $\overline{▾}\alpha =\overline{▾}\beta$ and $\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R.$

Proof. 

Assume that $\alpha =\beta \gamma$ for some $\gamma \in T^1$. If $\gamma =1$, then $\alpha =\beta$ and the theorem holds. Now, we prove for $\gamma \in T$. Let $a,b\in X$ be such that $a\beta =b\beta$. Then $a\alpha =a\beta \gamma =b\beta \gamma =b\alpha$. Thus, $ker\left(\beta \right)\subseteq ker\left(\alpha \right)$. For each $U\in \overline{▾}\beta$, $U\beta \subseteq r\widehat{\rho }$ for some $r\in R$ and so $U\alpha =U\beta \gamma \subseteq \left(r\widehat{\rho }\right)\gamma \subseteq s\widehat{\rho }$ for some $s\in R$. Thus, $U\subseteq \left(s\widehat{\rho }\right){\alpha }^{-1}\in \overline{▾}\alpha$, which implies that $\overline{▾}\beta \hookrightarrow \overline{▾}\alpha .$ Now, let $c\in \left(R{\beta }^{-1}\right)\alpha$. Then $c=d\alpha$ and $d\beta =t$ for some $d\in X$ and $t\in R$. Hence, $c=d\alpha =d\beta \gamma =t\gamma \in R$, that is, $\left(R{\beta }^{-1}\right)\alpha \subseteq R.$

Conversely, assume that $ker\left(\beta \right)\subseteq ker\left(\alpha \right),\overline{▾}\beta \hookrightarrow \overline{▾}\alpha$ and $\left(R{\beta }^{-1}\right)\alpha \subseteq R$. Let $r_0\in R$ be fixed, and define $\gamma \in T$ on each ρ-class as follows. Let $x\rho \in X/\rho$.

If $x\rho \cap X\beta =\varnothing$, then define $a\gamma =r_0$ for all $a\in x\rho$. Therefore $\left(x\rho \right)\gamma =\left\{r_0\right\}\subseteq r_0\widehat{\rho }$.

If $x\rho \cap X\beta \ne \varnothing$, then $x\rho \cap Y\ne \varnothing$. Let $x\rho \cap Y=r\widehat{\rho }$ for some $r\in R$. We obtain $\left(r\widehat{\rho }\right){\beta }^{-1}\ne \varnothing$. Since $\overline{▾}\beta \hookrightarrow \overline{▾}\alpha$, it follows that $\left(r\widehat{\rho }\right){\beta }^{-1}\subseteq \left(s\widehat{\rho }\right){\alpha }^{-1}$ for some $s\in R$. Now, let $a\in x\rho$ and consider two cases. If $a\notin X\beta$, then define $a\gamma =s\in s\widehat{\rho }$. If $a\in X\beta$, then $a\in x\rho \cap Y=r\widehat{\rho }$ and $a=b\beta$ for some $b\in X$. Thus, $b\in a{\beta }^{-1}\subseteq \left(r\widehat{\rho }\right){\beta }^{-1}\subseteq \left(s\widehat{\rho }\right){\alpha }^{-1}$ which implies $b\alpha \in s\widehat{\rho }$. Now, we define $a\gamma =b\alpha \in s\widehat{\rho }$ (this is well-defined since $ker\left(\beta \right)\subseteq ker\left(\alpha \right)$). We observe that $\left(x\rho \right)\gamma \subseteq s\widehat{\rho }$. To see that $r\gamma =s$, we get by the definition of γ that $r\gamma =s$ if $r\notin X\beta$. For $r\in X\beta$, we have $r=c\beta$ for some $c\in X$ and then $c\in r{\beta }^{-1}\subseteq \left(r\widehat{\rho }\right){\beta }^{-1}\subseteq \left(s\widehat{\rho }\right){\alpha }^{-1}$, hence $c\alpha \in s\widehat{\rho }\cap \left(R{\beta }^{-1}\right)\alpha \subseteq R$ since $\left(R{\beta }^{-1}\right)\alpha \subseteq R$. Thus, $r\gamma =c\alpha =s$.

To prove that $\alpha =\beta \gamma$, let $b\in X$. Then $b\beta \in X\beta$, and so $\left(b\beta \right)\rho \cap X\beta \ne \varnothing$. By the definition of γ, we obtain $b\beta \gamma =\left(b\beta \right)\gamma =b\alpha$. Therefore, $\alpha =\beta \gamma$. ☐

The $\mathcal{R}$-relation on $T(X,Y)$ is as follows.

Corollary 4. [6] [Theorem 3.3]

Let $\alpha ,\beta \in T(X,Y)$. Then $\alpha \mathcal{R}\beta$ if and only if $\pi \left(\alpha \right)=\pi \left(\beta \right)$.

Proof. 

If ρ is the identity relation, then $T=T(X,Y)$, $F=F(X,Y)$ and $R=Y$. Moreover, $\overline{▾}\alpha =\{\left(r\widehat{\rho }\right){\alpha }^{-1}:r\in Y\mathrm{and}\left(r\widehat{\rho }\right){\alpha }^{-1}\ne \varnothing \}=\{r{\alpha }^{-1}:r\in X\alpha \}=\pi \left(\alpha \right)$ for all $\alpha \in T(X,Y)$. In addition, $\left(R{\beta }^{-1}\right)\alpha \subseteq R$ always holds for all $\alpha ,\beta \in T(X,Y)$ since $\left(R{\beta }^{-1}\right)\alpha =\left(Y{\beta }^{-1}\right)\alpha =X\alpha \subseteq Y=R$. Since we have $ker\left(\alpha \right)=ker\left(\beta \right)$ if and only if $\pi \left(\alpha \right)=\pi \left(\beta \right)$, it follows from Theorem 2 that $\alpha \mathcal{R}\beta$ if and only if $\pi \left(\alpha \right)=\pi \left(\beta \right)$. ☐

As one might expect, there are two types of the $\mathcal{R}$-classes on T, the one that lies inside F and the other outside F. To see this, we need the two lemmas below.

