On Centralizers of Idempotents with Restricted Range

Abstract

This study delves into the structure and properties of left inverse zero divisor bands within semigroups, identifying their maximal forms and broadening the theoretical landscape of semigroup analysis. A significant focus is placed on the automorphisms of a semigroup S of centralizers of idempotent transformations with restricted range, revealing that these automorphisms are inner ones and induced by the units of S. Additionally, we establish that the automorphism group $\mathrm{Aut}\left(S\right)$ is isomorphic to $U_S$, the group of units of S. These findings extend previous results on semigroups of transformations, enhancing their applicability and providing a more unified theory. The practical implications of this work span multiple fields, including automata theory, coding theory, cryptography, and graph theory, offering tools for more efficient algorithms and models. By simplifying complex concepts and providing a solid foundation for future research, this study makes significant contributions to both theoretical and applied mathematics.

1. Introduction

The study of automorphisms of centralizers of idempotents in semigroups is a highly intricate and specialized field within algebra. The subject incorporates aspects of semigroup theory, ring theory, and linear algebra, uncovering complex structural characteristics and symmetries.

The study of idempotents in semigroups originated from the initial research on semigroup theory. In their influential work, Clifford and Preston [1] presented a thorough and fundamental exploration of idempotents and their centralizers in different forms of semigroups, elucidating their features and classifications.

Within the realm of semigroups, the structure of centralizers varies greatly from its equivalents in ring theory since it lacks supplementary algebraic structures, such as additive inverses. This disparity requires distinct approaches and viewpoints. Green’s relations, which were introduced by Green in [2], play a crucial role in comprehending the arrangement of elements around idempotents. They have a significant impact on the examination of centralizers and the automorphisms associated with them.

Munn [3] expanded upon these concepts by examining the semigroup of partial transformations and their idempotents. Munn’s research on the semigroup of partial transformations on a set X yielded significant insights into the topologies of centralizers in transformation semigroups. Recent research has further investigated these fundamental concepts, delving into more specialized categories of semigroups and examining the characteristics of their idempotent centralizers. The study of inverse semigroups, which extend the concept of groups by including partial symmetries, has been a central focus. Petrich [4] conducted a comprehensive investigation on inverse semigroups, emphasizing the significance of idempotents and their centralizers in these structures.

In [5], Howie presented a contemporary viewpoint on the categorization and examination of idempotents in different semigroups. He highlighted the significance of this research for centralizers and their automorphisms. Computational approaches have proven indispensable in recent studies. By employing algorithms and computational tools, researchers such as Easdown and Okninski [6] have made substantial progress in automating the process of identifying and categorizing automorphisms of centralizers in complicated semigroups. This advancement has greatly contributed to the comprehension of the structural characteristics of these semigroups.

The consequences of comprehending automorphisms of centralizers of idempotents in semigroups are significant. Within the realm of pure mathematics, these investigations contribute to the wider categorization of semigroups and their inherent symmetries. Within the field of applied mathematics, namely in the domains of computation theory and automata theory, the knowledge acquired about structures can have a significant impact on the development of efficient algorithms and the examination of computational processes. The investigation of automorphisms of centralizers of idempotents in semigroups is a developing area of study, originating from classical semigroup theory and growing with the use of contemporary computer methods.

The study of fixed points of principal bundles over algebraic curves represents a rich and intricate area of mathematical research. It bridges various domains such as algebraic geometry, topology, and mathematical physics, offering profound insights into the structure and behavior of moduli spaces. By examining these fixed points, one can uncover important invariants and geometric properties that are pivotal in understanding the underlying algebraic curves and the principal bundles themselves.

The concept of automorphisms plays a significant role in this context. Automorphisms of a principal bundle are transformations that map the bundle to itself while preserving its structure. Studying these automorphisms, particularly those that fix certain points, can reveal symmetries and invariants of the moduli space. These fixed points under automorphisms often correspond to special geometric configurations and can lead to a deeper understanding of the moduli space’s topology and geometry.

Additionally, studying automorphisms of the centralizers of idempotents with limited range holds significant importance in addressing fixed-point problems in principal and Higgs bundles over algebraic curves. These automorphisms serve as a key tool for systematically exploring invariant configurations within the fundamental objects of algebraic geometry and gauge theory. By focusing our investigation on the centralizers of idempotents, we can efficiently navigate through intricate algebraic interactions, thereby facilitating the analysis of fixed points and revealing profound insights into the symmetry, stability, and geometric properties of the bundles.

This approach not only enhances our understanding of these complex structures but also broadens the spectrum of potential applications in both theoretical and applied mathematics. By elucidating the role of automorphisms in the context of centralizers of idempotents, we can unlock new avenues for research and discovery, ultimately contributing to advancements in various mathematical disciplines and their practical implementations.

The works of Clifford, Preston, Green, and Munn have laid a strong foundation, while the recent improvements by Petrich, Howie, and Easdown have enhanced the comprehension and expanded the practical uses of these concepts. Further investigation in this field holds the potential to reveal additional knowledge about the structure and practical uses, hence strengthening its significance in modern algebra.

Consider an arbitrary non-empty set A and let $\mathcal{T}\left(A\right)$ be the semigroup that consists of maps from A to itself, where composition serves as the semigroup operation. In [5], it was established that any semigroup S can be embedded into a semigroup of transformations, making transformation semigroups prototypes for semigroups. The study of semigroups, specifically transformation semigroups, has a substantial impact on several mathematical disciplines, such as automata, language, combinatorics, and the theory of machines. An automorphism $\psi$ of a semigroup S is deemed inner if there is an $\alpha \in \mathcal{S}\left(A\right)$, the group of all bijections on A, such that $\psi \left(u\right)=\alpha u{\alpha }^{-1}$ for all $u\in S$. The study of automorphisms of transformation semigroups is crucial in the exploration of semigroup automorphisms, with significant prior research in this domain. Schreier [7] and Mal’cev [8] showed that every $\psi \in \mathrm{Aut}\left(\mathcal{T}\right(A\left)\right)$ is such that $\psi \left(u\right)=\alpha u{\alpha }^{-1}$ for all $u\in \mathcal{T}\left(A\right)$, where $\alpha \in \mathcal{S}\left(A\right)$, and $\mathrm{Aut}\left(\mathcal{T}\right(A\left)\right)\cong \mathcal{S}\left(A\right)$. Sutov [9] and Magil [10] obtained similar findings for the partial transformation semigroup $\mathcal{P}\left(A\right)$, while Liber [11] obtained similar results for $\mathcal{I}\left(A\right)$, which consists of all partial one–one mappings on A. Gluskin [12] presented more illustrations, elucidating the automorphisms of the endomorphism monoid of a vector space. Sullivan [13] and Levi [14,15] extended the findings of Liber, Schreier, Sutov, and Mal’cev by proposing the notion of $\mathcal{S}\left(A\right)$-normality. In a more recent study, Mir et al. [16] expanded upon Sullivan’s findings regarding ${\mathcal{P}}_{\mathcal{M}}\left(X\right)$, which refers to the partial monotone transformation posemigroups. Shah et al. [17] extended the inner automorphism theorem from groups to monoids, introducing the concept of nearly complete semigroups. They also provided several necessary and sufficient conditions for semigroups to be nearly complete.

Consider a non-empty subset B of A. Then, the set $\mathcal{T}(A,B)=\{\alpha \in T(A):\alpha (A)\subseteq B\}$ forms a subsemigroup of $\mathcal{T}\left(A\right)$. For an equivalence relation $\sigma$ on A, let $\widehat{\sigma }=\sigma \cap (B\times B)$ represent the restriction of $\sigma$ to B, and R be a transversal of the partition $B/\widehat{\sigma }$. Consider a semigroup

$$T(A,B,\sigma ,R)=\{\alpha \in \mathcal{T}(A,B):\alpha (R)\subseteq R\mathrm{and}(a,b)\in \sigma \Rightarrow (\alpha \left(a\right),\alpha \left(b\right))\in \sigma \},$$

with $\sigma$ denoting an equivalence relation on A, and R being the transversal of the partition $B/\widehat{\sigma }$.

In this paper, our aim is to describe the automorphism group of $T(A,B,\sigma ,R)$ (for arbitrary $\sigma$ and R). Our results extend the findings of Symons [18,19] and Araujo and Konieczny [20]. We determine the automorphisms of $\mathcal{T}(A,B,\sigma ,R)$ and prove that they are inner, where the conjugating element is chosen from $\mathcal{S}(A,B,\sigma ,R)$, where $\mathcal{S}(A,B,\sigma ,R)=\{g\in S(A,B):g(R)=R\mathrm{and}(a,b)\in \sigma \iff (g\left(a\right),g\left(b\right))\in \sigma \}$, with $S(A,B)=\{g\in S\left(A\right):g{|}_B\in S\left(B\right)\}$. Furthermore, we demonstrate that $\mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$ is isomorphic to $\mathcal{S}(A,B,\sigma ,R)$. These results extend those obtained for $\mathcal{T}(A,B)$ and $T(A,\sigma ,R)$, considering $\sigma$ as the identity relation on A and $A=B$, respectively.

The organization of the paper is as follows. In Section 1, we prove that the semigroup $\mathcal{T}(A,B,\sigma ,R)$ in general does not contain an identity. In Section 2, we describe the group of units of $\mathcal{S}(A,B,\sigma ,R)$, and we prove that the sets ${\mathcal{F}}^1$, ${\mathcal{F}}^2$ and $\mathcal{T}(A,2,\sigma ,R)$ are characteristic in $\mathcal{T}(A,B,\sigma ,R)$. In Section 3, we introduce left inverse zero divisor bands and determine the maximal left inverse zero divisor bands. Finally, in Section 4, we use the results of the previous sections to prove the main theorems of the paper: the automorphisms of $\mathcal{T}(A,B,\sigma ,R)$ are the inner automorphisms induced by the units of $\mathcal{T}(A,B,\sigma ,R)$, and $\mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$ is isomorphic to the group of units of $\mathcal{T}(A,B,\sigma ,R)$.

