Xiangjun Kong , Pei Wang and Jian Tang

# Quasi-ideal Ehresmann transversals: The spined product structure

De Gruyter | Published online: April 15, 2021

# Abstract

In any U-abundant semigroup with an Ehresmann transversal, two significant components R and L are introduced in this paper and described by Green’s -relations. Some interesting properties associated with R and L are explored and some equivalent conditions for the Ehresmann transversal to be a quasi-ideal are acquired. Finally, a spined product structure theorem is established for a U-abundant semigroup with a quasi-ideal Ehresmann transversal by means of R and L.

MSC 2010: 20M10

## 1 Introduction

Suppose that S is a regular semigroup and S o a subsemigroup of S. We denote the intersection of V ( a ) and S o by V S o ( a ) and that I = { a a o : a S , a o V S o ( a ) } and Λ = { a o a : a S , a o V S o ( a ) } . An inverse transversal of the semigroup S is a subsemigroup S o that contains exactly one inverse of every element of S, that is, S is regular and S o inverse with | V S o ( a ) | = 1 . This important concept was introduced by Blyth and McFadden [1]. Thereafter, this class of regular semigroups excited many semigroup researchers’ attention and a good deal of important results were obtained (see [1,2,3,4] and references therein). Tang [4] has shown that for S a regular semigroup with an inverse transversal S o , I and Λ are both bands with I left regular and Λ right regular. These two bands play an important role in the study of regular semigroups with inverse transversals. Other important subsets of S are R = { x S : x o x = x o x o o } and L = { x S : x x o = x o o x o } . They are subsemigroups with R and L left and right inverse, respectively.

To extend the class of regular semigroups and the class of abundant semigroups, a new relation ˜ U on a semigroup S is introduced in the following way. Let E ( S ) be the set of all idempotents of a semigroup S and suppose that U is a non-empty subset of E ( S ) . Then the relation ˜ U on S is defined as follows:

˜ U = { ( a , b ) S × S | ( e U ) a e = a b e = b } .
It is easy to see that ˜ U . In particular, if S is an abundant semigroup with U = E ( S ) , then ˜ U = . Furthermore, if S is a regular semigroup with U = E ( S ) , then = = ˜ U . Dually, a relation ˜ U is defined, and for any results concerning ˜ U , we also have the dual results for ˜ U . Recall that a semigroup S is a weakly U-abundant semigroup if each ˜ U -class and each ˜ U -class of S contain an idempotent from U, and we denote such a weakly U-abundant semigroup by ( S , U ) . In this case, we call the set U a set of distinguished idempotents of ( S , U ) , and an element u of U is called a distinguished idempotent of ( S , U ) . It is well known that the relations and are always right congruences on a semigroup S, but the same need not be true for ˜ U . A weakly U-abundant semigroup ( S , U ) is said to satisfy the congruence condition (C) if ˜ U and ˜ U are a right congruence and a left congruence on ( S , U ) , respectively. We call a weakly U-abundant semigroup having the congruence condition (C) a U-abundant semigroup. Clearly, regular semigroups and abundant semigroups are all U-abundant semigroups with U = E ( S ) . A U-abundant semigroup ( S , U ) is called an Ehresmann semigroup (quasi-Ehresmann semigroup) if U is a semilattice (band). The class of Ehresmann semigroups, named by Lawson [ 18], is an important subclass of the class of U-abundant semigroups, which are generalizations of inverse semigroups in the class of regular semigroups and adequate semigroups in the class of abundant semigroups.

The concept of an Ehresmann transversal was introduced almost simultaneously in three papers by Ma, Ren, Gong [19], Wang [20] and Yang [21] in the class of U-abundant semigroups as the generalisation of the concept of an inverse transversal and an adequate transversal. The aforementioned three kinds of Ehresmann transversals were basically the same, with only slight difference. To achieve interesting properties of Ehresmann transversals which parallel to those of adequate transversals, we give the following definition of Ehresmann transversals combining the aforementioned three kinds of Ehresmann transversals. In [19,20,21], some basic properties about Ehresmann transversals are given.

In the present paper, we continue along the lines of [3,15,16] by exploring the properties associated with R and L of Ehresmann transversals of U-abundant semigroups. The main result of this paper is to establish a spined product structure theorem for a U-abundant semigroup with a quasi-ideal Ehresmann transversal by means of R and L. The related results concerning adequate transversals are generalised and enriched.