Lemma 2.

Let $\alpha ,\beta \in F$ and $ker\left(\alpha \right)=ker\left(\beta \right)$. Then the following statements hold.

(i)

$\overline{▾}\alpha =\overline{▾}\beta$.

(ii)

$\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R$.

Proof. 

(i) Let $\left(r\widehat{\rho }\right){\alpha }^{-1}\in \overline{▾}\alpha$. Then $\left(r\widehat{\rho }\right){\alpha }^{-1}=\underset{i\in I}{\cup }{A}_i$, where $A_i$ is a ρ-class such that $A_i\alpha \subseteq r\widehat{\rho }$ for all $i\in I$. From $\alpha \in F$, there exists $A_{i_0}$ such that $A_{i_0}\cap Y\ne \varnothing$. Therefore there is $s\in A_{i_0}\cap R$ and $s\alpha =r$. Thus, $s\beta =t$ for some $t\in R$, which implies that $A_{i_0}\beta \subseteq t\widehat{\rho }$. We prove that $A_i\beta \subseteq t\widehat{\rho }$ for all $i\in I$. Let $i\in I$.

If $A_i\cap Y\ne \varnothing$, then there exists $u\in A_i\cap R$ and $u\alpha =r$. It follows that $u\alpha =s\alpha$, and so $u\beta =s\beta =t$ since $ker\left(\alpha \right)=ker\left(\beta \right)$. Thus, $A_i\beta \subseteq t\widehat{\rho }$.

If $A_i\cap Y=\varnothing$, then since $A_i\alpha \subseteq r\widehat{\rho }$, there exists $a\in A_i$ such that $a\alpha \in r\widehat{\rho }$. From $\alpha \in F$, we obtain that $b\alpha =a\alpha \in r\widehat{\rho }$ for some $b\in Y$. Hence, $b\in \left(r\widehat{\rho }\right){\alpha }^{-1}=\underset{i\in I}{\cup }{A}_i$, that is, $b\in A_j\cap Y\ne \varnothing$ for some $j\in I$, which implies $A_j\beta \subseteq t\widehat{\rho }$. Since $ker\left(\alpha \right)=ker\left(\beta \right)$, we obtain that $a\beta =b\beta \in A_j\beta \subseteq t\widehat{\rho }$, where $a\in A_i$, and it follows that $A_i\beta \subseteq t\widehat{\rho }$.

Therefore, $A_i\beta \subseteq t\widehat{\rho }$ for all $i\in I$, that is, $\left(r\widehat{\rho }\right){\alpha }^{-1}=\underset{i\in I}{\cup }{A}_i\subseteq \left(t\widehat{\rho }\right){\beta }^{-1}$ and $\overline{▾}\alpha \hookrightarrow \overline{▾}\beta$ as required. Similarly, since $\beta \in F$, we obtain $\overline{▾}\beta \hookrightarrow \overline{▾}\alpha .$ Thus, $\overline{▾}\alpha =\overline{▾}\beta$.

(ii) Let $a\in \left(R{\beta }^{-1}\right)\alpha$. Then $a=b\alpha$ and $b\beta =r$ for some $b\in X$ and $r\in R$. From $\beta \in F$, we get $b\beta =r=s\beta$ for some $s\in R$. Since $ker\left(\alpha \right)=ker\left(\beta \right)$, we obtain that $a=b\alpha =s\alpha \in R$, and thus $\left(R{\beta }^{-1}\right)\alpha \subseteq R$. Similarly, $\left(R{\alpha }^{-1}\right)\beta \subseteq R$. ☐

Lemma 3.

Let $\alpha ,\beta \in T$. If $ker\left(\alpha \right)=ker\left(\beta \right)$, then either both α and β are in F, or neither is in F.

Proof. 

Assume that $ker\left(\alpha \right)=ker\left(\beta \right)$. Suppose that one of α and β is not in F. Without loss of generality, assume that $\alpha \notin F$. Then there exists $x\in X\backslash Y$ such that $x\alpha \ne y\alpha$ for all $y\in Y$. Thus, $(x,y)\notin ker\left(\alpha \right)$, which implies that $x\beta \ne y\beta$ for all $y\in Y$. Hence, $Y\beta ⊊X\beta$, which leads to $\beta \notin F$. ☐

Using Theorem 2, Lemmas 2 and 3, we have the following corollary.

Corollary 5.

For $\alpha \in T$, the following statements hold.

(i)

If $\alpha \in F$, then $R_{\alpha }=\{\beta \in F:ker\left(\beta \right)=ker\left(\alpha \right)\}$.

(ii)

If $\alpha \in T\backslash F$, then $R_{\alpha }=\{\beta \in T\backslash F:ker\left(\beta \right)=ker\left(\alpha \right)$, $\overline{▾}\beta =\overline{▾}\alpha$ and $\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R\}$.

If $Y=X$, then $F=F(X,Y)\cap T=T\left(X\right)\cap T(X,\rho ,R)=T(X,\rho ,R)$, and so $T\backslash F=\varnothing$. Thus, Corollary 5 gives us a description of the $\mathcal{R}$-relation on $T(X,\rho ,R)$.

Corollary 6. [3] [Theorem 2.3]

Let $\alpha ,\beta \in T(X,\rho ,R)$. Then $\alpha \mathcal{R}\beta$ if and only if $ker\left(\alpha \right)=ker\left(\beta \right)$.

As direct consequences of Corollaries 2 and 5, we have the $\mathcal{H}$-relation on T as follows.

Corollary 7.

For $\alpha \in T$, the following statements hold.

(i)

If $\alpha \in F$, then $H_{\alpha }=\left\{\alpha \right\}\cup \{\beta \in F:▾\alpha \hookrightarrow {▾}^Y\beta ,▾\beta \hookrightarrow {▾}^Y\alpha \mathrm{and}ker\left(\alpha \right)=ker\left(\beta \right)\}$.

