On generalized Ehresmann semigroups

Shoufeng Wang 1
  • 1 Department of Mathematics, Yunnan Normal University, Kunming, 650500, Yunnan, China
Shoufeng Wang
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  • Department of Mathematics, Yunnan Normal University, Kunming, Yunnan, 650500, China
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Abstract

As a generalization of the class of inverse semigroups, the class of Ehresmann semigroups is introduced by Lawson and investigated by many authors extensively in the literature. In particular, Gomes and Gould construct a fundamental Ehresmann semigroup CE from a semilattice E which plays for Ehresmann semigroups the role that TE plays for inverse semigroups, where TE is the Munn semigroup of a semilattice E. From a varietal perspective, Ehresmann semigroups are derived from reduction of inverse semigroups. In this paper, from varietal perspective Ehresmann semigroups are extended to generalized Ehresmann semigroups derived instead from normal orthodox semigroups (i.e. regular semigroups whose idempotents form normal bands) with an inverse transversal. We present here a semigroup C(I,Λ,E) from an admissible triple (I, Λ, E) that plays for generalized Ehresmann semigroups the role that CE from a semilattice E plays for Ehresmann semigroups. More precisely, we show that a semigroup is a fundamental generalized Ehresmann semigroup whose admissible triple is isomorphic to (I, Λ, E) if and only if it is (2,1,1,1)-isomorphic to a quasi-full (2,1,1,1)-subalgebra of C(I,Λ,E). Our results generalize and enrich some results of Fountain, Gomes and Gould on weakly E-hedges semigroups and Ehresmann semigroups.

1 Introduction

Let S be a semigroup. We denote the set of all idempotents of S by E(S) and the set of all inverses of xS by V(x) . Recall that
V(x)={aS|xax=x,axa=a}
for all xS. A semigroup S is called regular if V(x) ≠ ∅ for any xS, and a regular semigroup S is called inverse if E(S) is a commutative subsemigroup (i.e. a subsemilattice) of S, or equivalently, the cardinal of V(x) is equal to 1 for all xS.

Recall that a regular semigroup S is fundamental if the largest congruence contained in 𝓗 on S is trivial. Structure theorems for certain important subclasses of the class of fundamental regular semigroups are already known. Munn [1] initiated the work in this direction. He proved that given a semilattice E, the Munn semigroup TE of all isomorphisms of principal ideals of E is “maximaltl” in the class of all fundamental inverse semigroups whose semilattices of idempotents are E, that is, every semigroup belonging to this class is isomorphic to a full inverse subsemigroup of TE. Further from Munn [1] if S is an inverse semigroup such that E(S) is isomorphic to a given semilattice E, then there exists a homomorphism f : STE and the kernel of f is the largest congruence contained in 𝓗 on S.

The pioneer work of Munn was generalized firstly by Hall in 1971 to orthodox semigroups (i.e. regular semigroups whose idempotents form subsemigroups) in [2] in which the Hall semigroup WB of a band B was constructed. Recall that a band is a semigroup in which every element is idempotent. The Hall semigroup WB has properties analogous to those described above for TE (see Hall [2] for details). As another direction, Fountain [3] generalized Munn’s result to a class of non-regular semigroup, namely ample semigroups. The next step in this direction was made by Fountain, Gomes and Gould in [4] where by using of a completely fresh technology they considered the class of weakly E-hedged semigroups which is a special class of Ehresmann semigroups first defined by Lawson in [5]. The natural next step in this direction is to look at the whole class of Ehresmann semigroups. This is done by Gomes and Gould in [6] in which they removed the “weakly E-hedgedtl” condition from [4] and considered the whole calss of Ehresmann semigroups. For a given semilattice E, in [6] the authors have constructed a semigroup CE which plays the TE role for Ehresmann semigroups and generalized the main results in [4]. Furthermore, El-Qallali-Fountain-Gould [7], Gomes-Gould [8] and Wang [9] went a step further to extend Hall’s approach for orthodox semigroups to some classes of non-regular semigroups having a band of idempotents. It is worth remarking that the class of Ehresmann semigroups and its subclasses are investigated extensively in literature by many famous semigroup researchers (see [10-13] for example). In particular, Jones [12] provided a common framework for Ehresmann semigroups and regular *-semigroups from varietal perspective. More recent developments in this area can be found in good survey articles by Gould [11, 14] and Hollings [15, 16].