## 2 Preliminaries

Throughout this paper, a U-abundant semigroup is always denoted as ( S , U ) and the ˜ U -class and the ˜ U -class of ( S , U ) containing the element a as L ˜ a U and R ˜ a U , respectively. Moreover, for any a ( S , U ) the distinguished idempotents in L ˜ a U U and R ˜ a U U are denoted by a and a + , respectively. In particular, when ( S , U ) is an Ehresmann semigroup, then there exists a unique distinguished idempotent in L ˜ a U U and a unique distinguished idempotent in R ˜ a U U . Therefore, as for Ehresmann semigroups, for all a , b ( S , U ) , a ˜ U b if and only if a + = b + ; a ˜ U b if and only if a = b , and consequently ( a b ) = ( a b ) and ( a b ) + = ( a b + ) + . We remark here that an excellent survey of investigations of restriction semigroups and Ehresmann semigroups was given by Gould [22] and the class of Ehresmann monoids were deeply investigated by Branco, Gomes and Gould [23,24].

## Lemma 2.1

Suppose that S is a semigroup and U a non-empty subset of E ( S ) . Let R e g U S = { a S | ( a V ( a ) ) a a , a a U } . Then

1. (i)

˜ U ( R e g U S × R e g U S ) = ( R e g U S × R e g U S ) .

2. (ii)

˜ U ( R e g U S × R e g U S ) = ( R e g U S × R e g U S ) .

If V U ( a ) = { a V ( a ) | a a , a a U } , then a R e g U S if and only if V U ( a ) . The following result is immediate from the definition of ˜ U , and there is a dual result for ˜ U .

## Lemma 2.2

Let a be an element of a weakly U-abundant semigroup ( S , U ) and e a distinguished idempotent of ( S , U ) . Then the following statements are equivalent:

1. (i)

e ˜ U a .

2. (ii)

a e = a , and for all f U , a f = a implies e f = e .

Let ( S , U ) be a U-abundant semigroup and ( T , V ) a U-abundant subsemigroup of ( S , U ) , that is, T S and V = U T . Then the U-abundant subsemigroup ( T , V ) is called a -subsemigroup of ( S , U ) if for any a T , there exist e L ˜ a U V and f R ˜ a U V , or equivalently, if ˜ V ( T ) = ˜ U ( S ) ( T × T ) and ˜ V ( T ) = ˜ U ( S ) ( T × T ) .

## Definition 2.3

An Ehresmann -subsemigroup ( T , V ) of a U-abundant semigroup ( S , U ) is called an Ehresmann transversal of ( S , U ) if U forms an order ideal of E ( S ) , and for any x S , there exist e , f U and a unique x ¯ T such that x = e x ¯ f , where e x ¯ + and f x ¯ . It can easily be shown that e and f are uniquely determined by x and T (see [15]). Hence, we denote e by e x and f by f x and we have e x ˜ U x , x ˜ U f x .

A subsemigroup ( T , V ) of a semigroup ( S , U ) is called a quasi-ideal if T S T T , and the corresponding Ehresmann transversal is called a quasi-ideal Ehresmann transversal. Let ( T , V ) be an Ehresmann transversal of the U-abundant semigroup ( S , U ) . Let

I = { e x U | x ( S , U ) } , Λ = { f x U | x ( S , U ) } .

## Lemma 2.4

Let ( S , U ) be a U-abundant semigroup with an Ehresmann transversal ( T , V ) . Then

1. (1)

I Λ = V ;

2. (2)

I = { e U | ( v V ) v e } , Λ = { f U | ( h V ) h f } ;

3. (3)

I V I , V Λ Λ ;

4. (4)

x ˜ U y if and only if e x = e y ; x ˜ U y if and only if f x = f y ;

5. (5)

If e , f I and e f , then e = f ; if e , f Λ and e f , then e = f ;

6. (6)

( x R e g U S ) | V U ( x ) T | 1 ;

7. (7)

If T is a quasi-ideal of S, then ( x R e g U S ) | V U ( x ) T | = 1 .

Similarly to the adequate transversal case, one can easily show the following decomposition formula. It is worth remarking that if the condition “quasi-ideal” is strengthened as “multiplicative,” the proof was obtained in [15].

## Lemma 2.5

Let ( S , U ) be a U-abundant semigroup with a quasi-ideal Ehresmann transversal ( T , V ) . For any x , y S , then x y ¯ = x ¯ f x e y y ¯ ; e x y = e x ( x ¯ f x e y ) + ; f x y = ( f x e y y ¯ ) f y .

The so-called Miller-Clifford theorem is crucial and frequently used in this paper.

## Lemma 2.6

[25]

1. (1)

Let e and f be D -equivalent idempotents of a semigroup S. Then each element a of R e L f has a unique inverse a in R f L e , such that a a = e and a a = f .

2. (2)

Let a , b S . Then a b R a L b if and only if L a R b contains an idempotent.