2. Semigroup $\mathit{T}(\mathit{A},\mathit{B},\mathit{\sigma },\mathit{R})$ and Centralizers of Idempotents

In general, the semigroup $T(A,B,\sigma ,R)$ does not contain an identity, as illustrated in the following example.

Example 1. 

Let $A=\{a_1,a_2,\dots ,a_6\}$, $B=\{a_1,a_3\}$, and $A/\sigma =\{\{a_1,a_2\},\{a_3,a_4\},\{a_5,a_6\}\}$. Additionally, let $R=\{a_1,a_3\}$, and $B/\widehat{\sigma }=\{\left\{a_1\right\},\left\{a_3\right\}\}$. Suppose ι is an identity element of $T(A,B,\sigma ,R)$. Consider an element α of $T(A,B,\sigma ,R)$ defined by

$$\alpha =\left(\begin{array}{cccccc}{a}_1& a_2& a_3& a_4& a_5& a_6\\ \\ a_1& a_1& a_1& a_1& a_3& a_3\end{array}\right).$$

We analyze the action of α on the identity element ι applied to $a_5$:

$$\alpha \left(\iota \left(a_5\right)\right)=\left(\alpha \iota \right)\left(a_5\right)=\alpha \left(a_5\right)=a_3.$$

This implies that $\iota \left(a_5\right)\in \{a_5,a_6\}$, since ι must map $a_5$ to an element within its equivalence class under σ which is $\{a_5,a_6\}$. However, this leads to a contradiction because $B=\{a_1,a_3\}$, and neither $a_5$ nor $a_6$ are elements of B. Hence, the assumption that $T(A,B,\sigma ,R)$ possesses an identity element must be reconsidered.

Therefore, we conclude that $T(A,B,\sigma ,R)$ is a semigroup with an adjoined identity element, implying that the identity element does not inherently exist within the structure and must be externally appended contradicts their absence in B. Hence, we assume that $T(A,B,\sigma ,R)$ is a semigroup with an adjoined identity.

Consider the subset of $S(A,B)$:

$$\mathcal{S}(A,B,\sigma ,R)=\{g\in S(A,B):g(R)=R\mathrm{and}(a,b)\in \sigma \iff (g\left(a\right),g\left(b\right))\in \sigma \}.$$

Clearly, id_A$\in \mathcal{S}(A,B,\sigma ,R)$. Also, $\mathcal{S}(A,B,\sigma ,R)$ is closed under composition and inverses, thus forming a group. Now, we prove the following theorem.

Theorem 1. 

For $\alpha \in \mathcal{T}(A,B,\sigma ,R)$ and $a\in A$ with ${\left[a\right]}_{\sigma }\cap B\ne \varnothing$, if $\alpha \in \mathcal{T}(A,B,\sigma ,R)\cap S(A,B)$, then $\alpha \in \mathcal{S}(A,B,\sigma ,R)$.

Proof. 

Suppose $\alpha \in \mathcal{T}(A,B,\sigma ,R)\cap S(A,B)$. Then, $\alpha \in \mathcal{T}(A,B,\sigma ,R)$ and $\alpha \in S(A,B)$, implying $\alpha \left(R\right)\subseteq R$ and $(a,a^{\prime })\in \sigma$, which further implies $(\alpha \left(a\right),\alpha \left(a^{\prime }\right))\in \sigma$. We need to show that $R\subseteq \alpha \left(R\right)$ and $(\alpha \left(a\right),\alpha \left(a^{\prime }\right))\in \sigma$ implies $(a,a^{\prime })\in \sigma$. Let $t\in R$. Since $\alpha \in S(A,B)$, ${\alpha |}_B$ is a bijection, so there exists some $s\in B$ such that $\alpha \left(s\right)=t\in R$. Clearly, $(s,r_s)\in \widehat{\sigma }$, implying $(\alpha \left(s\right), \alpha \left(r_s\right))\in \widehat{\sigma }\subseteq \sigma$, and thus $(\alpha \left(s\right), \alpha \left(r_s\right))\in \sigma$. As both $\alpha \left(s\right)$ and $\alpha \left(r_s\right)$ belong to R and are $\sigma$-related, we have $\alpha \left(s\right)=\alpha \left(r_s\right)$. Hence, $t=\alpha \left(s\right)=\alpha \left(r_s\right)\in \alpha \left(R\right)$. Therefore, $\alpha \left(R\right)=R$.

Let $a,a^{\prime }\in A$, and suppose $(\alpha \left(a\right), \alpha \left(a^{\prime }\right))\in \sigma$. Then,

$$\begin{array}{cc}\hfill & r_{\alpha \left(a\right)}=r_{\alpha \left(a^{\prime }\right)}\hfill \\ \hfill \Rightarrow & \alpha \left(r_a\right)=\alpha \left(r_{a^{\prime }}\right)\hfill \\ \hfill \Rightarrow & r_a=r_{a^{\prime }}\left(\mathrm{as}\alpha \mathrm{is} \mathrm{bijective}\right)\hfill \\ \hfill \Rightarrow & (a,a^{\prime })\in \sigma .\hfill \end{array}$$

□

3. Some Subsemigroups of $\mathcal{T}(\mathit{A},\mathit{B},\mathit{\sigma },\mathit{R})$

In this section, we establish key notations and essential facts for our investigation, beginning with the following lemma.

Lemma 1. 

Let S be a subsemigroup of $\mathcal{T}(A,B,\sigma ,R)$ containing a constant. For $h\in S$, h is constant ⇔$hfh=h$ for all $f\in S$.

Proof. 

Let g be a constant function defined on the set S. We can select an element $a^{\prime }\in A$ such that the function $h\left(a\right)$ maps every element $a\in A$ to $a^{\prime }$. Therefore, we may conclude that $hfh\left(a\right)=hf\left(a^{\prime }\right)=h\left(f\left(a^{\prime }\right)\right)=a^{\prime }=h\left(a\right)$, which means that $hfh=h$ for all $f\in S$. On the other hand, let us suppose that $h\in S$ meets the condition $hfh=h$ for every $f\in S$. Since this property applies to all functions f in the set S, including the constant mapping $f_{a^{\prime }}$, we can conclude that $h{f}_{a^{\prime }}h=h$. Hence, for any element a belonging to the set A, we have $h\left(a\right)=h{f}_{a^{\prime }}h\left(a\right)=h\left(a^{\prime }\right)=f_{h\left(a^{\prime }\right)}$, which demonstrates a fixed mapping as needed. □

By $\mathcal{K}\left(S\right)$, we mean the collection of all constant mappings of a semigroup S, and ${\alpha |}_U$ as the restriction of $\alpha$ to U. We now state the following lemma.

Lemma 2. 

Let S be a subsemigroup of $\mathcal{T}(A,B,\sigma ,R)$. For $\phi \in Aut\left(S\right)$, ${\phi |}_{\mathcal{K}\left(S\right)}=\mathcal{K}\left(S\right)$.

Proof. 

Consider $\phi \in \mathrm{Aut}\left(S\right)$ and by Lemma 1, we have for every $f\in S$:

$$\begin{array}{cc}\hfill f\in \mathcal{K}\left(S\right)& \iff f=fhf\mathrm{for}\mathrm{every}h\in S\hfill \\ & \iff \phi \left(f\right)=\phi \left(fhf\right)\mathrm{for}\mathrm{every}h\in S\hfill \\ & \iff \phi \left(f\right)=\phi \left(f\right)k\phi \left(f\right)\mathrm{for}\mathrm{every}k\in S\hfill \\ & \iff \phi \left(f\right)\in \mathcal{K}\left(S\right).\hfill \end{array}$$

Hence, ${\phi |}_{\mathcal{K}\left(S\right)}=\mathcal{K}\left(S\right)$. □

For any $k\ge 1$, we define $\mathcal{T}(A,k,\sigma ,R)$ as:

$$\mathcal{T}(A,k,\sigma ,R)=\{\alpha \in \mathcal{T}(A,B,\sigma ,R):|\mathrm{im}\left(\alpha \right)|=k\}.$$

Let $\alpha \in \mathcal{T}(A,2,\sigma ,R)$. Either $\mathrm{im}\left(\alpha \right)=\{r,b\}$ for some $r\in R$ and $b\in B\setminus R$ with $(r,b)\in \widehat{\sigma }$, or $\mathrm{im}\left(\alpha \right)=\{r,s\}$ for some $r,s\in R$. We define ${\mathcal{F}}^1=\{\alpha \in T(A,2,\sigma ,R):\mathrm{im}\left(\alpha \right)=\{r,b\}\}$ for $r\in R$ and $b\in B\setminus R$, and for $r,s\in R$, we define ${\mathcal{F}}^2=\{\alpha \in T(A,2,\sigma ,R):\mathrm{im}\left(\alpha \right)=\{r,s\}\}$. It is clear that ${\mathcal{F}}^1\cap {\mathcal{F}}^2=\varnothing$, and ${\mathcal{F}}^1\cup {\mathcal{F}}^2=T(A,2,\sigma ,R)$.

Define $\mathcal{L}\left(\alpha \right)=\{\beta \in S:\alpha =u\beta \mathrm{and}\beta =v\alpha \}$ for some $u,v\in S^1$, and $\mathcal{R}\left(\alpha \right)=\{\beta \in S:\alpha =\beta u^{\prime }\mathrm{and}\beta =\alpha v^{\prime }\}$ for some $u^{\prime },v^{\prime }\in S^1$. We state the following lemma.

Lemma 3. 

Let $\alpha \in {\mathcal{F}}^1$ and $\beta \in \mathcal{T}(A,B,\sigma ,R)$. Then,

$$\beta \in {\mathcal{F}}^1\iff \mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)\ne \varnothing .$$

Proof. 