On the other hand, Blyth-McFadden [17] introduced the concept of inverse transversals for regular semigroups. A subsemigroup S of a regular semigroup S is called an inverse transversal of S if V(x) ∩ S contains exactly one element for all xS. Clearly, in this case, S is an inverse subsemigroup of S. Since an inverse semigroup can be regarded as an inverse transversal of itself, the class of regular semigroups with inverse transversals contains the class of inverse semigroups as a proper subclass. Regular semigroups with inverse transversals are investigated extensively by many authors (see [18-21] and their references) and some generalizations of inverse transversals are proposed, see [22, 23] for example.

Inspired by the approach used in Jones [12], in this paper a common framework, termed generalized Ehresmann semigroups, for Ehresmann semigroups and normal orthodox semigroups with an inverse transversal is introduced from varietal perspective, where a normal orthodox semigroup means a regular semigroup whose idempotents form a normal band. We construct a semigroup C(I,Λ,E) from the so-called admissible triple (I, Λ, E) that plays for generalized Ehresmann semigroups the role that CE from a semilattice E plays for Ehresmann semigroups. More precisely, we show that a semigroup is a fundamental generalized Ehresmann semigroup whose admissible triple is isomorphic to (I, Λ, E) if and only if it is (2,1,1,1)-isomorphic to a so-called quasi-full (2,1,1,1)-subalgebra of C(I,Λ,E). This generalizes and enriches some results of Fountain, Gomes and Gould on weakly E-hedges semigroups and Ehresmann semigroups obtained in texts [4] and [6].

2 Generalized Ehresmann semigroups

In this section, after giving some preliminary results on Ehresmann semigroups and inverse transversals, we introduce generalized Ehresmann semigroups and consider some basic properties of this class of semigroups. Firstly, we consider Ehresmann semigroups. Let S be a semigroup and let EE(S). The relation R~E is defined on S by the rule that for any x, yS, we have x R~Ey if
ex=x if and only if ey=y for all eE.

Dually, we have the relation L~E on S. Observe that both R~E and L~E are equivalences on S but R~E (resp. L~E) may not be a left congruence (resp. a right congruence). From Lawson [5], a semigroup S is an Ehresmann semigroup with respect to E (or (S, E) is an Ehresmann semigroup) if

  1. E is a subsemilattice of S,
  2. every R~E-class contains a unique element of E and R~E is a left congruence,
  3. every L~E-class contains a unique element of E and L~E is a right congruence.

If this is the case, E is called the distinguished semilattice of S. We say that a semigroup S is Ehresmann in the sequel if (S, E) is an Ehresmann semigroup for some EE(S). From Lemma 2.2 and its dual in Gould [14], we have the following characterization of Ehresmann semigroups from a varietal perspective.

Lemma 2.1
A semigroup (S, ⋅) is Ehresmann if and only if there are two unary operations “+” and “*” on S such that the following identities hold:
x+x=x,x+y+=y+x+,(x+y+)+=x+y+,(xy)+=(xy+)+;xx=x,xy=yx,(xy)=xy,(xy)=(xy);(x+)=x+,(x)+=x.

In this case, S is an Ehresmann semigroup with distinguished semilattice {x+|xS} (={x*|xS}), and we shall call (S⋅, +, *) an Ehresmann semigroup.

Now we consider regular semigroups with an inverse transversal. Let S be a regular semigroup and S an inverse transversal of S. For any xS, we use x to denote the unique inverse of x in S and let x∘∘ = (x). We can consider the induced tri-unary semigroup (S, ⋅, +, *, −), where
x+=xx,x=xx,x¯=x.

Observe that x∘∘∘ = x and x∘∘x, xx∘∘E(S). Here we are only interested in normal orthodox semigroups with an inverse transversal. Recall that a normal orthodox semigroup means a regular semigroup whose idempotents form a normal band, and a band B is called left normal (resp. right normal, normal) if efg = egf (resp. efg = feg, efge = egfe) for all e, f, gB. Observe that a band B is normal if and only if efgh = egfh for all e, f, g, hB. The following proposition can be deduced from the text [20]. However, we give its direct proof for the sake of completeness.

Proposition 2.2

Let S be a normal orthodox semigroup with an inverse transversal S. The tri-unary semigroup (S, ⋅, +, *, −) satisfies the following identities:

Proof

By symmetry, we only need to show (1)–(8).