## 3 Some properties of R and L

The objective in this section is to give some properties about R and L. It is known that R and L play an important role in constructing regular semigroups with quasi-ideal inverse transversals and abundant semigroups with quasi-ideal adequate transversals. For any result concerning R, there is a dual result for L which we list but omit its proof.

## Proposition 3.1

Let ( S o , V ) be an Ehresmann transversal of a U-abundant semigroup ( S , U ) . If a , b S o and c S are such that a ˜ U c ˜ U b , then c S o .

## Proof

Since a , b S o and S o is an Ehresmann semigroup, there exist a + , b V , such that

a + ˜ U a ˜ U c ˜ U b ˜ U b .

From the definition of an Ehresmann transversal, we have c = e c c ¯ f c , where e c ˜ U c ¯ + and f c ˜ U c ¯ . Moreover, e c ˜ U c ˜ U f c . It follows that a + ˜ U c ˜ U e c ˜ U c ¯ + , consequently e c Λ I = V . Similarly, c ¯ ˜ U f c ˜ U c ˜ U b and f c V . Thus, c = e c c ¯ f c V S o V S o .□

## Proposition 3.2

Let ( S o , V ) be an Ehresmann transversal of a U-abundant semigroup ( S , U ) . Then for any x R e g U S , | V U ( x ) S o | = 1 and we call x o = V U ( x ) S o the unique U-inverse in S o of x.

## Proof

For every x R e g U S , since x , e x and f x are all U-regular, from e x ˜ U x ˜ U f x we deduce that e x x f x , so by Lemma 2.6 x has an inverse x R f x L e x and x x = e x , x x = f x . Thus, x ¯ f x x e x x ¯ + , so by Proposition 3.1 x S o . Now let ( x ) 1 denote the unique U-inverse of x in the Ehresmann semigroup S o . Then from x = x x ( x ) 1 x x and Definition 2.3, the uniqueness of x S o is obvious. That is | V U ( x ) S o | = 1 .□

## Proposition 3.3

Let ( S , U ) be a U-abundant semigroup with an Ehresmann transversal ( S o , V ) . If U = E ( S ) , then S is regular if and only if S o is an inverse semigroup; in this case, S o is an inverse transversal of S.

## Proof

Suppose that S is a regular semigroup. Then by U = E ( S ) and | V U ( x ) S o | = 1 , every regular element x S has a unique inverse in S o and, in particular, every element in S o has a unique U-inverse in itself. Hence, S o is an inverse semigroup and S o is an inverse transversal of S.

Conversely, assume that S o is an inverse semigroup. Then V = U E ( S o ) = E ( S ) E ( S o ) = E ( S o ) and for every x in S, there exists y S o such that x ¯ = x ¯ y x ¯ . Consequently,

x ( x ¯ y x ¯ + ) x = e x x ¯ ( f x x ¯ ) y ( x ¯ + e x ) x ¯ f x = e x x ¯ y x ¯ f x = e x x ¯ f x = x ,
and so x is regular in S. Thus, S is a regular semigroup.□

## Proposition 3.4

Let ( S , U ) be a U-abundant semigroup with an Ehresmann transversal ( S o , V ) . If S o is a right ideal of S, then f x V for every x S and U = I . Consequently, f x = x ¯ and thus x = e x x ¯ .

Dually, if S o is a left ideal of S, then e a V for every a S and U = Λ . Consequently, e a = a ¯ + and thus a = a ¯ f a .

## Proof

By Definition 2.3, for every x S , x = e x x ¯ f x , where e x x ¯ + and f x x ¯ . Since S o is a right ideal of S, f x = x ¯ f x S o and thus f x V . Let e U and e o denote the unique U-inverse in S o of e. Since e o e V ( e ) S o , we have e o e = e o . Thus, e e o = e for every e U . Consequently, U = I = { e x : x S } .□

## Proposition 3.5

Suppose that ( S , U ) is a U-abundant semigroup with an Ehresmann transversal ( S o , V ) . Let

R = { r S : f r V } a n d L = { a S : e a V } .

Then

R = { r S : ( v V ) r ˜ U v } a n d L = { a S : ( h V ) a ˜ U h } .

Consequently, R L = S o , U R = I and U L = Λ .

## Proof

It is clear that if r R , there exists v = f r V such that r ˜ U v .

Conversely, for r S , if there exists v V such that r ˜ U v , then f r ˜ U r ˜ U v . Hence, f r I . Therefore, f r I Λ = V .□

## Proposition 3.6

Suppose that ( S , U ) is a U-abundant semigroup with an Ehresmann transversal ( S o , V ) and S o is a quasi-ideal of S. Let R = { r S : f r V } and L = { a S : e a V } . Then ( R , I ) and ( L , Λ ) are both U-abundant semigroups with an Ehresmann transversal ( S o , V ) , where S o is a right ideal of R and a left ideal of L.