Suppose $\mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)\ne \varnothing$. Then, there exists $\gamma \in \mathcal{T}(A,B,\sigma ,R)$ such that $\gamma \in \mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)$. Hence, $\gamma \in \mathcal{R}\left(\alpha \right)$ and $\gamma \in \mathcal{L}\left(\beta \right)$. Therefore, there exist $u,v,u^{\prime },v^{\prime }\in {\mathcal{T}}^1(A,B,\sigma ,R)$ such that $\alpha =u\gamma$, $\gamma =v\alpha$, $\beta =\gamma u^{\prime }$, and $\gamma =\beta v^{\prime }$. This implies $\alpha =u\beta v^{\prime }$ and $\beta =v\alpha u^{\prime }$.

We have $|im\left(\alpha \right)|=|im\left(u\beta v^{\prime }\right)|\le |im\left(\beta \right)|$ and $|im\left(\beta \right)|=|im\left(v\alpha u^{\prime }\right)|\le |im\left(\alpha \right)|$, implying $|im(\alpha \left)\right|=|im(\beta \left)\right|=2$. Moreover, as $\alpha \in {\mathcal{F}}^1$, $im\left(\alpha \right)=\{r,b\}$, where $r\in R$ and $b\in B\setminus R$, and $(r,b)\in \widehat{\sigma }$. Thus, $\beta =v\alpha u^{\prime }$ for some $v,u^{\prime }\in \mathcal{T}{(A,B,\sigma ,R)}^1$, and $im\left(\beta \right)=im\left(v\alpha u^{\prime }\right)\subseteq \{v\left(r\right),v\left(b\right)\}$ (since $\alpha \in {\mathcal{F}}^1$). Since $v\in \mathcal{T}(A,B,\sigma ,R)$, $(v\left(r\right),v\left(b\right))\in \widehat{\sigma }$, implying $\left|\right\{v\left(r\right),v\left(b\right)\}\cap R|=1$, and so $\beta \in {\mathcal{F}}^1$.

Conversely, let $\beta \in {\mathcal{F}}^1$. Then, $im\left(\alpha \right)=\{r,b\}$ and $im\left(\beta \right)=\{s,b^{\prime }\}$ for $r,s\in R$ and $b,b^{\prime }\in B\setminus R$ with $(r,b)\in \widehat{\sigma }$ and $(s,b^{\prime })\in \widehat{\sigma }$. Let $X=\{z\in A\mid \beta (z)=s\}$ and $X^{\prime }=\{z^{\prime }\in A\mid \beta \left(z^{\prime }\right)=b^{\prime }\}$. Define $u,v\in \mathcal{T}(A,B)$ as follows: $u(A\setminus \left\{b^{\prime }\right\})=r$ and $u\left(b^{\prime }\right)=b$, $v(A\setminus \{b\left\}\right)=s$ and $v\left(b\right)=b^{\prime }$. Also, define d such that $d\left(X\right)=r$ and $d\left(X^{\prime }\right)=b$. By the definition of X and $X^{\prime }$ and the construction of u and v, we have $u,v\in \mathcal{T}(A,B,\sigma ,R)$. Now, if $z\in X$, then $\beta \left(z\right)=s$ and $vd\left(z\right)=v\left(r\right)=s$, and if $z\in X^{\prime }$, then $\beta \left(z\right)=b^{\prime }$ and $vd\left(z\right)=v\left(b\right)=b$. Thus, $\beta =vd$. Again, if $z\in X$, then $d\left(z\right)=r$ and $u\beta \left(z\right)=u\left(s\right)=r$. If $z\in X^{\prime }$, then $d\left(z\right)=b$ and $u\beta \left(z\right)=u\left(b^{\prime }\right)=b$. Thus, $d=u\beta$. Hence, $d\in \mathcal{L}\left(\beta \right)$.

Now, define the set $Y=\{z\in A\mid \alpha (z)=r\}$ and $Y^{\prime }=\{z^{\prime }\in A\mid \alpha \left(z^{\prime }\right)=b\}$. We define $c,c^{\prime }\in \mathcal{T}(A,B)$ as follows: $c\left(Y\right)=s$ and $c\left(Y^{\prime }\right)=x^{\prime }$ for some $x^{\prime }\in X^{\prime }$, $c^{\prime }\left(X\right)=r$ and $c^{\prime }\left(X^{\prime }\right)=y^{\prime }$ for some $y^{\prime }\in Y^{\prime }$. Clearly, $c,c^{\prime }\in \mathcal{T}(A,B,\sigma ,R)$ by definition of Y and $Y^{\prime }$. Now, let $z\in Y$. We have $\alpha \left(z\right)=r$ and $dc\left(z\right)=d\left(s\right)=r$. If $z\in Y^{\prime }$, then $\alpha \left(z\right)=b$ and $dc\left(z\right)=d\left(x^{\prime }\right)=b$. Thus, $\alpha =dc$. Again, for $z\in X$, we have $d\left(z\right)=r$ and $\alpha c^{\prime }\left(z\right)=\alpha \left(r\right)=r$. Otherwise, $d\left(z\right)=b$ and $\alpha c^{\prime }\left(z\right)=\alpha \left(y^{\prime }\right)=b$. Hence, $d=\alpha c^{\prime }$. Therefore, $d\in \mathcal{R}\left(\alpha \right)$. Thus, $d\in \mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)$, as required. □

Lemma 4. 

Let $\alpha \in {\mathcal{F}}^2$ and $\beta \in \mathcal{T}(A,B,\sigma ,R)$. Then,

$$\beta \in {\mathcal{F}}^2\iff \mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)\ne \varnothing .$$

Proof. 

Let $\mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)\ne \varnothing$. Then, there exists $\gamma \in \mathcal{T}(A,B,\sigma ,R)$ such that $\gamma \in \mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)$. This implies $\gamma \in \mathcal{R}\left(\alpha \right)$ and $\gamma \in \mathcal{L}\left(\beta \right)$. Thus, there exist $u,v,u^{\prime },v^{\prime }\in T^1(A,B,\sigma ,R)$ such that $\alpha =u\gamma$, $\gamma =v\alpha$, $\beta =\gamma u^{\prime }$, and $\gamma =\beta v^{\prime }$. Consequently, $\alpha =u\beta v^{\prime }$ and $\beta =v\alpha u^{\prime }$. Furthermore, $\left|\mathrm{im}\left(\alpha \right)\right|=|\mathrm{im}\left(u\beta v^{\prime }\right)|\le |\mathrm{im}\left(\beta \right)|$ and $\left|\mathrm{im}\left(\beta \right)\right|=|\mathrm{im}\left(v\alpha u^{\prime }\right)|\le |\mathrm{im}\left(\alpha \right)|$. This implies $\left|\mathrm{im}\right(\alpha \left)\right|=\left|\mathrm{im}\right(\beta \left)\right|=2$ since $\alpha \in {\mathcal{F}}^2$. Moreover, as $\alpha \in {\mathcal{F}}^2$, $\mathrm{im}\left(\alpha \right)=\{r,r^{\prime }\}$ where $r,r^{\prime }\in R$ and $(r,r^{\prime })\in \widehat{\sigma }$. Thus, $\mathrm{im}\left(\beta \right)=\mathrm{im}\left(v\alpha u^{\prime }\right)\subseteq \{v\left(r\right),v\left(r^{\prime }\right)\}$. Since $v\in \mathcal{T}(A,B,\sigma ,R)$, $(v\left(r\right),v\left(r^{\prime }\right))\in \widehat{\sigma }$, we have $\left|\right\{v\left(r\right),v\left(r^{\prime }\right)\}\cap R|=2$, meaning $\beta \in {\mathcal{F}}^2$.

Conversely, suppose $\beta \in {\mathcal{F}}^2$. Then, $\mathrm{im}\left(\alpha \right)=\{r,r^{\prime }\}$ and $\mathrm{im}\left(\beta \right)=\{s,s^{\prime }\}$ for some $r,r^{\prime },s,s^{\prime }\in R$. Let $X=\{z\in A|\beta \left(z\right)=s\}$ and $X^{\prime }=\{z\in A|\beta \left(z\right)=s^{\prime }\}$. Define $u,v,\in \mathcal{T}(A,B)$ as follows: $u(A\setminus \left\{s^{\prime }\right\})=r$ and $u\left(s^{\prime }\right)=r^{\prime }$; $v(A\setminus \left\{r^{\prime }\right\})=s$ and $v\left(r^{\prime }\right)=s^{\prime }$. Also, define $d\left(X\right)=r$ and $d\left(X^{\prime }\right)=r^{\prime }$. Clearly, $u,v\in T(A,B,\sigma ,R)$ by definition of X and $X^{\prime }$. Now, if $z\in X$, then $\beta \left(z\right)=s$ and $vd\left(z\right)=v\left(r\right)=s$. If $z\in X^{\prime }$, then $\beta \left(z\right)=s^{\prime }$ and $vd\left(z\right)=v\left(r^{\prime }\right)=s^{\prime }$, meaning $\beta =vd$. Also, if $z\in X$, then $d\left(z\right)=r$ and $u\beta \left(z\right)=u\left(s\right)=r$. If $z\in X^{\prime }$, then $d\left(z\right)=r^{\prime }$ and $u\beta \left(z\right)=u\left(s^{\prime }\right)=r^{\prime }$, implying $d=u\beta$. Hence, $d\in \mathcal{L}\left(\beta \right)$.