  1. This is equivalent to the equality (xx)x = x.
  2. As E(S) is a normal band, we have
    x+y+=(xx)[(yy)(yy)](yy)=(xx)[(yy)(yy)](yy)=xxyy.
    This implies that
    x+y+x+=x+(y+x+)=xx(yyoxx)=(xxoyy)(xx)=xx(yy)(xx)=xx(xx)(yy)=xxyy=x+y+.
  3. We first observe that
    (xy)=yx for all x,yS.
    In fact, since S is orthodox, yx is an inverse of xy. But yxS, so (xy) = yx. By (1) and (2), we have
    (x+y+)+=(xxyy)(xxyy)=xxyyyyxx=xxyyxx=xxxxyy=xxyy=x+y+.
  4. By (2), we have
    (xy+)+=(xyy)(xyy)=xyyyyx=xyyx=xy(xy)=(xy)+.
  5. From (2),
    (x+)=(xx)(xx)=xxxx=xx=xx=x¯+.
  6. It follows that x+¯ = (xx)∘∘ = x∘∘x = x∘∘(x∘∘) = x+ from (2).
  7. This is equivalent to the statement x = (xx)x∘∘(xx).
  8. Since E(S) is a normal band, using (1) and its dual, we have
    xy+=xxyy=xxxxyyyy=(xxyy)(xxyy)=(xxyy)(xxyy)=xxyy=((x)x)(y(y))=x¯y¯+,
    as required. □
    We shall term any tri-unary semigroup (S, ⋅, +, *, −) that satisfies the identities in Table 1 a generalized Ehresmann semigroup. By Proposition 2.2, any normal orthodox semigroup S with an inverse transversal S induces the generalized Ehresmann semigroup (S, ⋅, +, *, −) by setting x+ = xx, x* = xx and x = x∘∘. The following example gives a very special case of this kind of generalized Ehresmann semigroups.

Table 1

Generalized Ehresmann conditions

(1)x+x = x(1)′xx* = x
(2)x+y+x+ = x+y+(2)′x*y*x* = y*x*
(3)(x+y+)+ = x+y+(3)′(x*y*)* = x*y*
(4)(xy)+ = (xy+)+(4)′(xy)* = (x*y)*
(5)(x+)* = x+(5)′(x*)+ = x*
(6)x+ = x+(6)′x* = x*
(7)x = x+xx*(8)′x*y+ = x*y+

Example 2.3

Let S be a rectangular band. Fix an element u in S. Consider the tri-unary semigroup (S, ⋅, +, *, −) where x+ = xu, x* = ux, x = u. Then it is routine to check that the identities in Table 1 are satisfied and so (S, ⋅, +, *, −) is a generalized Ehresmann semigroup. In fact, S is indeed a normal band and {u} is an inverse transversal of S.

Example 2.4

Any Ehresmann semigroup (S, ⋅, +, *) also induces a generalized Ehresmann semigroup, which justifies our term “generalized Ehresmann semigroups” In fact, for an Ehresmann semigroup (S, ⋅, +, *), we define the third unary operation “−” on S by x = x. Then we have the tri-unary semigroup (S, ⋅, +, *, −) and it is easy to see that the identities in Table 1 are all satisfied by Lemma 2.1.

Since a rectangular band having more than one element must not be an Ehresmann semigroup, the class of generalized Ehresmann semigroups contains the class of Ehresmann semigroups and the class of rectangular bands as proper subclasses by the above two examples. We also observe that a generalized Ehresmann semigroup which is also regular may not contain any inverse transversal. In fact, any monoid S with the identity 1 is always a (generalized) Ehresmann semigroup by setting x+ = x* = 1 and x = x for all xS. Obviously, a regular monoid may not contain any inverse transversal. Here is an example.

Example 2.5

Let M = {1, b, c, x} (taken from Exercise 10 in Chapter VI of [24]) with the multiplication

M1bcx
11bcx
bbbbb
ccccc
xxcb1

Then M is a monoid and
E(M)={1,b,c},V(1)={1},V(b)={b,c}=V(c),V(x)={x}.

It is easy to check that M contains no inverse transversal.

In the remainder of this section, we consider some properties associated to generalized Ehresmann semigroups which will be used in the next sections. Let (S, ⋅, +, *, −) be a generalized Ehresmann semigroup. Denote
IS={x+|xS},ΛS={x|xS},ES={x¯+|xS}.
Lemma 2.6

Let (S, ⋅, +, *, −) be a generalized Ehresmann semigroup.

  1. i+ = i, i* = i and λ* = λ, λ+ = λ for all iIS and λ ∈ΛS.
  2. ES={x¯|xS}=ISΛS,
  3. x+𝓛x+ and x*𝓡x*.
  4. IS is a left normal band, ΛS is a right normal band and ES is a subsemilattice of S, respectively.