Moreover, for r , s R and a , b L , if r ¯ = a ¯ , s ¯ = b ¯ and e r a = e s b , then r = s and a = b .

## Proof

Obviously, S o R and S o L . For any r , s R , by Lemma 2.5, we have f r s = ( f r e s s ¯ ) f s = ( f r e s s ¯ ) s ¯ V and so R is a subsemigroup of ( S , U ) .

For any r R and s o S o , we have f r V and consequently s o r = s o r f r S o since S o is a quasi-ideal of S. Thus, S o is a right ideal of R and by Proposition 3.4, U ( R ) = I . Thus, ( S o , V ) is an Ehresmann transversal of ( R , I ) .

If r ¯ = a ¯ , s ¯ = b ¯ and e r a = e s b for r , s R and a , b L , then we have

e r a ¯ = e r ¯ f e r e a a ¯ ( by Lemma 2.5 ) = r ¯ + r ¯ + e a a ¯ ( by e r = e r r ¯ + r ¯ + ) = a ¯ + e a a ¯ = a ¯ + a ¯ = a ¯ .
Similarly, we have e s b ¯ = b ¯ . Thus, a ¯ = b ¯ , r ¯ = s ¯ and consequently f r = r ¯ = s ¯ = f s . Therefore,
r = e r r ¯ f r = e r a ¯ f r = e r ( a ¯ + a a ¯ ) f r = ( e r r ¯ + ) a ( r ¯ f r ) = e r a f r = e s b f r = e s b f s = e s b ¯ f s = e s s ¯ f s = s .

Similarly, we have a = b .□

## Proposition 3.7

Let ( S o , V ) be an Ehresmann transversal of a U-abundant semigroup ( S , U ) . Then the following statements are equivalent:

1. (1)

S o is a quasi-ideal of S;

2. (2)

Λ I S o ;

3. (3)

V I S o and Λ V S o ;

4. (4)

L R S o ;

5. (5)

S o is a left ideal of L and a right ideal of R;

6. (6)

S o I S o and Λ S o S o ;

7. (7)

S o I S o S o and S o Λ S o S o ;

8. (8)

S S o R , S o S L ;

9. (9)

R is a left ideal and L is a right ideal of S.

## Proof

(1) (2). For any f Λ and e I , there exist f + , e V such that f f + and e e . So we have f e = f + f e e V S V S o S S o S o .

(2) (3). This is trivial.

(3) (4). For any a L and r R , there exist h , v V such that a ˜ U h and r ˜ U v . Thus,

a r = h ( a r ) v = h e a r a r ¯ f a r v V I a r ¯ Λ V S o a r ¯ S o S o .

(4) (5). This is clear since S o L , R .

(5) (6). This is clear since I R and Λ L .

(6) (7). This is obvious.

(7) (8). If (7) holds, then for any a S , r o S o , we have

a r o = e a a ¯ f a r o ˜ U a ¯ + a ¯ f a r o = a ¯ f a r o ˜ U ( a ¯ f a r o ) V ,
since a ¯ f a r o S o Λ S o S o . Hence, a r o R by Proposition 3.5 and S S o R . Dually, S o S L .

(8) (9). For any a S , r R , r = e r r ¯ f r with f r V , we have a r = a e r r ¯ f r S S o R and R is a left ideal of S. Dually, S o S L implies that L is a right ideal of S.

(9) (1). For any s , t S o and a S , we have

s a t = ( s a ) t S S o S R R and s a t = s ( a t ) S o S L S L .

Consequently, s a t R L = S o and S o is a quasi-ideal of S.□

## 4 The main theorem

The main objective of this paper is to establish a structure theorem for U-abundant semigroups with quasi-ideal Ehresmann transversals. In what follows, ( R , I ) denotes a U-abundant semigroup with a right ideal Ehresmann transversal ( S o , V ) . Then, by Proposition 3.4, for every r R , f r = r ¯ V and r = e r r ¯ . And ( L , Λ ) denotes a U-abundant semigroup with a left ideal Ehresmann transversal ( S o , V ) , which has a dual result of ( R , I ) , especially, a = a ¯ f a for every a L . Then R and L will be the components in such construction, which satisfy some compatible conditions.

## Theorem 4.1

Let ( R , I ) and ( L , Λ ) be a pair of U-abundant semigroups sharing a common Ehresmann transversal ( S o , V ) . Suppose that S o is a right ideal of R and a left ideal of L. Let L × R S o described by ( a , r ) a r be a mapping such that for any r , s R and for any a , b L :

1. (1)

( a r ) s = a r s and b ( a r ) = b a r ;

2. (2)

If { r , a } S o , then a r = a r ;

3. (3)

For any a , b L , r , s R , if a ˜ Λ b , then ( a s ) ˜ V ( b s ) ; if r ˜ I s , then ( a r ) ˜ V ( a s ) .