Now, define the sets $Y=\{z\in A|\alpha \left(z\right)=r\}$ and $Y^{\prime }=\{z\in A|\alpha \left(z\right)=r^{\prime }\}$. Define $c,c^{\prime }\in \mathcal{T}(A,B)$ as follows: $c\left(Y\right)=s$ and $c\left(Y^{\prime }\right)=x^{\prime }$ for some $x^{\prime }\in X^{\prime }$; $c^{\prime }\left(X\right)=r$ and $c^{\prime }\left(X^{\prime }\right)=y^{\prime }$ for some $y^{\prime }\in Y^{\prime }$. Clearly, $c,d\in \mathcal{T}(A,B,\sigma ,R)$ by definition of Y and $Y^{\prime }$. Let $z\in Y$. We have $\alpha \left(z\right)=r$ and $dc\left(z\right)=d\left(s\right)=r$. If $z\in Y^{\prime }$, then $\alpha \left(z\right)=r^{\prime }$ and $dc\left(z\right)=d\left(x^{\prime }\right)=r^{\prime }$. Thus, $\alpha =dc$. Also, for $z\in X$, we have $d\left(z\right)=r$ and $\alpha c^{\prime }\left(z\right)=\alpha \left(r\right)=r$. Otherwise, $d\left(z\right)=r^{\prime }$ and $\alpha c^{\prime }\left(z\right)=\alpha \left(y^{\prime }\right)=r^{\prime }$ implies $d=\alpha c^{\prime }$. Hence, $d\in \mathcal{R}\left(\alpha \right)$. Thus, $d\in \mathcal{R}\left(\alpha \right)\cap \mathcal{L}\left(\beta \right)$. □

Remark 1. 

By Lemma 3, it follows that ${\mathcal{F}}^1$ is either ∅ or is a $\mathcal{D}$-class of $\mathcal{T}(A,B,\sigma ,R)$, and by Lemma 4, the same holds for ${\mathcal{F}}^2$.

Lemma 5. 

Let $D_1$ and $D_2$ be two $\mathcal{D}$-classes of $\mathcal{T}(A,B,\sigma ,R)$, and let $\phi \in \mathit{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$. If $\phi \left(D_1\right)\cap D_2\ne \varnothing$, then $\phi \left(D_1\right)=D_2$.

Proof. 

Let $\alpha ,\beta \in \mathcal{T}(A,B,\sigma ,R)$; then, we have

$$\alpha \mathcal{D}\beta \iff \phi \left(\alpha \right)\mathcal{D}\phi \left(\beta \right)\mathrm{for} \mathrm{all}\phi \in \mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right).$$

Suppose $\phi \left(D_1\right)\cap D_2\ne \varnothing$, that is, there exists $d_1\in D_1$ such that $\phi \left(d_1\right)\in D_2$. Let $d_2\in D_1$. Since $d_1\mathcal{D}{d}_2\iff \phi \left(d_1\right)\mathcal{D}\phi \left(d_2\right)$, we have $\phi \left(d_2\right)\in D_2$ as $\phi \left(d_1\right)\in D_2$. Thus, $\phi \left(D_1\right)\subseteq D_2$. For the reverse inclusion, $\phi \left(D_1\right)\cap D_2\ne \varnothing$ implies $D_1\cap {\phi }^{-1}\left(D_2\right)\ne \varnothing$. Let $d_1^{\prime }\in D_2$ be such that ${\phi }^{-1}\left(d_1^{\prime }\right)\in D_1$. Let $d_2^{\prime }\in D_2$. Since $d_1\mathcal{D}{d}_2\iff \phi \left(d_1\right)\mathcal{D}\phi \left(d_2\right)$, we have ${\phi }^{-1}\left(d_2^{\prime }\right)\in D_1$ as $\phi \left(d_1^{\prime }\right)\in D_1$. Thus, ${\phi }^{-1}\left(D_2\right)\subseteq D_1$. Hence, $\phi \left(D_1\right)=D_2$. □

Lemma 6. 

Let $\phi \in \mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$. Then, $\phi \left({\mathcal{F}}^1\right)\cap {\mathcal{F}}^2=\varnothing$ and $\phi \left({\mathcal{F}}^2\right)\cap {\mathcal{F}}^1=\varnothing$.

Proof. 

Assume that ${\mathcal{F}}^1\ne \varnothing$ and ${\mathcal{F}}^2\ne \varnothing$. Suppose $\phi \left({\mathcal{F}}^2\right)\cap {\mathcal{F}}^1\ne \varnothing$; then, there exists $\alpha \in {\mathcal{F}}^1$ such that $\phi \left(\alpha \right)\in \phi \left({\mathcal{F}}^2\right)$. Now, by Remark 1 and Lemma 5, we have $\phi \left({\mathcal{F}}^2\right)={\mathcal{F}}^1$. Let $r,r^{\prime }\in R$ with $r\ne r^{\prime }$, and choose $\beta \in \mathcal{T}(A,B,\sigma ,R)$ such that

$$\beta =\left(\begin{array}{cc}{\left[r\right]}_{\sigma }& A\setminus {\left[r\right]}_{\sigma }\\ \\ r^{\prime }& r\end{array}\right).$$

Since R contains a single element of each $\sigma -$class, $r^{\prime }\notin {\left[r\right]}_{\sigma }$. Clearly, $\beta \in {\mathcal{F}}^2$, and ${\beta }^2=\left(\begin{array}{cc}{\left[r\right]}_{\sigma }& A\setminus {\left[r\right]}_{\sigma }\\ r& r^{\prime }\end{array}\right)\in {\mathcal{F}}^2.$ Let $\gamma =\phi \left(\beta \right)$; then, $\gamma \in {\mathcal{F}}^1$, and ${\gamma }^2=\phi \left(\beta \right)\phi \left(\beta \right)=\phi \left({\beta }^2\right)\ne \phi \left(\beta \right)=\gamma$, meaning $\gamma$ is not idempotent, and ${\gamma }^2\in {\mathcal{F}}^1$. Since $\gamma \in {\mathcal{F}}^1$, we have $\gamma =\left(\begin{array}{cc}{A}_1& A\setminus A_1\\ \\ r^{″}& b\end{array}\right)$, where $A_1\subseteq A$, $r^{″}\in R$, and $b\in B\setminus R$ with $(r^{″},b)\in \widehat{\sigma }$. Since ${\gamma }^2\ne \gamma$, $(r^{″},b)\in \widehat{\sigma }$, and $\gamma \in \mathcal{T}(A,B,\sigma ,R)$, we have $b\in A_1$ and so ${\gamma }^2\left(a\right)=r^{″}$ for every $a\in A$, meaning $\left|\mathrm{im}\right({\gamma }^2\left)\right|=1$. This is a contradiction as ${\gamma }^2\in {\mathcal{F}}^1$. Hence, $\phi \left({\mathcal{F}}^2\right)\cap {\mathcal{F}}^1=\varnothing$. Similarly, we can show that $\phi \left({\mathcal{F}}^1\right)\cap {\mathcal{F}}^2=\varnothing$. □

Lemma 7. 

For $\alpha \in {\mathcal{F}}^1$, $\beta \in {\mathcal{F}}^2$, and $\gamma \in T(A,B,\sigma ,R)$, we have $\gamma \alpha \notin {\mathcal{F}}^1$ and $\gamma \beta \notin {\mathcal{F}}^2$.

Proof. 

For $\alpha \in {\mathcal{F}}^1$ with $\mathrm{im}\left(\alpha \right)=\{r,b\}$, where $r\in R$ and $(r,b)\in \sigma$, we observe that $\mathrm{im}\left(\gamma \alpha \right)=\left\{\gamma \right(r),\gamma (b\left)\right\}$ and, given $\gamma \in \mathcal{T}(A,B,\sigma ,R)$, $\left(\gamma \right(r),\gamma (b\left)\right)\in \sigma$. Thus, $\left|\mathrm{im}\right(\gamma \alpha )\cap R|=1$, leading to $\gamma \alpha \notin {\mathcal{F}}^2$. Similarly, $\gamma \beta \notin {\mathcal{F}}^2$. □

Theorem 2. 

For $\phi \in Aut\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$, the following properties hold:

(i) 

$\phi \left({\mathcal{F}}^1\right)={\mathcal{F}}^1$;

(ii) 

$\phi \left({\mathcal{F}}^2\right)={\mathcal{F}}^2$;

(iii) 

$\phi \left(T\right(A,2,\sigma ,R\left)\right)=T(A,2,\sigma ,R).$

Proof. 

Assuming ${\mathcal{F}}^1\ne \varnothing$, let $\phi \in \mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R)$. Consider $\beta \in {\mathcal{F}}^1$. Since $\phi$ is surjective, $\beta =\phi \left(\alpha \right)$ for some $\alpha \in \mathcal{T}(A,B,\sigma ,R)$. By Lemma 3, $\left|\mathrm{im}\right(\alpha \left)\right|\ge 2$, so we can choose $\gamma \in \mathcal{T}(A,B,\sigma ,R)$ such that $\gamma \alpha \in T(A,2,\sigma ,R)$. We then have

$$\left|\mathrm{im}\right(\phi \left(\gamma \alpha \right)\left)\right|=\left|\mathrm{im}\right(\phi \left(\gamma \right)\phi \left(\alpha \right)\left)\right|\le \left|\mathrm{im}\right(\phi \left(\alpha \right)\left)\right|=\left|\mathrm{im}\right(\beta \left)\right|=2.$$

Therefore, $\phi \left(\gamma \alpha \right)\in T(A,2,\sigma ,R)$. As $\phi \left(\alpha \right)=\beta \in {\mathcal{F}}^1$ and $\phi \left(\gamma \alpha \right)=\phi \left(\gamma \right)\phi \left(\alpha \right)\notin {\mathcal{F}}^2$, we have $\phi \left(\gamma \alpha \right)\in {\mathcal{F}}^1$. Consequently, since $\phi \left({\mathcal{F}}^2\right)\cap {\mathcal{F}}^1=\varnothing$, $\gamma \alpha \notin {\mathcal{F}}^2$. This implies $\gamma \alpha \notin {\mathcal{F}}^1$ and $\phi \left({\mathcal{F}}^1\right)\cap {\mathcal{F}}^1\ne \varnothing$. By Lemma 6, we have $\phi \left({\mathcal{F}}^1\right)={\mathcal{F}}^1$. Similarly, by exchanging ${\mathcal{F}}^1$ and ${\mathcal{F}}^2$, we can show that $\phi \left({\mathcal{F}}^2\right)={\mathcal{F}}^2$.