Proof

  1. Using identities (1),(4) and (3) in Table 1, we have
    x+=(x+x)+=(x+x+)+=x+x+
    and so x+E(S). Take iIS. Then i = x+ for some xS. This implies that
    i+=(x+)+=(x+x+)+=x+x+=x+
    by x+E(S) and the identity (3). Moreover,
    i=(x+)=x¯+=x+¯=i¯
    by the identities (5) and (6). By symmetry, λ* = λ and λ+ = λ for all λ ∈ ΛS.
  2. By identity (5), x+ = (x+)*IS ∩ ΛS for all xS. Now let u = x+IS ∩ ΛS for some xS. Using item (a) and the identity (5), we have
    u=u=(x+)=x¯+ES.
    Thus ES = IS ∩ ΛS. Dually, {x*|xS} = IS ∩ ΛS.
  3. Using identities (7), (4), (3), (4), (5)′ and (1)′ in Table 1 in that order, we have
    x+=(x+x¯x)+=(x+(x¯x)+)+=x+(x¯x)+=x+(x¯(x)+)+=x+(x¯x¯)+=x+x¯+.
    Since x+ ∈ ΛS by (b), we have (x+)* = x+ by (a). Using the identities (5), (8) and (6), we have
    x¯+x+=(x+)x+=(x+¯)x¯+=(x¯+)x¯+=x¯+x¯+=x¯+.
    This shows that x+𝓛x+. Dually, x*𝓡x*
  4. In view of the proof of item (a), every element in IS is idempotent. By the identity (3), x+y+ = (x+y+)+IS for all x+, y+IS. So IS is a subband of S. By the identity (2) and (2)' and item (b), ES is a subsemilattice of S. Moreover, for x, y, zS, by identities (5), (8) and (6) and items (b) and (a), we have
    x¯+y+=(x+)y+=x+¯y¯+=(x¯+)y¯+=x¯+y¯+.
    Similarly, y+z+ = y+z+. In view of item (c) and the fact that ES is a subsemilattice,
    x+y+z+=x+x¯+y+z+=x+x¯+y¯+z+=x+x¯+y¯+z¯+=x+x¯+y¯+z¯+=x+y¯+z¯+=x+z¯+y¯+.
    Similarly, we have x+z+y+ = x+z+y+. This yields that IS is a left normal band. Dually, ΛS is a right normal band. □Let S be a semigroup. Recall that the natural partial order “≤” on E(S) is defined as follows:
    ef if and only if ef=fe=e for all e,fE(S).

Lemma 2.7

Let (S, ⋅, +, *, −) be a generalized Ehresmann semigroup and x, yS.

  1. (xy)+ = x+(xy)+, (xy)* = (xy)*y*.
  2. (xy)+x+, (xy)* ≤ y*.
  3. xy = xy.

Proof

  1. Using the identities (4), (7), (8), (1)′, (4) and (3) in that order, we have
    (xy)+=(xy+)+=(x+x¯xy+)+=(x+x¯x¯y¯+)+=(x+x¯y¯+)+=(x+x¯y¯)+=(x+(x¯y¯)+)+=x+(x¯y¯)+.
    Dually, (xy)* = (xy)*y*.
  2. Since (xy)+ = x+(xy)+ by item (a), we have x+(xy)+ = (xy)+. Moreover, by the identity (2),
    (xy)+x+=x+(x¯y¯)+x+=x+(x¯y¯)+=(xy)+.
    So (xy)+x+. Dually, (xy)* ≤ y*.
  3. In view of the identity (8) and items (b), (d) in Lemma 2.6, we have (x+)(x¯y¯)+=x+¯x¯y¯¯+ES so that ((x+)*(xy)+)* = (x+)*(xy)+ by Lemma 2.6 (a). Using the identity (5), item (a) of this lemma, the identities (4)′, (5), (3), (4), (1) in that order, we obtain that
    xy¯+=((xy)+)=(x+(x¯y¯)+)=((x+)(x¯y¯)+)=(x+)(x¯y¯)+=x¯+(x¯y¯)+=(x¯+(x¯y¯)+)+=(x¯+x¯y¯)+=(x¯y¯)+.
    Dually, xy* = (xy)*. Using the identities (1) and (1)′, Lemma 2.6 (c), the identities (7), (8), (1), (1)′, item (b) of this lemma, Lemma 2.6 (c), the identities (1) and (1)′ in that order, we get
    xy¯=xy¯+xy¯xy¯=xy¯+(xy)+xy¯(xy)xy¯=xy¯+xyxy¯=xy¯+x+x¯xy+y¯yxy¯=xy¯+x+x¯x¯y¯+y¯yxy¯=xy¯+x+x¯y¯yxy¯=(x¯y¯)+x+x¯y¯y(x¯y¯)=(x¯y¯)+x¯+x+x¯y¯yy¯(x¯y¯)=(x¯y¯)+x¯+x¯y¯y¯(x¯y¯)=(x¯y¯)+x¯y¯(x¯y¯)=x¯y¯,
    as required. □