Define a multiplication on the set

Γ = R | × | L = { ( r , a ) R × L : r ¯ = a ¯ }
by
( r , a ) ( s , b ) = ( e r ( a s ) , ( a s ) f b ) .

Then ( Γ , E ( Γ ) ¯ ) is a U-abundant semigroup with a quasi-ideal Ehresmann transversal which is isomorphic to ( S o , V ) .

Conversely, every U-abundant semigroup with a quasi-ideal Ehresmann transversal can be constructed in this manner.

To prove this theorem, in the following we will give a sequence of lemmas.

## Lemma 4.1

The multiplication on Γ is well-defined.

## Proof

We only need to prove ( e r ( a s ) , ( a s ) f b ) Γ . It follows from a s S o that e r ( a s ) R and ( a s ) f b L . It is obvious that both a right ideal and a left ideal are quasi-ideals, and so we have

e r ( a s ) ¯ = e r ¯ f e r e ( a s ) a s ¯ ( by Lemma 2 .5 ) = r ¯ + r ¯ + ( a s ) + ( a s ) ( a s S o ) = r ¯ + ( a s ) = a ¯ + ( a s ) ( r ¯ = a ¯ ) = e a ( a s ) ( since e a V ) = a s .
On the other hand, we have
( a s ) f b ¯ = a s ¯ f a s e f b f b ¯ = ( a s ) ( a s ) b ¯ b ¯ = ( a s ) b ¯ = ( a s ) s ¯ ( s ¯ = b ¯ ) = ( a s ) f s ( since f s V ) = a s .
Therefore, e r ( a s ) ¯ = ( a s ) f b ¯ and ( e r ( a s ) , ( a s ) f b ) Γ .□

## Lemma 4.2

Γ is a semigroup.

## Proof

Let ( r , a ) , ( s , b ) , ( t , c ) Γ . Then we have

[ ( r , a ) ( s , b ) ] ( t , c ) = [ e r ( a s ) , ( a s ) f b ] ( t , c ) = ( e e r ( a s ) { [ ( a s ) f b ] t } , { [ ( a s ) f b ] t } f c ) = ( e r ( a s ) + [ ( a s ) ( f b t ) ] , [ ( a s ) ( f b t ) ] f c ) = ( e r ( a s ) ( f b t ) , ( a s ) ( f b t ) f c ) .
On the other hand, we have
( r , a ) [ ( s , b ) ( t , c ) ] = ( r , a ) [ e s ( b t ) , ( b t ) f c ] = ( e r { a [ e s ( b t ) ] } , { a [ e s ( b t ) ] } f ( b t ) f c ) = ( e r [ ( a e s ) ( b t ) ] , [ ( a e s ) ( b t ) ] ( b t ) f c ) = ( e r ( a e s ) ( b t ) , ( a e s ) ( b t ) f c ) = ( e r ( a s ) ( f b t ) , ( a s ) ( f b t ) f c ) ,
since e s b = e s e b b ¯ f b = e s b ¯ + b ¯ f b = e s b ¯ f b = e s s ¯ f b = s f b . Thus, [ ( r , a ) ( s , b ) ] ( t , c ) = ( r , a ) [ ( s , b ) ( t , c ) ] and so Γ is a semigroup.□

## Lemma 4.3

Let ( r , a ) Γ . Then ( r , a ) E ( Γ ) if and only if a r = r ¯ = a ¯ .

## Proof

Since ( r , a ) ( r , a ) = ( e r ( a r ) , ( a r ) f a ) , and note that r R , a L and a r S o , it is clear that if a r = r ¯ = a ¯ , then

( e r ( a r ) , ( a r ) f a ) = ( e r r ¯ , a ¯ f a ) = ( r , a ) .
Thus, ( r , a ) E ( Γ ) . Conversely, if ( r , a ) E ( Γ ) , then e r ( a r ) = r = e r r ¯ . It follows from e r r ¯ + that r ¯ = r ¯ + r ¯ = r ¯ + ( a r ) = a ¯ + ( a r ) = ( e a a r ) = ( a r ) .□

## Lemma 4.4

Let E ( Γ ) ¯ = { ( r , a ) Γ | r I , a Λ a n d a r = a ¯ = r ¯ } E ( Γ ) be set of distinguished idempotents of Γ . Then Γ is a weakly abundant semigroup.

