Quasi-ideal Ehresmann transversals: The spined product structure

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

2. ℛ ˜ U ∩ ( R e g U S × R e g U S ) = ℛ ∩ ( 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 ℒ ˜ U , and there is a dual result for ℛ ˜ 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. e ℒ ˜ U a .

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

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 ℒ x ¯ + and f ℛ 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 ℛ ˜ U x , x ℒ ˜ U f x .

Lemma 2.4

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

1. I ∩ Λ = V ;

2. I = { e ∈ U | ( ∃ v ∈ V ) v ℒ e } , Λ = { f ∈ U | ( ∃ h ∈ V ) h ℛ f } ;

3. I V ⊆ I , V Λ ⊆ Λ ;

4. x ℛ ˜ U y if and only if e x = e y ; x ℒ ˜ U y if and only if f x = f y ;

5. If e , f ∈ I and e ℛ f , then e = f ; if e , f ∈ Λ and e ℒ f , then e = f ;

6. ( ∀ x ∈ R e g U S ) | V U ( x ) ∩ T | ≤ 1 ;

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

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 .

Lemma 2.6

[25]

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

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 ℛ ˜ U c ℒ ˜ 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

From the definition of an Ehresmann transversal, we have c = e c c ¯ f c , where e c ℒ ˜ U c ¯ + and f c ℛ ˜ U c ¯ ∗ . Moreover, e c ℛ ˜ U c ℒ ˜ U f c . It follows that a + ℛ ˜ U c ℛ ˜ U e c ℒ ˜ U c ¯ + , consequently e c ∈ Λ ∩ I = V . Similarly, c ¯ ∗ ℛ ˜ U f c ℒ ˜ U c ℒ ˜ 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 ℛ ˜ U x ℒ ˜ U f x we deduce that e x ℛ x ℒ 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 ¯ ∗ ℛ f x ℛ x ′ ℒ e x ℒ 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,

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 ℒ x ¯ + and f x ℛ 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

Then

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 ℒ ˜ U v .

Conversely, for r ∈ S , if there exists v ∈ V such that r ℒ ˜ U v , then f r ℒ ˜ U r ℒ ˜ 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

Similarly, we have e s b ¯ = b ¯ . Thus, a ¯ = b ¯ , r ¯ = s ¯ and consequently f r = r ¯ ∗ = s ¯ ∗ = f s . Therefore,

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. S o is a quasi-ideal of S;

2. Λ I ⊆ S o ;

3. V I ⊆ S o and Λ V ⊆ S o ;

4. L R ⊆ S o ;

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

6. S o I ⊆ S o and Λ S o ⊆ S o ;

7. S o I S o ⊆ S o and S o Λ S o ⊆ S o ;

8. S S o ⊆ R , S o S ⊆ L ;

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 ℛ f + and e ℒ 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 ℛ ˜ U h and r ℒ ˜ U v . Thus,

(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

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

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

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

2. If { r , a } ∩ S o ≠ ∅ , then a ∗ r = a r ;

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

Define a multiplication on the set

by

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.

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

On the other hand, we have

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

On the other hand, we have

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

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 ℒ 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,

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 ) ℛ ˜ E ( Γ ) ¯ ( e r , r ¯ + ) .