(ii)

If $\alpha \in T\backslash F$, then $H_{\alpha }=\left\{\alpha \right\}$.

3.2. $\mathcal{D}$-relation and $\mathcal{J}$-relation

Let $\varphi :B\to C$ be a function from a set B to a set C. For a family $\mathcal{A}$ of subsets of B, $\left(\mathcal{A}\right)\varphi$ denotes the family $\left\{\right(A)\varphi :A\in \mathcal{A}\}$ of subsets of C.

The main results used for characterizing the Green’s $\mathcal{D}$-relation on T below are Corollaries 5 and 2. Moreover, the technique for defining such a function $\varphi$ in (i) is taken from Reference [3] [Theorem 2.6].

Theorem 3.

For $\alpha \in T$, the following statements hold.

(i)

If $\alpha \in F$, then $D_{\alpha }=\{\beta \in F:$$ker\left(\beta \right)=ker\left(\alpha \right)$; or there exists a bijection $\varphi :X\alpha \to X\beta$ such that $(R\cap X\alpha )\varphi \subseteq R,$$(▾\alpha )\varphi \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\varphi \}$.

(ii)

If $\alpha \in T\backslash F$, then $D_{\alpha }=\{\beta \in T\backslash F:ker\left(\beta \right)=ker\left(\alpha \right),\overline{▾}\beta =\overline{▾}\alpha$ and $\left(R{\beta }^{-1}\right)\alpha ,$$\left(R{\alpha }^{-1}\right)\beta \subseteq R\}$.

Proof. 

Let α be any element in T and let $\beta \in D_{\alpha }$. Then $\alpha \mathcal{R}\gamma$ and $\gamma \mathcal{L}\beta$ for some $\gamma \in T$.

(i) Assume that $\alpha \in F$. By Corollary 5, $\gamma \in F$ and $ker\left(\gamma \right)=ker\left(\alpha \right)$. By Corollary 2, $\beta \in F$; also, $\beta =\gamma$ or $▾\beta \hookrightarrow {▾}^Y\gamma ,$$▾\gamma \hookrightarrow {▾}^Y\beta$. If $\beta =\gamma$, then $ker\left(\beta \right)=ker\left(\alpha \right)$. Now, assume that $▾\beta \hookrightarrow {▾}^Y\gamma$ and $▾\gamma \hookrightarrow {▾}^Y\beta$. Define $\varphi :X\alpha \to X\beta$ by $\left(a\alpha \right)\varphi =a\gamma$. We have $a\gamma \in \left(a\rho \right)\gamma \subseteq \left(r\widehat{\rho }\right)\beta \subseteq X\beta$ for some $r\in R$, since $▾\gamma \hookrightarrow {▾}^Y\beta$, so $a\gamma \in X\beta$. Since $ker\left(\gamma \right)=ker\left(\alpha \right)$, we obtain that φ is well-defined and injective. To see that φ is surjective, let $b\beta \in X\beta$. Then $b\beta \in \left(b\rho \right)\beta \subseteq \left(s\widehat{\rho }\right)\gamma$ for some $s\in R$, since $▾\beta \hookrightarrow {▾}^Y\gamma$. It follows that $b\beta =c\gamma$ for some $c\in s\widehat{\rho }$, and so $\left(c\alpha \right)\varphi =c\gamma =b\beta$, hence φ is surjective. To show that $(R\cap X\alpha )\varphi \subseteq R$, let $t\in R\cap X\alpha$. Since $t\in X\alpha$ and $\alpha \in F$, there exists $p\in R$ such that $t\in \left(p\widehat{\rho }\right)\alpha$, thus $t=p\alpha$ and $t\varphi =\left(p\alpha \right)\varphi =p\gamma \in R$. Moreover, by the definition of φ, $(▾\alpha )\varphi =▾\gamma$ and $\left({▾}^Y\alpha \right)\varphi ={▾}^Y\gamma$. Hence, $(▾\alpha )\varphi \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\varphi$.

Conversely, assume that $\lambda \in F$. If $ker\left(\lambda \right)=ker\left(\alpha \right)$, then $\lambda \mathcal{R}\alpha$, and it follows that $\lambda \in D_{\alpha }$. If there exists a bijection $\varphi :X\alpha \to X\lambda$ such that $(R\cap X\alpha )\varphi \subseteq R,(▾\alpha )\varphi \hookrightarrow {▾}^Y\lambda$ and $▾\lambda \hookrightarrow \left({▾}^Y\alpha \right)\varphi$, then we define $\gamma :X\to X$ by $a\gamma =\left(a\alpha \right)\varphi$ for all $a\in X$. From $(R\cap X\alpha )\varphi \subseteq R$, we obtain that $r\gamma =\left(r\alpha \right)\varphi \in R$ for all $r\in R$. Thus, $R\gamma \subseteq R$. To show that $\left(x\rho \right)\gamma \subseteq s\widehat{\rho }$ for some $s\in R$, considering $\left(x\rho \right)\gamma =\left(\right(x\rho \left)\alpha \right)\varphi \in (▾\alpha )\varphi$ and $(▾\alpha )\varphi \hookrightarrow {▾}^Y\lambda$, we get $\left(x\rho \right)\gamma \subseteq \left(t\widehat{\rho }\right)\lambda \subseteq s\widehat{\rho }$ for some $s,t\in R$. Thus, $\left(x\rho \right)\gamma \subseteq s\widehat{\rho }$, that is, $\gamma \in T$. To see that $\gamma \in F$, let $a\in X$. Then $a\gamma =\left(a\alpha \right)\varphi =\left(b\alpha \right)\varphi =b\gamma$ for some $b\in Y$, since $\alpha \in F$. It follows that $\gamma \in F(X,Y)\cap T=F$. Since φ is an injective map, we obtain that $ker\left(\gamma \right)=ker\left(\alpha \right)$. By the definition of γ, $▾\gamma =(▾\alpha )\varphi$ and ${▾}^Y\gamma =\left({▾}^Y\alpha \right)\varphi$. Hence, $▾\gamma \hookrightarrow {▾}^Y\lambda$ and $▾\lambda \hookrightarrow {▾}^Y\gamma$. By Corollaries 5 and 2, $\alpha \mathcal{R}\gamma$ and $\gamma \mathcal{L}\lambda$. Therefore, $\lambda \in D_{\alpha }$.