Since $\mathcal{T}(A,2,\sigma ,R)={\mathcal{F}}^1\cup {\mathcal{F}}^2$, we have

$$\begin{array}{cc}\hfill \phi \left(\mathcal{T}\right(A,2,\sigma ,R\left)\right)& =\phi ({\mathcal{F}}^1\cup {\mathcal{F}}^2)\hfill \\ \hfill & =\phi \left({\mathcal{F}}^1\right)\cup \phi \left({\mathcal{F}}^2\right)\hfill \\ \hfill & ={\mathcal{F}}^1\cup {\mathcal{F}}^2\hfill \\ \hfill & =\mathcal{T}(A,2,\sigma ,R).\hfill \end{array}$$

Hence the theorem. □

4. Left Inverse Zero Divisor Bands

A band is a semigroup in which every element is idempotent. In a band $\mathcal{B}$ with a zero element, 0, it is termed a zero divisor band if for all $l,m\in \mathcal{B}$, $l\ne m\Rightarrow lm=0$.

For $\alpha ,\beta \in {\mathcal{F}}^1$, we define $\alpha {\mathcal{L}}^{{\mathcal{F}}^1}\beta$ if and only if there exist $\gamma ,\delta \in {\mathcal{F}}^1$ such that $\alpha =\gamma \beta$ and $\beta =\delta \alpha$.

Lemma 8. 

For idempotents $e,e^{\prime }\in {\mathcal{F}}^1$, $e{\mathcal{L}}^{{\mathcal{F}}^1}{e}^{\prime }\iff Ker\left(e\right)=Ker\left(e^{\prime }\right)$.

Proof. 

If $e{\mathcal{L}}^{{\mathcal{F}}^1}{e}^{\prime }$, then there exist $u,v\in {\mathcal{F}}^1$ such that $e=u{e}^{\prime }$ and $e^{\prime }=ve$. Thus, $\mathrm{Ker}\left(e\right)=\mathrm{Ker}\left(u{e}^{\prime }\right)\subseteq \mathrm{Ker}\left(e^{\prime }\right)$. Similarly, $\mathrm{Ker}\left(e^{\prime }\right)=\mathrm{Ker}\left(ve\right)\subseteq \mathrm{Ker}\left(e\right)$, hence $\mathrm{Ker}\left(e\right)=\mathrm{Ker}\left(e^{\prime }\right)$.

Conversely, if $\mathrm{Ker}\left(e\right)=\mathrm{Ker}\left(e^{\prime }\right)$, since $e,e^{\prime }\in {\mathcal{F}}^1$ are idempotents, then $\mathrm{im}\left(e\right)=\{b,r\}$ and $\mathrm{im}\left(e^{\prime }\right)=\{b^{\prime },r^{\prime }\}$, where $b,b^{\prime }\in B\setminus R$, $r,r^{\prime }\in R$, and $e\left(b\right)=b$, $e^{\prime }\left(b^{\prime }\right)=b^{\prime }$. Suppose $e\left(r\right)=b^{\prime }$, then $e\left(b^{\prime }\right)=e\left(r\right)$, implying $(b^{\prime },r)\in \mathrm{Ker}\left(e\right)=\mathrm{Ker}\left(e^{\prime }\right)$. This implies $b^{\prime }=e^{\prime }\left(b^{\prime }\right)=e^{\prime }\left(r\right)\in R$, a contradiction. Thus, the partition induced by $\mathrm{Ker}\left(e\right)=\mathrm{Ker}\left(e^{\prime }\right)$ must have the form $\{\{b,b^{\prime },\cdots \},\{r,r^{\prime },\cdots \}\}$, leading to:

$$e=\left(\begin{array}{cc}\{b,b^{\prime },...\}& \{r,r^{\prime },...\}\\ \\ b& r\end{array}\right)\mathrm{and}{e}^{\prime }=\left(\begin{array}{cc}\{b,b^{\prime },\cdots \}& \{r,r^{\prime },\cdots \}\\ \\ b^{\prime }& r^{\prime }\end{array}\right).$$

Consider $\alpha =\left(\begin{array}{cc}\left\{b\right\}& A\setminus \left\{b\right\}\\ \\ b^{\prime }& r^{\prime }\end{array}\right)$. Clearly, $\alpha \in {\mathcal{F}}^1$; thus, $e^{\prime }=\alpha e$. Similarly, consider $\beta =\left(\begin{array}{cc}\left\{b^{\prime }\right\}& A\setminus \left\{b^{\prime }\right\}\\ \\ b& r\end{array}\right)$. It is evident that $\beta \in {\mathcal{F}}^1$; thus, $e=\beta e^{\prime }$. Hence, $e{\mathcal{L}}^{{\mathcal{F}}^1}{e}^{\prime }$. □

Consider a zero divisor band $\mathcal{B}$ with a zero element denoted as 0. A zero divisor band in ${\mathcal{F}}^1$ is defined as a set $\mathcal{B}$ such that $\mathcal{B}\setminus \left\{0\right\}$ is a subset of ${\mathcal{F}}^1$ and 0 belongs to the semigroup $\mathcal{T}(A,1,\sigma ,R)$. A zero divisor band $\mathcal{B}$ is considered a left inverse if, for every non-zero element e in $\mathcal{B}$, the ${\mathcal{L}}^{{\mathcal{F}}^1}$-class of e in ${\mathcal{F}}^1$ contains just one idempotent, which is e itself.

For every $b\in B\setminus R$ and $r\in R$ such that $b\widehat{\sigma }r$, ${\phi }_{\{b,r\}}$ denotes the element of $T(A,B,\sigma ,R)$ defined by ${\phi }_{\{b,r\}}\left(b\right)=b$ and ${\phi }_{\{b,r\}}(A\setminus \left\{b\right\})=\left\{r\right\}$. For $r\in R$, $A_r$ represents the constant element of $\mathcal{T}(A,B,\sigma ,R)$. Clearly, ${\phi }_{\{b,r\}}^2={\phi }_{\{b,r\}}$ in ${\mathcal{F}}^1$. For $r\in R$, define the set

$${\mathcal{B}}_r=\{{\phi }_{\{b,r\}};b\in {\left[r\right]}_{\widehat{\sigma }}\}\cup \left\{A_r\right\}.$$

Note that if ${\left[r\right]}_{\widehat{\sigma }}=\left\{r\right\}$ then ${\mathcal{B}}_r=\left\{A_r\right\}$.

Lemma 9. 

For $r\in R$, ${\mathcal{B}}_r$ is a maximal left inverse zero divisor band in ${\mathcal{F}}^1$.

Proof. 

If ${\phi }_{\{b,r\}},{\phi }_{\{b^{\prime },r\}}\in {\mathcal{B}}_r$, then ${\phi }_{\{b,r\}}{\phi }_{\{b,r\}}={\phi }_{\{b,r\}}$, and if $b\ne b^{\prime }$, then ${\phi }_{\{b,r\}}{\phi }_{\{b^{\prime },r\}}=A_r$. Thus, ${\mathcal{B}}_r$ is a zero divisor band with zero $A_r$. Since every ${\phi }_{\{b,r\}}\in {\mathcal{F}}^1$, ${\mathcal{B}}_r$ is a zero divisor band in ${\mathcal{F}}^1$. Now, we prove that ${\mathcal{B}}_r$ is left inverse.

Note that $\ker \left({\phi }_{\{b,r\}}\right)=\{\left\{b\right\},A\setminus \left\{b\right\}\}$. Suppose $e\in {\mathcal{F}}^1$ is an idempotent such that $e{\mathcal{L}}^{{\mathcal{F}}^1}{\phi }_{\{b,r\}}$. Then, by Lemma 8, $\ker \left(e\right)=\ker \left({\phi }_{\{b,r\}}\right)$, and so

$$e=\left(\begin{array}{cc}\left\{b\right\}& A\setminus \left\{b\right\}\\ \\ a& s\end{array}\right)$$

for some $s\in R$ and $a\in B\setminus R$. Since e is idempotent, $e\left(b\right)=b$ implies $a=b$. Moreover, since $a\in {\left[s\right]}_{\widehat{\sigma }}$ and $a=b\in {\left[r\right]}_{\widehat{\sigma }}$, we have ${\left[s\right]}_{\widehat{\sigma }}={\left[r\right]}_{\widehat{\sigma }}$, and so $r=s$. Thus, ${\mathcal{B}}_r$ is left inverse.

Let ${\mathcal{B}}_r$ be contained in a left inverse zero divisor band $\mathcal{B}$ of ${\mathcal{F}}^1$. Then, $A_r\in \mathcal{B}$, so $A_r$ is zero in $\mathcal{B}$. Let $A_r\ne e\in \mathcal{B}$. Then, $e\in {\mathcal{F}}^1$, so

$$e=\left(\begin{array}{cc}M& N\\ \\ b& s\end{array}\right)$$

for some $s\in R$ and $b\in B\setminus R$. Since e is idempotent, $b\in M$ and $s\in N$. Suppose there is $a\in M$ such that $a\ne b$. Since $e\left(R\right)\subseteq R$ and $b\notin R$, this implies $M\cap R=\varnothing$. Thus, $r_a\in N$, so

$$e^{\prime }=\left(\begin{array}{cc}M& N\\ \\ a& r_a\end{array}\right)$$

is an idempotent in ${\mathcal{F}}^1$ such that $\mathrm{Ker}\left(e\right)=\mathrm{Ker}\left(e^{\prime }\right)$. This is a contradiction since $e\ne e^{\prime }$ and B is left inverse. It follows that $M=\left\{b\right\}$, and so $e={\phi }_{\{b,r\}}$. Since $A_r$ is the zero in B, ${\phi }_{\{b,r\}}{A}_r=A_r$. Thus, $s=r$ and so $e={\phi }_{\{b,r\}}\in {\mathcal{B}}_r$. Therefore, $B={\mathcal{B}}_r$, demonstrating that ${\mathcal{B}}_r$ is a maximal left inverse zero divisor band in ${\mathcal{F}}^1$. □

Lemma 10. 