3 The semigroup C(I,Λ,E)

We call a generalized Ehresmann semigroup (S,⋅, +,*,−) fundamental if the largest semigroup congruence μS contained in the equivalence
{(a,b)S×S|a+=b+,a=b}
is the identity relation on S. In this section, we shall construct a fundamental generalized Ehresmann semigroup which plays the similar role in the class of generalized Ehresmann semigroups as the Munn semigroup of a semilattice in the class of inverse semigroups. To do this, we need to introduce the notion of admissible triples, which is motivated by Lemma 2.6.
Definition 3.1

Let I (resp. Λ) be a left normal band (resp. a right normal band), E = I ∩ Λ a subsemilattice of I and Λ. The triple (I, Λ, E) is called admissible if for all gI and f ∈ Λ, there exist g, fE such that g𝓛g and f𝓡f.

Remark 3.2

Let (I, Λ,E) be an admissible triple. Since E is a subsemilattice, the elements g and f in Definition 3.1 are uniquely determined by g and f, respectively. In particular, iE if and only if i = i.

Remark 3.3

Let (S, .,+,*,−) be a generalized Ehresmann semigroup. By Lemma 2.6 (b), (c) and (d), (IS, ΛS, ES) is an admissible triple which will be called the admissible triple of S. In this case, for all iIS and λ ∈ ΛS, we have i = i* and λ = λ+. In fact, if x+IS, then by the identity (5) in Table 1 and Lemma 2.6 (c), we have x+𝓛x+ = (x+)*. The case for λ ∈ ΛS can be showed dually.

To construct the semigroup C(I,Λ,E) of an admissible triple (I, Λ, E), we need some preliminaries. First, we have the following basic facts on admissible triples.

Lemma 3.4
Let (I, Λ,E) be an admissible triple and e,gI,f,h ∈ Λ. Then
eg=eg,(eg)=eg,fh=fh,(fh)=fh.

Moreover, we have eEe = eE and fEf = Ef, which are subsemilattices of I and Λ, respectively.

Proof
Since g𝓛g and I is a left normal band, we have eg = egg = egg = eg. This implies that eg = eg, and so eg𝓛eg = egE by the fact that e𝓛e. This yields that (eg) = eg. Finally, it follows that eEe = eE by the fact that I is a left normal band. Moreover, for i, jE, we have
(ei)(ej)=(eie)j=eij=eji=(eje)i=(ej)(ei),
whence eE is a subsemilattice of I. The remaining facts of this lemma can be proved by symmetry. □
Let (I, Λ, E) be an admissible triple. We use I1 (resp. Λ1) to denote I (resp. Λ) with the identity adjoined. Furthermore, we always assume that 1 = 1 for the adjoined identity 1 on both I and Λ. A function η from I1 to Λ is called order-preserving if in Λ for all x, yI1 with xy in I1, where ≤ is defined in (3). Denote the set of order-preserving functions from I1 to Λ by 𝓞(I1 → Λ). Dually, we also have 𝓞(Λ1I). Similarly, we have 𝓞(I1I) and 𝓞(Λ1 →Λ). Moreover, we denote the set of the morphisms from I1 to Λ by End (I1 → Λ). Dually, we have End (Λ1I). Obviously,
End(I1Λ)O(I1Λ),End(Λ1I)O(Λ1I).
For every eI (resp. f ∈ Λ), define
ρe:I1I,xex(resp.σf:Λ1Λ,xxf).

Then it is easy to see that ρe ∈ 𝓞(I1I) and σf ∈ 𝓞(Λ1 → Λ) for all eI and f ∈ Λ as I is a left normal band and Λ is a right normal band. Moreover, we use ρ1 and σ1 to denote the identity maps on I and Λ, respectively.