(ii) Assume that $\alpha \in T\backslash F$. Corollaries 5 and 2 imply that $\beta =\gamma \in T\backslash F$. Thus, $\alpha \mathcal{R}\beta$. Again by Corollary 5, we have $ker\left(\beta \right)=ker\left(\alpha \right)$, $\overline{▾}\beta =\overline{▾}\alpha \mathrm{and}\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R$. Therefore, $D_{\alpha }\subseteq \{\beta \in T\backslash F:ker\left(\beta \right)=ker\left(\alpha \right),\overline{▾}\beta =\overline{▾}\alpha$ and $\left(R{\beta }^{-1}\right)\alpha ,$$\left(R{\alpha }^{-1}\right)\beta \subseteq R\}$. The other containment is clear since $\mathcal{R}\subseteq \mathcal{D}$. ☐

The two corollaries below are the $\mathcal{D}$-relations on $T(X,\rho ,R)$ and $T(X,Y)$, respectively.

Corollary 8. [3] [Theorem 2.6]

Let $\alpha ,\beta \in T(X,\rho ,R)$. Then $\alpha \mathcal{D}\beta$ if and only if there is a bijection $\varphi :X\alpha \to X\beta$ such that $(R\cap X\alpha )\varphi \subseteq R,$$(▾\alpha )\varphi \hookrightarrow ▾\beta$ and $▾\beta \hookrightarrow (▾\alpha )\varphi$.

Proof. 

If we replace Y with X in Theorem 3, then $T\backslash F=\varnothing$. Therefore we have that: For $\alpha \in T(X,\rho ,R),$$D_{\alpha }=\{\beta \in T(X,\rho ,R):$$ker\left(\beta \right)=ker\left(\alpha \right)$; or there is a bijection $\varphi :X\alpha \to X\beta$ such that $(R\cap X\alpha )\varphi \subseteq R,$$(▾\alpha )\varphi \hookrightarrow ▾\beta$ and $▾\beta \hookrightarrow (▾\alpha )\varphi \}$.

Now, we assert that $ker\left(\beta \right)=ker\left(\alpha \right)$ implies that there is a bijection $\varphi :X\alpha \to X\beta$ such that $(R\cap X\alpha )\varphi \subseteq R,$$(▾\alpha )\varphi \hookrightarrow ▾\beta$ and $▾\beta \hookrightarrow (▾\alpha )\varphi$. Assume that $ker\left(\beta \right)=ker\left(\alpha \right)$ and define $\varphi :X\alpha \to X\beta$ by $\left(a\alpha \right)\varphi =a\beta$ for all $a\in X$. Then φ is a well-defined injective map, since $ker\left(\beta \right)=ker\left(\alpha \right)$. It is obvious that φ is surjective. By the definition of φ, $\left[\right(a\rho \left)\alpha \right]\varphi =\left(a\rho \right)\beta$. Thus, $(▾\alpha )\varphi \hookrightarrow ▾\beta$ and $▾\beta \hookrightarrow (▾\alpha )\varphi$. Finally, $(R\cap X\alpha )\varphi \subseteq R\beta \subseteq R$. Hence, we have our assertion, and, therefore, $D_{\alpha }=\{\beta \in T(X,\rho ,R):$ There is a bijection $\varphi :X\alpha \to X\beta$ such that $(R\cap X\alpha )\varphi \subseteq R,$$(▾\alpha )\varphi \hookrightarrow ▾\beta$ and $▾\beta \hookrightarrow (▾\alpha )\varphi \}$, as required. ☐

Corollary 9. [6] [Theorem 3.7]

For $\alpha \in T(X,Y)$, the following statements hold.

(i)

If $\alpha \in F(X,Y)$, then $D_{\alpha }=\{\beta \in F(X,Y):|X\beta |=\left|X\alpha \right|\}$.

(ii)

If $\alpha \in T(X,Y)\backslash F(X,Y)$, then $D_{\alpha }=\{\beta \in T(X,Y)\backslash F(X,Y):\pi \left(\beta \right)=\pi \left(\alpha \right)\}$.

Proof. 

As in the proof of Corollary 4, if we replace ρ with the identity relation, then (ii) of Theorem 3 is as follows. If $\alpha \in T(X,Y)\backslash F(X,Y)$, then $D_{\alpha }=\{\beta \in T(X,Y)\backslash F(X,Y):\pi \left(\beta \right)=\pi \left(\alpha \right)\}$.

Now, we claim that the conditions $ker\left(\beta \right)=ker\left(\alpha \right)$; or there is a bijection $\varphi :X\alpha \to X\beta$ such that $(R\cap X\alpha )\varphi \subseteq R,$$(▾\alpha )\varphi \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\varphi$ in (i) of Theorem 3 is equivalent to $\left|X\alpha \right|=\left|X\beta \right|$ for all $\alpha ,\beta \in F(X,Y)$. It is clear that the above conditions imply $\left|X\alpha \right|=\left|X\beta \right|$. Now, let $\alpha ,\beta \in F(X,Y)$ and $\left|X\alpha \right|=\left|X\beta \right|$. Then there is a bijection $\varphi :X\alpha \to X\beta$ and $(R\cap X\alpha )\varphi \subseteq X\beta \subseteq Y=R$. To see that the remaining conditions hold, we observe that $\left(Y\alpha \right)\varphi =\left(X\alpha \right)\varphi =X\beta =Y\beta$, since $\alpha ,\beta \in F(X,Y)$. From $\left(X\alpha \right)\varphi =Y\beta$ and ρ as the identity relation, we obtain $(▾\alpha )\varphi =\{\left\{\left(x\alpha \right)\varphi \right\}:x\in X\}=\{\left\{r\beta \right\}:r\in Y\}={▾}^Y\beta$, hence, $(▾\alpha )\varphi \hookrightarrow {▾}^Y\beta$. Similarly, from $\left(Y\alpha \right)\varphi =X\beta$, we obtain $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\varphi$. Therefore, we have our claim. ☐

To characterize the $\mathcal{J}$-relation on T, we need the terminology below. For each $\alpha \in T$, we define

$$R\left(\alpha \right)=\{r\in R:r\widehat{\rho }\cap X\alpha \ne \varnothing \}.$$

The following example shows that $R\left(\alpha \right)$ is necessary.