## Proof

Let ( r , a ) Γ . Then ( e r , r ¯ + ) Γ and r ¯ + e r = r ¯ + e r = r ¯ + = e r ¯ = r ¯ + ¯ and so ( e r , r ¯ + ) E ( Γ ) ¯ . Thus,

( e r , r ¯ + ) ( r , a ) = ( e r r ¯ + r , r ¯ + r f a ) = ( e r r , r ¯ f a ) = ( r , a ¯ f a ) = ( r , a ) .
Suppose that ( s , b ) E ( Γ ) ¯ such that ( s , b ) ( r , a ) = ( r , a ) . Then, from ( s , b ) ( r , a ) = ( e s ( b r ) , ( b r ) f a ) , we have e s ( b r ) = r and ( b r ) f a = a . From e s ( b r ) = r , we deduce that r ¯ = e s ( b r ) ¯ = e s ¯ f e s e ( b r ) b r ¯ = s ¯ + s ¯ + ( b r ) + ( b r ) = s ¯ + ( b r ) = b ¯ + ( b r ) = ( e b b ) r = b r . Thus, r ¯ = b r = ( b e r ) r ¯ and so r ¯ + = ( b e r ) r ¯ + = b ( e r r ¯ + ) = b e r . Therefore, ( s , b ) ( e r , r ¯ + ) = ( e s ( b e r ) , ( b e r ) r ¯ + ) = ( e s r ¯ + , r ¯ + r ¯ + ) = ( e r , r ¯ + ) and ( r , a ) ˜ E ( Γ ) ¯ ( e r , r ¯ + ) .

For ( r , a ) Γ , we have ( a ¯ , f a ) Γ with f a a ¯ = f a a ¯ = a ¯ = a ¯ ¯ = f a ¯ and so ( a ¯ , f a ) E ( Γ ) ¯ . Thus,

( r , a ) ( a ¯ , f a ) = ( e r ( a a ¯ ) , ( a a ¯ ) f f a ) = ( e r a a ¯ , a a ¯ f a ) = ( e r a ¯ , a f a ) = ( e r r ¯ , a ) = ( r , a ) .
Suppose that ( s , b ) E ( Γ ) ¯ such that ( r , a ) ( s , b ) = ( r , a ) . Then ( e r ( a s ) , ( a s ) f b ) = ( r , a ) and e r ( a s ) = r , ( a s ) f b = a . That e r ( a s ) = r implies that r ¯ = e r ( a s ) ¯ = e r ¯ f e r e a s a s ¯ = r ¯ + r ¯ + ( a s ) + ( a s ) = r ¯ + ( a s ) = ( a ¯ + a ) s = ( a ¯ + e a a ¯ f a ) s = ( a ¯ f a ) s = a s . Thus, a ¯ = r ¯ = a s = ( a ¯ f a ) s = a ¯ ( f a s ) and so a ¯ = a ¯ ( f a s ) = ( a ¯ f a ) s = f a s . Therefore, ( a ¯ , f a ) ( s , b ) = ( a ¯ ( f a s ) , ( f a s ) f b ) = ( a ¯ , f a ) and ( r , a ) ˜ E ( Γ ) ¯ ( a ¯ , f a ) .□

## Lemma 4.5

Let ( r 1 , a 1 ) , ( r 2 , a 2 ) Γ . Then

1. (1)

( r 1 , a 1 ) ˜ U ( r 2 , a 2 ) if and only if e r 1 = e r 2 if and only if r 1 ˜ I r 2 .

2. (2)

( r 1 , a 1 ) ˜ U ( r 2 , a 2 ) if and only if f a 1 = f a 2 if and only if a 1 ˜ Λ a 2 .

## Proof

1. (1)

To prove the first part of this Lemma, by Lemma 4.4, it then suffices to show that ( e r 1 , r 1 ¯ + ) ˜ E ( Γ ) ¯ ( e r 2 , r 2 ¯ + ) if and only if e r 1 = e r 2 . If e r 1 = e r 2 , then r 1 ¯ + = r 2 ¯ + and ( e r 1 , r 1 ¯ + ) = ( e r 2 , r 2 ¯ + ) . Conversely, if u 1 = ( e r 1 , r 1 ¯ + ) ˜ E ( Γ ) ¯ ( e r 2 , r 2 ¯ + ) = u 2 , then u 1 u 2 = u 2 and u 2 u 1 = u 1 , and this implies

e r 2 = e r 1 e r 2 , r 2 ¯ + = r 1 ¯ + e r 2 ;
e r 1 = e r 2 e r 1 , r 1 ¯ + = r 2 ¯ + e r 1 .
Thus, e r 1 e r 2 , and from Lemma 2.4 we deduce that e r 1 = e r 2 . That e r 1 = e r 2 if and only if r 1 ˜ I r 2 is obvious.