Suppose $\mathcal{B}$ is a maximal left inverse zero divisor band in ${\mathcal{F}}^1$. Then, there exists $r\in R$ such that $\mathcal{B}={\mathcal{B}}_r$.

Proof. 

Let $r\in R$ be such that $A_r$ is zero in $\mathcal{B}$. Assume $A_r\ne e\in \mathcal{B}$ with im$\left(e\right)=\{b,s\}$ ($b\in B\setminus R,s\in R$). By Lemma 9, $e={\phi }_{\{b,r\}}$. Hence, $\mathcal{B}\subseteq {\mathcal{B}}_r$, and thus ${\mathcal{B}}_r=\mathcal{B}$, as ${\mathcal{B}}_r$ is maximal. □

Given that $\left|\mathrm{im}\right(\alpha \beta \left)\right|\le \left|\mathrm{im}\right(\beta \left)\right|$ for all $\alpha ,\beta \in \mathcal{T}(A,B)$, by Lemma 7, we conclude that ${\mathcal{F}}^1\cup T(A,1,\sigma ,R)$ is a subsemigroup of $\mathcal{T}(A,B,\sigma ,R)$. Now, we introduce the following lemma.

Lemma 11. 

For every automorphism ψ of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ and $r\in R$, there exists $s\in R$ such that $\psi \left({\mathcal{B}}_r\right)={\mathcal{B}}_s$.

Proof. 

Let $\psi \in \mathrm{Aut}({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))$ and ${\mathcal{B}}_r$ be a zero divisor band. Since $A_r$ is zero in ${\mathcal{B}}_r$, for any $\alpha \in {\mathcal{B}}_r$, we have $\psi \left(\alpha \right)\psi \left(A_r\right)=\psi \left(\alpha A_r\right)=\psi \left(A_r\right)$. Thus, $\psi \left(A_r\right)$ is zero in $\psi \left({\mathcal{B}}_r\right)$. By Lemma 2, $\psi \left(A_r\right)\in \mathcal{T}(A,1,\sigma ,R)$, so $\psi \left(A_r\right)=A_{r^{\prime }}$ for some $r^{\prime }\in R$. Suppose $A_s\in \psi \left({\mathcal{B}}_r\right)$ where $s\in R$. Then, $A_s{A}_{r^{\prime }}=A_{r^{\prime }}$, as $A_{r^{\prime }}$ is zero in ${\mathcal{B}}_r$. Hence, $s=A_s{A}_{r^{\prime }}\left(s\right)=A_{r^{\prime }}\left(s\right)=r^{\prime }$, and so $A_{r^{\prime }}=A_s$. Consequently, $\psi \left({\mathcal{B}}_r\right)\setminus A_{r^{\prime }}\in {\mathcal{F}}^1$, and thus $\psi \left({\mathcal{B}}_r\right)$ is a zero divisor band in ${\mathcal{F}}^1$.

Since $\psi \in \mathrm{Aut}({\mathcal{F}}^1\cup T(A,1,\sigma ,R))$ and $\psi \left(\mathcal{T}\right(A,1,\sigma ,R\left)\right)=\mathcal{T}(A,1,\sigma ,R)$, by the definition of ${\mathcal{L}}^{{\mathcal{F}}^1}$, we have $\alpha {\mathcal{L}}^{{\mathcal{F}}^1}\beta$ if and only if $\psi \left(\alpha \right){\mathcal{L}}^{{\mathcal{F}}^1}\psi \left(\beta \right)$ for all $\alpha ,\beta \in {\mathcal{F}}^1$. For every $e\in {\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$, e is idempotent in ${\mathcal{F}}^1$ if and only if $\psi \left(e\right)$ is idempotent in ${\mathcal{F}}^1$. Hence, for every $0\ne e^{\prime }\in \psi \left({\mathcal{B}}_r\right)$, the ${\mathcal{L}}^{{\mathcal{F}}^1}$ -class of $e^{\prime }$ has just one idempotent. Consequently, $\psi \left({\mathcal{B}}_r\right)$ is a left inverse zero divisor band in ${\mathcal{F}}^1$.

As $A_s$ is zero in $\psi \left({\mathcal{B}}_r\right)$, by the proof of Lemma 10, $\psi \left({\mathcal{B}}_r\right)\subseteq {\mathcal{B}}_s$. Hence, ${\mathcal{B}}_r\subseteq {\psi }^{-1}\left({\mathcal{B}}_s\right)$. Conversely, we have ${\psi }^{-1}\left({\mathcal{B}}_s\right)$ is a left inverse zero divisor band in ${\mathcal{F}}^1$, as by Lemma 9, ${\mathcal{B}}_r$ is a maximal left inverse zero divisor band in ${\mathcal{F}}^1$. Thus, ${\mathcal{B}}_r={\psi }^{-1}\left({\mathcal{B}}_s\right)$, and so ${\psi }^{-1}\left({\mathcal{B}}_r\right)={\mathcal{B}}_s$. □

5. Automorphism Group of $\mathcal{T}(\mathit{A},\mathit{B},\mathit{\sigma },\mathit{R})$

Let $\psi \in \mathrm{Aut}({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))$, where ${\mathcal{F}}^1$ is the set containing those elements $\alpha \in \mathcal{T}(A,B,\sigma ,R)$ with $\left|\mathrm{im}\right(\alpha \left)\right|=2$. Then, by Lemma 11, for every $b\in B\setminus R$, there exists $b^{\prime }\in B\setminus R$ such that $\psi \left({\phi }_{\{b,r_b\}}\right)={\phi }_{\{b^{\prime },r_{b^{\prime }}\}}$, and for every $r\in R$, there exists $r^{\prime }\in R$ such that $\psi \left(A_r\right)=A_{r^{\prime }}$.

We can thus define ${\alpha }^{\psi }\in \mathcal{T}\left(A\right)$ as follows:

$${\alpha }^{\psi }\left(a\right)=\left\{\begin{array}{cc}a,\hfill & \mathrm{if}a\in A\setminus B;\hfill \\ a^{\prime },\hfill & \mathrm{if}\psi \left({\phi }_{\{a,r_a\}}\right)={\phi }_{\{a^{\prime },r_{a^{\prime }}\}}\mathrm{for}a,a^{\prime }\in B\setminus R;\hfill \\ r^{\prime },& \mathrm{if}\psi \left(A_r\right)=A_{r^{\prime }}\mathrm{for}r,r^{\prime }\in R\mathrm{and}a\in R.\end{array}\right.$$

It is clear that ${\alpha }^{\psi }\in S\left(A\right)$.

Note that if $\psi \left({\phi }_{\{a,r_a\}}\right)={\phi }_{\{a^{\prime },r_{a^{\prime }}\}}$, then $\psi \left(A_{r_a}\right)=A_{r_{a^{\prime }}}$. Thus, $({\alpha }^{\psi }\left(a\right),{\alpha }^{\psi }\left(r_a\right))\in \sigma$.

Lemma 12. 

Let $\psi \in \mathrm{Aut}({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))$, then ${\alpha }^{\psi }\in \mathcal{S}(A,B,\sigma ,R)$.

Proof. 

We show that ${\alpha }^{\psi }{|}_B\in S\left(B\right)$. For this, let $b,b^{\prime }\in B\setminus R$. Suppose

$$\begin{array}{cc}\hfill b=b^{\prime }& \iff {\phi }_{\{b,r_b\}}={\phi }_{\{b^{\prime },r_{b^{\prime }}\}}\hfill \\ \hfill & \iff \psi \left({\phi }_{\{b,r_b\}}\right)=\psi \left({\phi }_{\{b^{\prime },r_{b^{\prime }}\}}\right)\hfill \\ \hfill & \iff {\phi }_{\{{\alpha }^{\psi }\left(b\right),r_{{\alpha }^{\psi }\left(b\right)}\}}={\phi }_{\{{\alpha }^{\psi }\left(b^{\prime }\right),r_{{\alpha }^{\psi }\left(b^{\prime }\right)}\}}\hfill \\ \hfill & \iff {\alpha }^{\psi }\left(b\right)={\alpha }^{\psi }\left(b^{\prime }\right).\hfill \end{array}$$

Thus, ${\alpha }^{\psi }$ is well-defined and injective. Since ${\psi }^{-1}$ is also an automorphism, there is $b\in B$ such that ${\psi }^{-1}\left({\phi }_{\{b^{\prime },r_{b^{\prime }}\}}\right)={\phi }_{\{b,r_b\}}$, implying $\psi \left({\phi }_{\{b,r_b\}}\right)={\phi }_{\{b^{\prime },r_{b^{\prime }}\}}$, that is, ${\alpha }^{\psi }\left(b\right)=b^{\prime }$. Therefore, ${\alpha }^{\psi }$ is surjective. Thus, ${\alpha }^{\psi }\in S(A,B)$.