Example 5.

Let $X=\{1,2,3,4,5,6,7,8\}$ and $Y=\{3,4,5,6,7,8\}$. Let

$$X/\rho =\{\{1,2\},\{3,4\},\{5,6,7\},\left\{8\right\}\},Y/\widehat{\rho }=\{\{3,4\},\{5,6,7\},\left\{8\right\}\}andR=\{3,5,8\}.$$

Define $\alpha ,\beta \in T$ as follows:

$$\alpha =\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 4& 4& 5& 6& 5& 6& 6& 5\end{array}\right)and\beta =\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 7& 7& 5& 6& 3& 4& 4& 8\end{array}\right).$$

Then $R\left(\alpha \right)=\{3,5\}⊈X\alpha$ and $R\left(\beta \right)=\{3,5,8\}\subseteq X\beta$. Moreover,

$$▾\alpha =\left\{\right\{4\},\{5,6\},\{5\left\}\right\}and{▾}^Y\beta =\{\{3,4\},\{5,6\},\left\{8\right\}\}.$$

We show further that $\alpha =\lambda \beta \mu$ for some $\lambda ,\mu \in T$, but there is no function $\varphi :Y\beta \to X\alpha$ such that $(R\cap Y\beta )\varphi$$\subseteq R$ and $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$. Define $\lambda ,\mu \in T$ by

$$\lambda =\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 7& 7& 3& 4& 3& 4& 4& 3\end{array}\right)and\mu =\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 3& 3& 3& 4& 5& 6& 5& 3\end{array}\right).$$

We see that $\alpha =\lambda \beta \mu$. Suppose that there is such a function φ, which would imply $\left\{4\right\}=\{1,2\}\alpha \subseteq A\varphi \subseteq \{3,4\}$ for some $A\in {▾}^Y\beta$, and $\varnothing \ne (R\cap A)\varphi \subseteq A\varphi \cap R=\left\{3\right\}$, since $\varnothing \ne R\cap A\subseteq Y\beta$. Thus, $3\in (R\cap A)\varphi \subseteq X\alpha$, which is a contradiction. However, since $3\in R\left(\alpha \right)\backslash X\alpha$, we can define $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ satisfying the conditions $(R\cap Y\beta )\varphi$$\subseteq R$ and $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$ as follows:

$$\varphi =\left(\begin{array}{ccccc}3& 4& 5& 6& 8\\ 3& 4& 5& 6& 5\end{array}\right).$$

Note that if $Y=X$ or $\rho$ is the identity relation, then $R\left(\alpha \right)\subseteq X\alpha$ for all $\alpha \in T$.

The following result is the key lemma in characterizing the $\mathcal{J}$-relation on T. The outline of the proof is the same as Theorem 2.7 in [3], but there are differences in detail, for example, to prove the “only if” part, the function $\varphi$ has to be defined from $Y\beta$ into $X\alpha \cup R\left(\alpha \right)$ in order to make $r_a\mu$ in (2) well-defined. In addition, because of the restricted range of T, the function $\mu$ defined in (3) of the “if” part is greatly different from that defined in Theorem 2.7 [3]. Moreover, each step of the proof, the restricted range is involved.

Lemma 4.

Let $\alpha ,\beta \in T$. Then $\alpha =\lambda \beta \mu$ for some $\lambda ,\mu \in T$ if and only if there exists $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ such that $(R\cap Y\beta )\varphi$$\subseteq R$ and $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$.

Proof. 

Assume that $\alpha =\lambda \beta \mu$ for some $\lambda ,\mu \in T$. We define $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ such that $(R\cap Y\beta )\varphi \subseteq R$ and $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$ as follows.

Fix $r_0\in X\alpha \cap R$. Let $a\in Y\beta$, and define φ in three steps.

(1)

If $a\in X\left(\lambda \beta \right)\subseteq Y\beta$, we define $a\varphi =a\mu \in X\alpha$.

(2)

If $a\in Y\beta \backslash X\left(\lambda \beta \right)$ and $a\widehat{\rho }\cap X\left(\lambda \beta \right)\ne \varnothing$, then define $a\varphi =r_a\mu$.

(3)

If $a\in Y\beta \backslash X\left(\lambda \beta \right)$ and $a\widehat{\rho }\cap X\left(\lambda \beta \right)=\varnothing$, then define $a\varphi =r_0\in X\alpha$.

We observe that $r_a\mu$ in (2) belongs to $R\left(\alpha \right)$, since $r_a\mu \in R$ and $r_a\mu \widehat{\rho }b\mu \in X\alpha$ for some $b\in a\widehat{\rho }\cap X\left(\lambda \beta \right)$.