2. (2)

is dual to (1).□

## Lemma 4.6

( Γ , E ( Γ) ¯ ) is an E ( Γ ) ¯ -abundant semigroup.

## Proof

We need only to prove that ( Γ , E ( Γ ) ¯ ) satisfies the congruence conditions. Suppose that ( r , a ) , ( s , b ) Γ with ( r , a ) ˜ E ( Γ ) ¯ ( s , b ) . It follows from Lemma 4.5 that a ˜ Λ b and f a = f b , a ¯ = b ¯ . For any ( t , c ) Γ , we have

( r , a ) ( t , c ) = ( e r ( a t ) , ( a t ) f c ) ,
( s , b ) ( t , c ) = ( e s ( b t ) , ( b t ) f c ) .
Since a ˜ Λ b and so ( a t ) ˜ V ( b t ) by (3), thus ( a t ) f c ˜ V ( b t ) f c . Consequently, by Lemma 4.5 again ( r , a ) ( t , c ) ˜ E ( Γ ) ¯ ( s , b ) ( t , c ) . Therefore, ˜ E ( Γ ) ¯ is a right congruence on ( Γ , E ( Γ ) ¯ ) , and dually, ˜ E ( Γ ) ¯ is a left congruence. That is, ( Γ , E ( Γ ) ¯ ) is an E ( Γ ) ¯ -abundant semigroup.□

## Lemma 4.7

Let W = { ( r , r ) : r ( S o , V ) } with the distinguished idempotent set W ¯ = { ( r , r ) : r V } . Then ( W , W ¯ ) is an Ehresmann -subsemigroup of ( Γ , E ( Γ ) ¯ ) , and ( W , W ¯ ) is isomorphic to ( S o , V ) .

## Proof

Clearly, ( W , W ¯ ) ( Γ , E ( Γ ) ¯ ) . For any ( r , r ) , ( s , s ) ( W , W ¯ ) , we have ( r , r ) ( s , s ) = ( e r ( r s ) , ( r s ) f s ) = ( e r r s , r s f s ) = ( r s , r s ) ( W , W ¯ ) and ( W , W ¯ ) is a subsemigroup. For any s ( S o , V ) , we naturally define a mapping φ : ( S o , V ) ( W , W ¯ ) by s φ = ( s , s ) . It is easy to see that φ is bijective. For any s , t ( S o , V ) , s φ t φ = ( s , s ) ( t , t ) = ( s t , s t ) = ( s t ) φ . Thus, φ is a homomorphism and ( S o , V ) ( W , W ¯ ) . By Lemmas 4.4 and 4.5, ( r + , r + ) , ( r , r ) W ¯ with ( r + , r + ) ˜ E ( Γ ) ¯ ( r , r ) ˜ E ( Γ ) ¯ ( r , r ) . Therefore, ( W , W ¯ ) is a -subsemigroup of ( Γ , E ( Γ ) ¯ ) .□

## Lemma 4.8

( W , W ¯ ) is an Ehresmann transversal of ( Γ , E ( Γ ) ¯ ) .

## Proof

For any λ = ( r , a ) Γ , let λ ¯ = ( r ¯ , r ¯ ) W , e λ = ( e r , r ¯ + ) , f λ = ( a ¯ , f a ) . It is obvious that λ = e λ λ ¯ f λ with e λ λ ¯ + = ( r ¯ + , r ¯ + ) , f λ λ ¯ = ( a ¯ , a ¯ ) .

If λ can be written in another form λ = e λ λ ¯ f λ , where e λ = ( s 1 , b 1 ) E ( Γ ) ¯ , f λ = ( s 2 , b 2 ) E ( Γ ) ¯ , λ ¯ = ( s ¯ , s ¯ ) W and e λ λ ¯ + = ( s ¯ + , s ¯ + ) , f λ λ ¯ = ( s ¯ , s ¯ ) . By Lemma 4.7, we have f b 1 = s ¯ + , b 1 ¯ = s ¯ + and a ¯ = e s 2 , s 2 ¯ + = a ¯ . From e λ ˜ U λ ˜ U e λ , we deduce that e r = e s 1 , r ¯ + = s 1 ¯ + . Similarly, we have f a = f b 2 , a ¯ = b 2 ¯ . Therefore,