Let $(b,b^{\prime })\in \sigma$. If $b,b^{\prime }\in R$, then $b=b^{\prime }$, so $({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(b^{\prime }\right))\in \sigma$. If $b\in A\setminus R$ and $b^{\prime }\in R$, then $b^{\prime }=r_b$, so $({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(b^{\prime }\right))=({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(r_b\right))\in \sigma$. Now, assume that $b,b^{\prime }\in B\setminus R$. Since $(b,b^{\prime })\in \sigma$, then $r_b=r_{b^{\prime }}$. By the observation before the lemma $({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(r_b\right))\in \sigma$ and $({\alpha }^{\psi }\left(b^{\prime }\right),{\alpha }^{\psi }\left(r_b\right))\in \sigma$. Since $\sigma$ is an equivalence, $({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(b^{\prime }\right))\in \sigma$. Let $b,b^{\prime }\in B$, and suppose $({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(b^{\prime }\right))\in \sigma$. Then,

$$\begin{array}{cc}\hfill & r_{{\alpha }^{\psi }\left(b\right)}=r_{{\alpha }^{\psi }\left(b^{\prime }\right)}\hfill \\ \hfill & \Rightarrow {\alpha }^{\psi }\left(r_b\right)={\alpha }^{\psi }\left(r_{b^{\prime }}\right)\hfill \\ \hfill & \Rightarrow r_b=r_{b^{\prime }}\left(\mathrm{as}{\alpha }^{\psi }\mathrm{is}\mathrm{bijective}\right)\hfill \\ \hfill & \Rightarrow (b,b^{\prime })\in \sigma .\hfill \end{array}$$

Thus, we have $(b,b^{\prime })\in \sigma \iff ({\alpha }^{\psi }\left(b\right),{\alpha }^{\psi }\left(b^{\prime }\right))\in \sigma$.

Now, we show that ${\alpha }^{\psi }\left(R\right)=R$. Let $r\in R$. Since $\psi \left(\mathcal{T}\right(A,1,\sigma ,R\left)\right)=T(A,1,\sigma ,R)$, $\psi \left(A_r\right)=A_{r^{\prime }}$ for some $r^{\prime }\in R$. Thus, ${\alpha }^{\psi }\left(r\right)=r^{\prime }\in R$, and so ${\alpha }^{\psi }\left(R\right)\subseteq R$.

Now, we have to show that $R\subseteq {\alpha }^{\psi }\left(R\right)$. For this, let $t\in R$. Since ${\alpha }^{\psi }{|}_B$ is a bijection, there exists some $s\in R$ such that ${\alpha }^{\psi }\left(s\right)=t\in R$. As $r_b\in R$, it implies ${\alpha }^{\psi }\left(r_b\right)\in R$. Moreover, $(s,r_s)\in \sigma$, which implies $({\alpha }^{\psi }\left(s\right),{\alpha }^{\psi }\left(r_s\right))\in \sigma$. Thus, ${\alpha }^{\psi }\left(s\right)={\alpha }^{\psi }\left(r_s\right)$. Hence, $t={\alpha }^{\psi }\left(s\right)={\alpha }^{\psi }\left(r_s\right)\in {\alpha }^{\psi }\left(R\right)$. Therefore, ${\alpha }^{\psi }\left(R\right)=R$. Hence, ${\alpha }^{\psi }\in \mathcal{S}(A,B,\sigma ,R)$. □

Lemma 13. 

Let S be a subsemigroup of $\mathcal{T}(A,B,\sigma ,R)$ which contains ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$, and $F^1$ is characteristic in S. If $\psi ,\phi \in Aut\left(S\right)$ and ${\psi |}_{B_r}{=\phi |}_{B_r}\forall r\in R$, then $\psi =\phi$.

Proof. 

Let $\gamma ={\phi }^{-1}\psi$. Then, $\gamma \in \mathrm{Aut}\left(S\right)$ and ${\gamma |}_{B_r}={id}_{B_r}$ for every $r\in R$. Note that $\psi =\phi \iff \gamma ={\mathrm{id}}_S.$ To prove $\gamma ={\mathrm{id}}_S$, let $u\in S$ and $a\in A$. We aim to show that $\gamma \left(u\right)\left(a\right)=u\left(a\right)$. If $a=r\in R$ and $u\left(r\right)=r^{\prime }$, then for every $a^{\prime }\in A$, $u{A}_r\left(a^{\prime }\right)=u\left(r\right)=r^{\prime }=A_{r^{\prime }}\left(a^{\prime }\right)$, implying $\gamma \left(u\right)\left(A_r\right)=\gamma \left(A_{r^{\prime }}\right)=A_{r^{\prime }}$. Thus,

$$\gamma \left(u\right)\left(r\right)=\gamma \left(u\right)\left(A_r\left(r\right)\right)=\gamma \left(u{A}_r\right)\left(r\right)=A_{r^{\prime }}\left(r\right)=r^{\prime }=u\left(r\right).$$

If $a\in B\setminus R$ and $v=u{\phi }_{\{a,r_a\}}$, we claim that $\gamma \left(v\right)=v$. Now, we have $v\left(a\right)=u{\phi }_{\{a,r_a\}}\left(a\right)=u\left(a\right)$ and $v\left(z\right)=u{\phi }_{\{a,r_a\}}\left(z\right)=u\left(r_a\right)=r_{u\left(a\right)}$ for all $z\in A\setminus \left\{a\right\}$. If $u\left(a\right)=r_{u\left(a\right)}$, then $v\left(a\right)=u\left(a\right)=r_{u\left(a\right)}$ and hence $v=A_{u\left(a\right)}$, implying $\gamma \left(v\right)=v$. If $u\left(a\right)\ne r_{u\left(a\right)}$, then $\mathrm{im}\left(v\right)=\{u\left(a\right),r_{u\left(a\right)}\}$ and so $v\in {\mathcal{F}}^1$. Moreover, $v{\phi }_{\{a,r_a\}}=v$ and ${\phi }_{\{u\left(a\right),r_{u\left(a\right)}\}}v=v$. Thus, since $\gamma$ fixes ${\phi }_{\{a,r_a\}}$ and ${\phi }_{\{u\left(a\right),r_{u\left(a\right)}\}}$, we have $\gamma \left(v\right){\phi }_{\{a,r_a\}}=\gamma \left(v\right)$ and ${\phi }_{\{u\left(a\right),r_{u\left(a\right)}\}}\gamma \left(v\right)=\gamma \left(v\right)$. If $\gamma \left(v\right)\left(a\right)\ne u\left(a\right)$, then $\gamma \left(v\right)\left(a\right)={\phi }_{\{u\left(a\right),r_{u\left(a\right)}\}}\gamma \left(v\right)\left(a\right)=r_{u\left(a\right)}$, and for every $z\in A\setminus \left\{a\right\}$, $\gamma \left(v\right)\left(z\right)=\gamma \left(v\right){\phi }_{\{a,r_a\}}\left(z\right)=\gamma \left(v\right)\left(r_a\right)\in R$. Thus, $\mathrm{im}\left(\gamma \right(v\left)\right)\subseteq R$ and so $\gamma \left(v\right)\notin {\mathcal{F}}^1$. This is a contradiction since $v\in {\mathcal{F}}^1$ and $\gamma \left({\mathcal{F}}^1\right)={\mathcal{F}}^1$. Thus, $\gamma \left(v\right)\left(a\right)=u\left(a\right)$. For every $z\in A\setminus \left\{a\right\}$, we have $\gamma \left(v\right)\left(z\right)=\gamma \left(v\right){\phi }_{\{a,r_a\}}\left(z\right)=\gamma \left(v\right)\left(r_a\right)={\phi }_{\{u\left(a\right),r_{u\left(a\right)}\}}\gamma \left(v\right)\left(r_a\right)=r_{u\left(a\right)}$, since $\gamma \left(v\right)\left(r\right)\in R$ and ${\phi }_{\{u\left(a\right),r_{u\left(a\right)}\}}\left(R\right)=r_{u\left(a\right)}$.

Thus, $\gamma \left(v\right)=v$. Hence,

$$\gamma \left(u\right)\left(a\right)=\left(\gamma \left(u\right){\phi }_{\{a,r_a\}}\right)\left(a\right)=\gamma \left(u{\phi }_{\{a,r_a\}}\right)\left(a\right)=\gamma \left(v\right)\left(a\right)=v\left(a\right)=u\left(a\right).$$

Thus, $\gamma \left(u\right)=u$, and $\gamma ={\mathrm{id}}_S={\phi }^{-1}\psi$ implies that $\psi =\phi$. □

Let $\alpha \in \mathcal{S}(A,B,\sigma ,R)$. We denote by ${\delta }_{\alpha }$ the inner automorphism of $\mathcal{T}(A,B,\sigma ,R)$ induced by $\alpha$, i.e., ${\delta }_{\alpha }\left(\beta \right)=\alpha \beta {\alpha }^{-1}$ for every $\beta \in \mathcal{T}(A,B,\sigma ,R)$. Let ${\tau }_{\alpha }={\delta }_{\alpha }{|}_{({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))}$. Since $\phi ({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))={\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ for every automorphism $\phi$ of $\mathcal{T}(A,B,\sigma ,R)$, ${\tau }_{\alpha }$ is an induced automorphism of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ induced by $\alpha$, i.e., ${\tau }_{\alpha }\left(\beta \right)=\alpha \beta {\alpha }^{-1}$ for every $\beta \in {\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$.

The following lemma demonstrates that each automorphism of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ is an inner automorphism induced by an element of $\mathcal{S}(A,B,\sigma ,R)$.

Lemma 14. 

Every automorphism of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ is inner, induced by an element of $\mathcal{S}(A,B,\sigma ,R)$, and $Aut({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))=\{{\tau }_{\alpha }:\alpha \in \mathcal{S}(A,B,\sigma ,R)\}$.

Proof. 