By the definition of φ and the fact that $R\mu \subseteq R$, we obtain that $(R\cap Y\beta )\varphi \subseteq R$. To see that $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi$, let $\left(x\rho \right)\alpha \in ▾\alpha$. Then there exists $r\in R$ such that $\left(x\rho \right)\lambda \subseteq r\widehat{\rho }$. We have $\left(x\rho \right)\alpha =\left(x\rho \right)\lambda \beta \mu \subseteq [\left(r\widehat{\rho }\right)\beta \cap X\left(\lambda \beta \right)]\mu \subseteq \left[\left(r\widehat{\rho }\right)\beta \right]\varphi \in$$\left({▾}^Y\beta \right)\varphi$. Thus, $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi$. To show that $\left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$, let $\left(r\widehat{\rho }\right)\beta \in {▾}^Y\beta$. By the definition of φ, either $\left[\left(r\widehat{\rho }\right)\beta \right]\varphi \subseteq \left(\left(r\widehat{\rho }\right)\beta \right)\mu \subseteq s\widehat{\rho }$ for some $s\in R$ (if $\left(r\beta \right)\widehat{\rho }\cap X\left(\lambda \beta \right)\ne \varnothing$) or $\left[\left(r\widehat{\rho }\right)\beta \right]\varphi =\left\{r_0\right\}\subseteq r_0\widehat{\rho }$ (if $\left(r\beta \right)\widehat{\rho }\cap X\left(\lambda \beta \right)=\varnothing$). Therefore, $\left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$.

Conversely, assume that there exists $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ such that $(R\cap Y\beta )\varphi$$\subseteq R$ and $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$. Let $A\in ▾\alpha$. Then there is a unique $r_A\in R$ such that $A\subseteq r_A\widehat{\rho }$ (note that $r_A$ may not belong to $X\alpha$ but $r_A\in R\left(\alpha \right)$). Since $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$, there exists $C_A\in {▾}^Y\beta$ such that $A\subseteq C_A\varphi \subseteq r_A\widehat{\rho }$. From $C_A\in {▾}^Y\beta$, there is $s_A\in R$ such that $C_A=\left(s_A\widehat{\rho }\right)\beta$. Let $t_A=s_A\beta$. Then $C_A=\left(s_A\widehat{\rho }\right)\beta \subseteq t_A\widehat{\rho }$. For every $a\in A$, we choose $u_a^A\in C_A$ such that $a=u_a^A\varphi$ (since $t_A\varphi =r_A$, we may assume that $u_a^A=t_A$ if $a=r_A$). Let $C_A^{\prime }=\{u_a^A:a\in A\}.$ Then $C_A^{\prime }\subseteq C_A$. For every $b\in C_A^{\prime }$, we choose $v_b^A\in s_A\widehat{\rho }$ such that $b=v_b^A\beta$ (since $s_A\beta =t_A$, we may assume that $v_b^A=s_A$ if $b=t_A$).

We aim to define $\lambda ,\mu \in T$ such that $\alpha =\lambda \beta \mu$. We first define λ. Let $x\in X$, $A=\left(x\rho \right)\alpha$ and $a=x\alpha$. Then $x\alpha =a\in A$, so there exists $b=u_a^A$ such that $b=v_b^A\beta$, thus we define $x\lambda =v_b^A$. By the definition of λ, $\left(x\rho \right)\lambda \subseteq s_A\widehat{\rho }$. If $p\in x\rho \cap R$, then $a^{\prime }=p\alpha =r_A\in R\cap A$, and so $b^{\prime }=u_{a^{\prime }}^A=t_A$. Hence $v_{b^{\prime }}^A=s_A$, which implies that $p\lambda =v_{b^{\prime }}^A=s_A\in R$. Thus, $\lambda \in T$.

To define μ, fix $r_0\in R$ and let $x\in X$.

(1)

If $x\in C_A^{\prime }\mathrm{for}\mathrm{some}A\in ▾\alpha$, define $x\mu =x\varphi$.

(2)

If $x\notin C_B^{\prime }\mathrm{for}\mathrm{all}B\in ▾\alpha \mathrm{and}{C}_A\subseteq x\rho \mathrm{for}\mathrm{some}A\in ▾\alpha$, define $x\mu =r_A$.

(3)

If $x\rho \cap C_A=\varnothing$ for all $A\in ▾\alpha$, define $x\mu =r_0$.

To see that the definition of μ in (2) does not depend on the choice of A, we suppose that there are $A,B\in ▾\alpha$ such that $C_A,C_B\subseteq x\rho$. Since $C_A=\left(s_A\widehat{\rho }\right)\beta \subseteq t_A\widehat{\rho }\subseteq x\rho$ and $C_B=\left(s_B\widehat{\rho }\right)\beta \subseteq t_B\widehat{\rho }\subseteq x\rho$, we obtain $t_A=t_B$, and thus $r_A=t_A\varphi =t_B\varphi =r_B$. Next, we prove that $\mu \in T$. By the definition of μ, we see that $R\mu \subseteq R$. Now, if $x\rho \cap C_A=\varnothing$ for all $A\in ▾\alpha$, then $\left(x\rho \right)\mu =\left\{r_0\right\}\subseteq r_0\widehat{\rho }$. For $C_A\subseteq x\rho$ for some $A\in ▾\alpha$, we have $x\mu =x\varphi \in r_B\widehat{\rho }=r_A\widehat{\rho }$ if $x\in C_B^{\prime }$ for some $B\in ▾\alpha$ and $x\mu =r_A\in r_A\widehat{\rho }$ if $x\notin C_B^{\prime }$ for all $B\in ▾\alpha$. Thus, $\left(x\rho \right)\mu \subseteq r_A\widehat{\rho }$.

To prove that $\alpha =\lambda \beta \mu$, let $x\in X$, $A=\left(x\rho \right)\alpha$ and $a=x\alpha$. Let $b=u_a^A$ (note that $u_a^A\in C_A^{\prime }$ was selected such that $u_a^A\varphi =a$). By the definitions of λ and μ, we have $x\lambda =v_b^A$ (recall that $v_b^A$ was chosen so that $v_b^A\beta =b$) and $b\mu =u_a^A\mu =u_a^A\varphi =a=x\alpha$. Thus, $x\lambda \beta \mu =v_b^A\beta \mu =b\mu =x\alpha$, as required. ☐

If we take $Y=X$ in Lemma 4, then $T=T(X,\rho ,R)$, which contains an identity element, the identity map. Thus, we obtain the $\mathcal{J}$-relation on $T(X,\rho ,R)$.