b 1 = e b 1 b 1 ¯ f b 1 = b 1 ¯ b 1 ¯ = b 1 ¯ S o ,
s 2 = e s 2 s 2 ¯ f s 2 = e s 2 s 2 ¯ = s 2 ¯ + s 2 ¯ = s 2 ¯ S o ,
and so e λ = ( s 1 , b 1 ¯ ) , f λ = ( s 2 ¯ , b 2 ) . It follows from e λ , f λ E ( Γ ) ¯ and Lemma 4.3 that b 1 ¯ s 1 = s 1 ¯ , b 2 s 2 ¯ = b 2 ¯ . From b 1 ¯ s 1 = s 1 ¯ , we deduce that b 1 ¯ e s 1 s 1 ¯ = s 1 ¯ since s 1 = e s 1 s 1 ¯ . Thus, b 1 ¯ e s 1 s 1 ¯ + = s 1 ¯ + since s 1 ¯ ˜ V s 1 ¯ + . Note that s 1 ¯ = b 1 ¯ and s 1 I , we have b 1 ¯ = s 1 ¯ = e s 1 ¯ = s 1 ¯ + = r ¯ + and similarly s 2 ¯ = a ¯ . Consequently, e λ = ( s 1 , b 1 ¯ ) = ( e s 1 , r ¯ + ) = ( e r , r ¯ + ) = e λ and similarly f λ = f λ . Hence,
λ = e λ λ ¯ f λ = ( e r , r ¯ + ) ( s ¯ , s ¯ ) ( a ¯ , f a ) = ( e r s ¯ , r ¯ + s ¯ f a ) .

This implies r = e r s ¯ , and combining with r = e r r ¯ implies that e r s ¯ = e r r ¯ . Premultiplying by r ¯ + , we obtain r ¯ + s ¯ = r ¯ . Note that r ¯ + = b 1 ¯ = b 1 ¯ = s ¯ + . Therefore, r ¯ = r ¯ + s ¯ = s ¯ + s ¯ = s ¯ and by Definition 2.3, ( W , W ¯ ) is an Ehresmann Transversal of ( Γ , E ( Γ ) ¯ ) .□

## Lemma 4.9

W is a quasi-ideal of Γ .

## Proof

For any ( s , s ) , ( t , t ) W , ( r , a ) Γ , we have

( s , s ) ( r , a ) ( t , t ) = ( e s s r f a t , s r f a t f t ) = ( s r f a t , s r f a t ) W ,
since both a right ideal and a left ideal are quasi-ideals.□

Now we shall prove the converse half of Theorem 4.1. Assume that ( S , U ) is a U-abundant semigroup with a quasi-ideal Ehresmann transversal ( S o , V ) . Let

R = { r S : f r V } , L = { a S : e a V } .
Then by Proposition 3.6, R and L are both U-abundant semigroups with an Ehresmann transversal ( S o , V ) , where S o is a right ideal of R and a left ideal of L.

For every ( a , r ) L × R , let a r = a r . Then e a , f r V and so a r = a r = e a a r f r S o since S o is a quasi-ideal of S. Obviously, the map satisfies (1), (2) and (3). Therefore, we may acquire a U-abundant semigroup Γ in the way of the direct part of Theorem 4.1. In the following, we prove that Γ is isomorphic to ( S , U ) .

For every ( r , a ) Γ , we define θ : Γ S by ( r , a ) θ = e r a . It is clear that θ is well-defined and by Proposition 3.6, θ is injective.

For any ( r , a ) , ( s , b ) Γ , we have

[ ( r , a ) ( s , b ) ] θ = [ e r ( a s ) , ( a s ) f b ] θ = e e r ( a s ) ( a s ) f b = e r ( e r ¯ f e r ( a s ) + ) + ( a s ) f b = e r r ¯ + ( a s ) + ( a s ) f b = e r ( a s ) f b = e r a s f b = e r a e s b ( s f b = e s s ¯ f b = e s b ¯ f b = e s b ) = ( r , a ) θ ( s , b ) θ ,
and so θ is a homomorphism.

For any a S , it is obvious that a a ¯ R and a ¯ + a L with a a ¯ ¯ = a ¯ f a a ¯ a ¯ = a ¯ a ¯ = a ¯ and a ¯ + a ¯ = a ¯ + a ¯ + e a a ¯ = a ¯ + a ¯ = a ¯ . Consequently, ( a a ¯ , a ¯ + a ) Γ and ( a a ¯ , a ¯ + a ) θ = e a a ¯ a ¯ + a = e a a ¯ + a = e a a = a , that is, θ is surjective. Therefore, θ is an isomorphism.

Funding information: This research was supported by the Project Higher Educational Science and Technology Program of Shandong Province (J18KA248), Shandong Province NSF (ZR2020MA002, ZR2016AM02) and the NSFC (11801081, 11871301).

Author contributions: Xiangjun Kong contributed in an overall design, put forward the main results and gave most of the proof; Pei Wang gave the idea of writing and specific writing of this paper, using latex to compile this paper and give some proof; Jian Tang optimized the structure of this paper, polished the language, and gave the valuable suggestion for Theorem 4.1 which greatly improved the readability of the paper.

Conflict of interest: Authors state no conflict of interest.

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