Lemma 3.5
Let (I, Λ,E) be an admissible triple. Define a multiplication “⋄” on 𝓞(I1 → Λ)as follows: for all α,β ∈ 𝓞(I1 → Λ),
αβ:I1Λ,x(xα)β.
Then 𝓞(I1 → Λ) is a semigroup with respect to “⋄”. Dually, for α, β ∈ 𝓞(Λ1I), define
βα:Λ1I,x(xβ)α,
then 𝓞(Λ1I) forms a semigroup with respect to “⋆”.
Proof
Since ∅ ≠ End (I1 → Λ) ⊆ 𝓞(I1 → Λ), it follows that 𝓞(I1 → Λ) ≠ ∅. Observe that
xy for all x,yI1(or x,yΛ1)with xy
by Lemma 3.4. Let x, yI1 and xy. Then , ∈ Λ and as α is order-preserving. This implies that () ≤ (). Observe that β is also order-preserving, it follows that (())β ≤ (())β. Thus αβ ∈ 𝓞(I1 → Λ). Now let α, β, γ ∈ 𝓞(I1 → Λ) and x, yI1. Then
x[(αβ)γ]=[(x(αβ))]γ=[(xα)β]γ=(xα)(βγ)=x[α(βγ)].

This implies that 𝓞(I1 → Λ) is a semigroup with respect to “⋄”. Dually, 𝓞(Λ1I) forms a semigroup with respect to “⋆”. □

Corollary 3.6
Let (I, Λ,E) be an admissible triple, and “⋄” and “⋆” be defined as in Lemma 3.5. Then 𝓞(I1 → Λ) × 𝓞(Λ1I) forms a semigroup by defining
(α,β)(γ,δ)=(αγ,δβ)
for all (α,β),(γ, δ) ∈ 𝓞(I1 → Λ) × 𝓞(Λ1I).
The semigroup 𝓞(I1 → Λ) is partially ordered by “≤” where for all α, β ∈ 𝓞(I1 → Λ),
αβ if and only if xαxβ in Λ for all xI1.

Similarly, the semigroups 𝓞(Λ1I), 𝓞(I1I) and 𝓞(Λ1 → Λ) can be partially ordered, respectively.

Consider the subset
C(I,Λ,E)={(α,β)|(xI1)(yΛ1)xα=xα,yβ=yβ,σxαβρxα,ρyβασyβ}
of the product semigroup 𝓞(I1 → Λ) × 𝓞(Λ1I).
Lemma 3.7
Let (I, Λ, E) be an admissible triple. For eI, define θe and τe as follows:
θe:I1Λ,xex,τe:Λ1I,xex.
Then (θe, τe) ∈ C(I,Λ,E). Dually, for f ∈ Λ, define η f and ξf as follows:
ηf:I1Λ,xxf,ξf:Λ1I,xxf.

Then (ηf, ξf) ∈ C(I,Λ,E). Moreover, if eI ∩ Λ = E, then (θe, τe) = (ηe, ξe).

Proof
Firstly, by Lemma 3.4, we have ex = exE ⊆ Λ for all xI, and xE, exI for all x ∈ Λ. Therefore θe and τe are well-defined. Secondly, since I is a left normal band, we have
(xθe)(yθe)=(ex)(ey)=exey=exy=(xy)θe
for all x, yI1. On the other hand, by Lemma 3.4, we have
(xτe)(yτe)=exey=exy=e(xy)=(xy)τe
for all x, y ∈ Λ1. This shows that θe and τe are morphisms and so order-preserving. Thirdly, if xI, then e = ex = ex = xθe. Similarly, e = ey = e(y) = yτe for all y ∈ Λ. Finally, let xI1 and u ∈ Λ1. Then by Lemma 3.4,
uσxθe=uσex=u(ex)=u(ex)=uex
and
u(τeρxθe)=e(x(eu))=e(x(eu))=e(xeu)=exu=exu=uex.
This implies that σe = τeρxθe for all xI1. Similarly, ρe = θeσyτe for all y ∈ Λ1. Thus (θe, τe) ∈ C(I,Λ,E). Dually, (ηf, ξf) ∈ C(I,Λ,E). If eE, then e = e and so
xθe=ex=ex=ex=xe=xηe
for all xI1. This shows that θe = ηe. Dually, τe = ξe. □
Lemma 3.8

C(I,Λ,E) is a subsemigroup of 𝓞(I1 → Λ)× 𝓞(Λ1I).