Corollary 10. [3] [Theorem 2.8]

Let $\alpha ,\beta \in T(X,\rho ,R)$. Then $\alpha \mathcal{J}\beta$ if and only if there exist $\varphi :X\beta \to X\alpha$ and $\phi :X\alpha \to X\beta$ such that $(R\cap X\beta )\varphi \subseteq R$, $▾\alpha \hookrightarrow (▾\beta )\varphi \hookrightarrow X/\rho$; also $(R\cap X\alpha )\phi \subseteq R$, $▾\beta \hookrightarrow (▾\alpha )\phi \hookrightarrow X/\rho$.

Now, we are ready to prove the $\mathcal{J}$-relation on T.

Theorem 4.

Let $\alpha ,\beta \in T$. Then $\alpha \mathcal{J}\beta$ if and only if one of the following conditions holds:

(i)

$ker\left(\alpha \right)=ker\left(\beta \right)$, $\overline{▾}\alpha =\overline{▾}\beta and\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R$;

(ii)

there exist $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ such that $(R\cap Y\beta )\varphi \subseteq R$, $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$ and $\phi :Y\alpha \to X\beta \cup R\left(\beta \right)$ such that $(R\cap Y\alpha )\phi \subseteq R$, $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\phi \hookrightarrow Y/\widehat{\rho }$.

Proof. 

Assume that $\alpha \mathcal{J}\beta$. Then $\alpha =\sigma \beta \delta$ and $\beta ={\sigma }^{\prime }\alpha {\delta }^{\prime }$ for some $\sigma ,{\sigma }^{\prime },\delta ,{\delta }^{\prime }\in T^1$. If $\sigma =1={\sigma }^{\prime }$, then $\alpha =\beta \delta$ and $\beta =\alpha {\delta }^{\prime }$, which implies that $\alpha \mathcal{R}\beta$, and so $ker\left(\alpha \right)=ker\left(\beta \right)$, $\overline{▾}\alpha =\overline{▾}\beta \mathrm{and}\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R$. If $\delta =1={\delta }^{\prime }$, then $\alpha =\sigma \beta$ and $\beta ={\sigma }^{\prime }\alpha$, which implies that $\alpha \mathcal{L}\beta$, and so $\alpha =\beta$; or $▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha$. If $\alpha =\beta$, then (i) holds. If $▾\alpha \hookrightarrow {▾}^Y\beta$ and $▾\beta \hookrightarrow {▾}^Y\alpha$, then we define φ and ϕ to be the identity maps on $Y\beta$ and $Y\alpha$, respectively. It follows that $(R\cap Y\beta )\varphi \subseteq R$, $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho },(R\cap Y\alpha )\phi \subseteq R$ and $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\phi \hookrightarrow Y/\widehat{\rho }$. That is, (ii) holds. For the other cases, we can conclude that $\alpha =\lambda \beta \mu$ and $\beta ={\lambda }^{\prime }\alpha {\mu }^{\prime }$ for some $\lambda ,{\lambda }^{\prime },\mu ,{\mu }^{\prime }\in T$ (for example, if $\sigma =1$ and ${\sigma }^{\prime }\in T$, then $\alpha =\beta \delta$ and $\beta ={\sigma }^{\prime }\alpha {\delta }^{\prime }$ imply $\alpha =\beta \delta =\left({\sigma }^{\prime }\alpha {\delta }^{\prime }\right)\delta ={\sigma }^{\prime }\alpha \left({\delta }^{\prime }\delta \right)={\sigma }^{\prime }\left(\beta \delta \right){\delta }^{\prime }\delta ={\sigma }^{\prime }\beta \left(\delta {\delta }^{\prime }\delta \right)$). Thus, Lemma 4 gives that (ii) holds in all the remaining cases.

Conversely, assume that the statement holds. If $ker\left(\alpha \right)=ker\left(\beta \right),\overline{▾}\alpha =\overline{▾}\beta$ and $\left(R{\beta }^{-1}\right)\alpha ,\left(R{\alpha }^{-1}\right)\beta \subseteq R$, then $\alpha \mathcal{R}\beta$, and so $\alpha \mathcal{J}\beta$. If there exist $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ such that $(R\cap Y\beta )\varphi \subseteq R$, $▾\alpha \hookrightarrow \left({▾}^Y\beta \right)\varphi \hookrightarrow Y/\widehat{\rho }$ and $\phi :Y\alpha \to X\beta \cup R\left(\beta \right)$ such that $(R\cap Y\alpha )\phi \subseteq R$, $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\phi \hookrightarrow Y/\widehat{\rho }$, then $\alpha =\lambda \beta \mu$ and $\beta ={\lambda }^{\prime }\alpha {\mu }^{\prime }$ for some $\lambda ,{\lambda }^{\prime },\mu ,{\mu }^{\prime }\in T$ by Lemma 4. Therefore, $\alpha \mathcal{J}\beta$, as required. ☐

By setting $\rho$ to be the identity relation in Theorem 4, we obtain the $\mathcal{J}$-relation on $T(X,Y)$ as follows.

Corollary 11. [6] [Theorem 3.9]

Let $\alpha ,\beta \in T(X,Y)$. Then $\alpha \mathcal{J}\beta$ if and only if $\pi \left(\alpha \right)=\pi \left(\beta \right)$ or $\left|X\alpha \right|=\left|Y\alpha \right|=\left|Y\beta \right|=\left|X\beta \right|$.

Proof. 