Let $\psi \in Aut({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))$. Then, by Lemma 12, $\alpha ={\alpha }^{\psi }\in \mathcal{S}(A,B,\sigma ,R)$. We claim that $\psi ={\tau }_{\alpha }$. By Lemma 2, for every automorphism $\phi$ of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$, we have $\phi \left(\mathcal{T}\right(A,1,\sigma ,R\left)\right)=\mathcal{T}(A,1,\sigma ,R)$; thus, $\phi \left({\mathcal{F}}^1\right)={\mathcal{F}}^1$, and by Lemma 13, it remains to show that ${\psi |}_{B_r}={\tau }_{\alpha }{|}_{B_r}$ for every $r\in R$. Let $r\in R$. Then, $\psi \left(A_r\right)=A_{r^{\prime }}$ for some $r^{\prime }\in R$. By definition of ${\alpha }^{\psi }$, we have $\alpha \left(r\right)=r^{\prime }$. Thus, for every $a\in A$, we have ${\tau }_{\alpha }\left(A_r\right)\left(a\right)=\left(\alpha A_r{\alpha }^{-1}\right)\left(a\right)=\alpha \left(r\right)=r^{\prime }$, and so ${\tau }_{\alpha }\left(A_r\right)=A_{r^{\prime }}$. By Lemma 11, $\psi \left(B_r\right)=B_{r^{\prime }}={\tau }_{\alpha }\left(B_r\right)$, and so ${\psi |}_{B_r}={\tau }_{\alpha }{|}_{B_r}$. Hence, by Lemma 13, we have $\psi ={\tau }_{\alpha }$. Thus, for every $\beta \in {\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ we have $\psi \left(\beta \right)=\alpha \beta {\alpha }^{-1}$ for some $\alpha \in \mathcal{S}(A,B,\sigma ,R)$. That is, every automorphism of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$ is inner, induced by an element of $S(A,B,\sigma ,R)$, and $\mathrm{Aut}({\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R))=\{{\tau }_{\alpha }:\alpha \in \mathcal{S}(A,B,\sigma ,R)\}$. Hence the lemma. □

Now, we show that every automorphism of $\mathcal{T}(A,B,\sigma ,R)$ is inner, induced by elements of $S(A,B,\sigma ,R).$

Theorem 3. 

Every automorphism of $\mathcal{T}(A,B,\sigma ,R)$ is inner, induced by elements of $S(A,B,\sigma ,R)$, and $Aut\left(\mathcal{T}(A,B,\sigma ,R)\right)=\{{\delta }_{\alpha }:\alpha \in \mathcal{S}(A,B,\sigma ,R)\}$.

Proof. 

Let $\psi \in Aut\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$. Then, by Theorem 2, we have ${\psi |}_{{\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)}$ is an automorphism of ${\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)$. Thus, by Lemma 14, we have ${\psi |}_{{\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)}={\tau }_{\alpha }$ for some $\alpha \in \mathcal{S}(A,B,\sigma ,R)$. Hence, ${\psi |}_{{\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)}={\delta }_{\alpha }{|}_{{\mathcal{F}}^1\cup \mathcal{T}(A,1,\sigma ,R)}$, and so by Lemma 14, we have $\psi ={\delta }_{\alpha }$. That is,

$$\mathrm{Aut}\left(\mathcal{T}(A,B,\sigma ,R)\right)=\{{\delta }_{\alpha }:\alpha \in \mathcal{S}(A,B,\sigma ,R)\}.$$

Thus, for every $\beta \in \mathcal{T}(A,B,\sigma ,R)$, we have $\psi \left(\beta \right)=\alpha \beta {\alpha }^{-1}$ for some $\alpha \in \mathcal{S}(A,B,\sigma ,R)$. That is, every automorphism of $\mathcal{T}(A,B,\sigma ,R)$ is inner, induced by elements of $S(A,B,\sigma ,R)$, and $\mathrm{Aut}\left(\mathcal{T}(A,B,\sigma ,R)\right)=\{{\delta }_{\alpha }:\alpha \in \mathcal{S}(A,B,\sigma ,R)\}$. □

Now, we show that the automorphism group of $\mathcal{T}(A,B,\sigma ,R)$ is isomorphic to the group $\mathcal{S}(A,B,\sigma ,R)$.

Theorem 4. 

$Aut\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)\cong \mathcal{S}(A,B,\sigma ,R)$.

Proof. 

Since we know that $\mathrm{Aut}\left(\mathcal{T}(A,B,\sigma ,R)\right)=\{{\delta }_{\alpha }:\alpha \in \mathcal{S}(A,B,\sigma ,R)\}$, define $\Theta :\mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)\to \mathcal{S}(A,B,\sigma ,R)$ by $\Theta \left({\delta }_{\alpha }\right)=\alpha$. Now, for any ${\delta }_{{\alpha }_1},{\delta }_{{\alpha }_2}\in \mathrm{Aut}\left(\mathcal{T}(A,B,\sigma ,R)\right)$, we have

$$\begin{array}{c}\hfill {\delta }_{{\alpha }_1}={\delta }_{{\alpha }_2}\iff {\alpha }_1={\alpha }_2\iff \Theta \left({\delta }_{{\alpha }_1}\right)=\Theta \left({\delta }_{{\alpha }_2}\right).\end{array}$$

That is, $\Theta$ is an injective map. By Theorem 3, $\Theta$ is clearly onto and hence a bijection from $\mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)$ to $\mathcal{S}(A,B,\sigma ,R)$.

Let $u\in \mathcal{T}(A,B,\sigma ,R)$; we have

$${\delta }_{\alpha }{\delta }_{\beta }\left(u\right)={\delta }_{\alpha }\left(\beta u{\beta }^{-1}\right)=\alpha \left(\beta u{\beta }^{-1}\right){\alpha }^{-1}=\alpha \beta u{\left(\alpha \beta \right)}^{-1}={\delta }_{\alpha \beta }\left(u\right).$$

That is, ${\delta }_{\alpha }{\delta }_{\beta }={\delta }_{\alpha \beta }$.

Now, for any ${\delta }_{{\alpha }_1},{\delta }_{{\alpha }_2}\in \mathrm{Aut}\left(T(A,B,\sigma ,R)\right)$, we have

$$\begin{array}{c}\hfill \Theta \left({\delta }_{{\alpha }_1}{\delta }_{{\alpha }_2}\right)=\Theta \left({\delta }_{{\alpha }_1{\alpha }_2}\right)={\alpha }_1{\alpha }_2=\Theta \left({\delta }_{{\alpha }_1}\right)\Theta \left({\delta }_{{\alpha }_2}\right).\end{array}$$

That is, $\Theta$ is a homomorphism. Thus, we have $\Theta$ is an isomorphism from $\mathrm{Aut}\left(\mathcal{T}\right(A,B,$$\sigma ,R\left)\right)$ to $\mathcal{S}(A,B,\sigma ,R)$. Hence, $\mathrm{Aut}\left(\mathcal{T}\right(A,B,\sigma ,R\left)\right)\cong \mathcal{S}(A,B,\sigma ,R)$. □

6. Conclusions

In this study, we explored $\mathcal{T}(A,B,\sigma ,R)$, a semigroup composed of full maps from a set A with a restricted range preserving an equivalence relation $\sigma$ and a transversal R. Our investigation yielded several key results:

• We established that $\mathcal{T}(A,B,\sigma ,R)$ generally lacked an identity element.
• We demonstrated the characteristics of the sets ${\mathcal{F}}^1$, ${\mathcal{F}}^2$, and $\mathcal{T}(A,2,\sigma ,R)$ within $\mathcal{T}(A,B,\sigma ,R)$.
• We identified the maximal left inverse zero divisor bands within $\mathcal{T}(A,B,\sigma ,R)$.
• We delved into the automorphisms of $\mathcal{T}(A,B,\sigma ,R)$, revealing that each automorphism was inner, induced by elements of $\mathcal{S}(A,B,\sigma ,R)$.
• Furthermore, we established an isomorphism between the automorphism group of $\mathcal{T}(A,B,\sigma ,R)$ and $\mathcal{S}(A,B,\sigma ,R)$.

These findings significantly contribute to the understanding of $\mathcal{T}(A,B,\sigma ,R)$ and extend previous results concerning $\mathcal{T}(A,B)$ and $\mathcal{T}(A,\sigma ,R)$, where $\sigma$ represents the identity relation and $A=B$, respectively.

In conclusion, the research presented in this paper holds significant importance across a spectrum of mathematical and applied fields. It provides foundational insights into algebraic structures, aiding in their analysis and comprehension. Moreover, it contributes to our understanding of fundamental groups in algebraic topology, shedding light on their invariance properties. Furthermore, the study of inner automorphisms enriches character theory and module actions in representation theory, offering valuable tools for exploring the behavior of group actions on vector spaces. Beyond theoretical realms, the implications extend to practical domains such as cryptographic protocol design, where inner automorphisms play a crucial role in ensuring security and robustness. Moreover, the insights gleaned from this research deepen our understanding of Lie algebras and their symmetries, opening avenues for further exploration in mathematical physics. Applications in mapping class groups and fiber bundles within geometry and topology further underscore the breadth of its impact. Crucially, the results established in this paper hold significant applications in mathematical physics, particularly in the realms of quantum groups and gauge theory, underscoring their relevance in advancing our understanding of complex physical phenomena. By harnessing the power of inner automorphisms, researchers can delve deeper into the intricacies of symmetry, structure, and properties inherent in both mathematical abstractions and real-world systems, thus highlighting their broad relevance and importance in diverse fields.

Future studies in this area have the potential to significantly deepen our understanding of these structures and their properties. Here are several promising avenues for further research:

• We established that $\mathcal{T}(A,B,\sigma ,R)$ generally lacked an identity element. It is natural to seek conditions under which $\mathcal{T}(A,B,\sigma ,R)$ can admit an identity element, potentially by relaxing certain constraints or introducing additional structure.
• Finding the necessary and sufficient conditions under which $\mathcal{T}(A,B,\sigma ,R)$ is a $BQ$-semigroup, i.e., a semigroup whose bi-ideals and quasi-ideals coincide.
• It is well known that $\mathcal{T}\left(X\right)\cong \mathcal{T}\left(Y\right)$ if and only if $\left|X\right|=\left|Y\right|$. Investigating analogous results for $\mathcal{T}(A,B,\sigma ,R)$ is worthwhile.
• Exploring the maximal left inverse zero divisor bands within other semigroups of this type, such as the semigroup $\mathcal{T}(A,B,\sigma ,\rho ,R)$, where the two equivalences $\rho$ and $\sigma$ are preserved along with a cross section.

By pursuing these directions, future studies can further advance our understanding of $\mathcal{T}(A,B,\sigma ,R)$ and its broader implications in mathematics and beyond.