Proof
Denote C = C(I,Λ,E). We have seen that C is non-empty by Lemma 3.7. Let (α, β), (γ, δ) ∈ C. Then
(α,β)(γ,δ)=(αγ,δβ).
We first show that σx(αγ) ≤ (δβ)ρx(αγ) for all xI1. Let u ∈ Λ1 and xI1. Then uδI and (uδ)I ∩ Λ. Since σβρxα by the fact that (α, β) ∈ C, we have ()σ≤()βρxα whence ()()≤(x(()β))α. In view of (4) and Lemma 3.4, we can obtain that
(xα)(uδ)=(xα)(uδ)=((uδ)(xα))((x((uδ)β))α).
Since γ is order-preserving, it follows that
u(δρ(xα)γ)=((xα)(uδ))γ(((x((uδ)β))α))γ=u(δβ)ρx(αγ).
This shows that δρ()γ≤ (δβ)ρx(αγ). Since σδρvγ for all vI1 by the fact (γ, δ) ∈ C and ()I, it follows that
σx(αγ)=σ(xα)γδρ(xα)γ(δβ)ρx(αγ).
Finally, let xI. Since xα = by the fact that (α, β) ∈ C, we get
x(αy)=(xα)γ=(xα)γ=x(αγ).

By symmetry, we can obtain ρy(δβ) ≤ (αγ)σy (δβ) and y(δβ) = y(δβ) for all y ∈ Λ1. Thus (α, β)(γ, δ) ∈ C. So C is a subsemigroup. □

Now, we are in a position to state our main result of this section.

Theorem 3.9
Define three unary operations on the semigroup C = C(I,Λ,E) as follows:
(α,β)+=(α+,β+),(α,β)=(α,β),(α,β)¯=(α¯,β¯),
where
α¯:I1Λ,x(xα),β¯:Λ1I,x(xβ),α+=θ1β:I1Λ,x(1β)x,β+=τ1β:Λ1I,x(1β)xα=η1α:I1Λ,xx(1α)β=ξ1α:Λ1I,xx(1α).

Then (C, ., +, *, ) is a generalized Ehresmann semigroup.

Proof
Let (α, β) ∈ C. By Lemma 3.7 and the fact that 1βI, 1α ∈ Λ, it follows that
(α+,β+)=(θ1β,τ1β)C,(α,β)=(η1α,ξ1α)C.

This shows that “+″ and “*″ are well-defined.

Now, let (α, β) ∈ C. Then
xα=xα,yβ=yβ,σxαβρxα,ρyβασyβ
for all xI1 and y ∈ Λ1. Let x, yI1 and xy. Since α is order-preserving, . It follows that xα = () ≤ () = yα by (4). This shows that α is order-preserving. Dually, β is also order-preserving. Moreover, for all xI1, we have
xα¯=(xα)=(xα)=xα¯
as xα = for all xI1. Dually, we have yβ = yβ for all y ∈ Λ1. Now let xI1. For u ∈ Λ1, since σβρxα, we have x αρxα. That is, u()≤(x())α. By (4), (u(x α))≤((x())α). This implies that
uσxα¯=u(xα¯)=u(xα)=u(xα)=(u(xα))((x(uβ))α)=((x(uβ))α)=(x(uβ))α¯=(uβ)(ρxα¯)=u(β¯ρxα¯)
by Lemma 3.4. This shows that σxαβρxα for all xI1. Dually, ρyβα σy β for all y ∈ Λ1. Thus (α, β) ∈ C. This implies that “−″ is also well-defined.

We next show that the identities (1-8) in Table 1 are satisfied. By symmetry, the identities (1)′-(6)′ also hold. Let (α, β), (γ, δ) ∈ C.