If ρ in Theorem 4 is the identity relation, then $T=T(X,Y)$. By the same proof as given for Corollary 4, we have that (i) of Theorem 4 is equivalent to $\pi \left(\alpha \right)=\pi \left(\beta \right)$. Now, we claim that (ii) is equivalent to $\left|X\alpha \right|=\left|Y\alpha \right|=\left|Y\beta \right|=\left|X\beta \right|$. If (ii) holds, then $\varphi :Y\beta \to X\alpha$ is onto, since $▾\alpha =\{\left\{x\alpha \right\}:x\in X\}\hookrightarrow \{\left\{\left(r\beta \right)\varphi \right\}:r\in Y\}=\left({▾}^Y\beta \right)\varphi$ implies that for each $x\alpha \in X\alpha$, $x\alpha =\left(r\beta \right)\varphi$ for some $r\in Y$. Similarly, from $\phi :Y\alpha \to X\beta$ with $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\phi$, we obtain that ϕ is onto. Thus, $\left|X\alpha \right|\le \left|Y\beta \right|$ and $\left|X\beta \right|\le \left|Y\alpha \right|$, and so $\left|Y\alpha \right|\le \left|X\alpha \right|\le \left|Y\beta \right|\le \left|X\beta \right|\le \left|Y\alpha \right|$. Therefore, $\left|X\alpha \right|=\left|Y\alpha \right|=\left|Y\beta \right|=\left|X\beta \right|$. Conversely, if $\left|X\alpha \right|=\left|Y\alpha \right|=\left|Y\beta \right|=\left|X\beta \right|$, then there exist bijections $\varphi :Y\beta \to X\alpha$ and $\phi :Y\alpha \to X\beta$. Since φ is onto, it follows that $▾\alpha =\{\left\{x\alpha \right\}:x\in X\}\hookrightarrow \{\left\{\left(r\beta \right)\varphi \right\}:r\in Y\}=\left({▾}^Y\beta \right)\varphi \hookrightarrow \{\left\{y\right\}:y\in Y\}=Y/\widehat{\rho }$. Similarly, as ϕ is onto, we have $▾\beta \hookrightarrow \left({▾}^Y\alpha \right)\phi \hookrightarrow Y/\widehat{\rho }$. Moreover, $(R\cap Y\beta )\varphi \subseteq \left(Y\beta \right)\varphi =X\alpha \subseteq Y=R$, and in the same way, $(R\cap Y\alpha )\phi \subseteq R$. Therefore, we have our claim. ☐

Recall that $\mathcal{D}\subseteq \mathcal{J}$ on any semigroup and $\mathcal{D}=\mathcal{J}$ on $T\left(X\right)$, but in T it is not always true, so we end this section with an example showing that $\mathcal{D}\ne \mathcal{J}$ on T.

Example 6.

Let X be the set of all positive integers and $Y=X\backslash \{1,2\}$. Let

$$\begin{array}{cc}\hfill X/\rho & =\left\{\right\{1,2\},\{3,4,5\left\}\right\}\cup \left\{\right\{2n+4,2n+5\}:n\in X\},\hfill \\ \hfill Y/\widehat{\rho }& =\left\{\right\{3,4,5\left\}\right\}\cup \left\{\right\{2n+4,2n+5\}:n\in X\}and\hfill \\ \hfill R& =\{3,6,8,10,\dots \}.\hfill \end{array}$$

Then we define

$$\alpha =\left(\begin{array}{cccccccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& \dots \\ 4& 4& 8& 9& 9& 12& 13& 16& 17& 20& 21& 24& 25& \dots \end{array}\right),$$

$$\beta =\left(\begin{array}{cccccccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& \dots \\ 9& 9& 6& 7& 7& 3& 4& 3& 5& 6& 7& 10& 11& \dots \end{array}\right).$$

Thus, $\alpha ,\beta \in T\backslash F$ and

$$\begin{array}{cc}\hfill ▾\alpha & =\left\{\right\{4\},\{8,9\},\{12,13\},\{16,17\},\{20,21\},\{24,25\},\dots \},\hfill \\ \hfill {▾}^Y\alpha & =\left\{\right\{8,9\},\{12,13\},\{16,17\},\{20,21\},\{24,25\},\dots \},\hfill \\ \hfill ▾\beta & =\left\{\right\{9\},\{6,7\},\{3,4\},\{3,5\},\{10,11\},\{14,15\},\{18,19\},\dots \},\hfill \\ \hfill {▾}^Y\beta & =\left\{\right\{6,7\},\{3,4\},\{3,5\},\{10,11\},\{14,15\},\{18,19\},\dots \}.\hfill \end{array}$$

It is clear that $ker\left(\alpha \right)\ne ker\left(\beta \right)$. Therefore, α and β are not $\mathcal{D}$-related by Theorem 3. However, we can define $\varphi :Y\beta \to X\alpha \cup R\left(\alpha \right)$ and $\phi :Y\alpha \to X\beta \cup R\left(\beta \right)$ as follows:

$$\begin{array}{ccc}\left(\right\{3,4\left\}\right)\varphi =\{3,4\},\hfill & \left(\right\{3,5\left\}\right)\varphi =\{3,4\},\hfill & \left(\right\{6,7\left\}\right)\varphi =\{8,9\},\hfill \\ \left(\right\{10,11\left\}\right)\varphi =\{12,13\},\hfill & \left(\right\{14,15\left\}\right)\varphi =\{16,17\},\hfill & \left(\right\{18,19\left\}\right)\varphi =\{20,21\},\dots \hfill \\ \left(\right\{8,9\left\}\right)\phi =\{8,9\},\hfill & \left(\right\{12,13\left\}\right)\phi =\{6,7\},\hfill & \left(\right\{16,17\left\}\right)\phi =\{3,4\},\hfill \\ \left(\right\{20,21\left\}\right)\phi =\{3,5\},\hfill & \left(\right\{24,25\left\}\right)\phi =\{10,11\},\hfill & \left(\right\{28,29\left\}\right)\phi =\{14,15\},\dots .\hfill \end{array}$$

Both φ and ϕ satisfy the required properties of Theorem 4. Therefore, α and β are $\mathcal{J}$-related.

However, if X is a finite set, then T is a finite semigroup and it is periodic. Hence $\mathcal{D}=\mathcal{J}$ in this case.