  1. For xI1, we have σβρxα by the fact that (α, β) ∈ C. So
    xα=1σxα1(βρxα)=(x(1β))α.
    Since I is a left normal band, we have x(1β) ≤ x. This shows that (x(1β))α as α is order-preserving. So = (x(1β))α. Observe that vα = for all vI1 by the fact that (α, β) ∈ C, it follows that (x(1β))α = (x(1β))α. Thus
    xα=(x(1β))α=(x(1β))α=(x(1β))α=((1β)x)α=x(α+α)
    by lemma 3.4. We have shown that α+α = α. Dually, ββ+ = β. Thus
    (α,β)+(α,β)=(α+,β+)(α,β)=(α+α,ββ+)=(α,β).
  2. For xI1, by Lemma 3.4 we have
    x(α+γ+)=(xα+)γ+=((1β)x)γ+=((1β)x)γ+=(1δ)((1β)x)=(1δ)(1β)x
    and
    x((α+γ+)α+)=(x(α+γ+))α+=((1δ)(1β)x)α+=((1δ)(1β)x)α+=(1β)(1δ)(1β)x=(1δ)(1β)x.
    This implies that α+γ+ = α+γ+α+. Dually, β+δ+β+ = δ+β+. Thus
    (α,β)+(γ,δ)+(α,β)+=(α,β)+(γ,δ)+.
  3. For all xI1, we have x(α+0γ+) = (1δ)(1β)x by (7) and
    x(α+γ+)+=(1(δ+β+))x=((1β)(1δ))x=((1δ)(1β))x=(1δ)0(1β)x
    by Lemma 3.4. This yields that α+γ+ = (α+γ+)+. Dually, (δ+β+)+ = δ+β+. Thus ((α, β)+(γ, δ)+)+ = (α, β)+(γ, δ)+.
  4. For all xI1, we have x(αγ)+ = (1 (δβ))x = ((1δ)β)x and
    x(αγ+)+=(1(δ+β))x=((1δ+)β)x=((1δ)β)x.
    This implies that (αγ)+ = (αγ+)+. Dually, (δβ)+ = (δ+β)+. Thus
    ((α,β)(γ,δ))+=((α,β)(γ,δ)+)+.
  5. For all xI1, by Lemma 3.4 we have
    x(α+)=x(1α+)=x(1β)=(1β)x=(1β)x=(1β¯)x=(1β¯)x=xα¯+.
    This shows that (α+)* = α+. Dually, (β+)* = β+. Thus ((α, β)+)* = (α, β)+
  6. For all xI1, by Lemma 3.4 and (8) we have
    xα¯+=(1β)x=((1β)x)=(xα+)=xα+¯.
    So α¯+=α+¯.Dually,β¯+=β+¯.Thus α,β¯+=α,β+¯.
  7. For all xI1, by Lemma 3.4 and (6), we have
    x(α+α¯α)=(((1β)x)α)(1α)=(((1β)x)α)(1α)=(xα)(1α)=(xα)(1α).
    Since x ≤ 1 and α is order-preserving, we have (x α)(1α) = . This shows that x(α+αα*) = . So α+αα* = α. Dually, β*ββ+ = β. Thus
    (α,β)=(α,β)+(α,β)¯(α,β).
  8. For all xI1, by Lemma 3.4 we have
    x(αγ+)=(xα)γ+=(x(1α))γ+=(1δ)x(1α)
    and
    x(α¯γ¯+)=(xα¯)γ¯+=(x(1α¯))γ¯+=(x(1α))γ¯+=(x(1α))γ¯+=(1δ¯)x(1α)=(1δ)x(1α).
    This shows that α*γ+ = α*γ+. Dually, δ+β* = δ+β*. Thus
    (α,β)(γ,δ)+=(α,β)¯(γ,δ)¯+.

We have shown that (C, ⋅, +, *, ) is a generalized Ehresmann semigroup. □

Remark 3.10

In the case that I = Λ = E is a semilattice, the above semigroup C(I,Λ,E) is exactly the Ehresmann semigroup CE constructed in [6].

Corollary 3.11
On the semigroup (C, ., +, *, ), we have
IC={(θe,τe)|eI},ΛC={(ηf,ξf)|fΛ},EC={(θe,τe)|eE}.
Proof
In view of (5), it follows that
(α,β)+=(α+,β+)=(θ1β,τ1β)
for all (α, β) ∈ C. This gives IC ⊆ {(θe, τe) | eI}. Conversely, for eI, we have (θe, τe) ∈ C by Lemma 3.7. So
(θe,τe)=(θ1τe,τ1τe)=(θe,τe)+IC.

Thus IC = {(θe, τe)|eI}. Dually, ΛC = {(ηf, ξf)|f ∈ Λ}.

For (α, β) ∈ C, we have 1βI and (1β)E. By (5), we have
(α,β)¯+=(α¯,β¯)+=(α¯+,β¯+)=(θ1β¯,τ1β¯)=(θ(1β),τ(1β)){(θe,τe)|eE}.
This shows that EC ⊆ {(θe, τe)|eE}. Conversely, for eE, by (5) again we have
(θe,τe)=(θe,τe)=(θ(1τe),τ(1τe))=(θ1τe¯,τ1τe¯)=(θe¯+,τe¯+)=(θe¯,τe¯)+=(θe,τe)¯+EC.

Thus EC = {(θe, τe)|eE}. □

We say that two admissible triples (I, Λ, E) and (J, Π, F) are isomorphic if there exist an isomorphism φ from I onto J and an isomorphism ψ from Λ onto Π such that
φ|E=ψ|E,Eφ=F.

If this is the case, then one can easily show that C(I,Λ,E) is (2,1,1,1)-isomorphic to C(J, Π, F). Moreover, we have the following.

Corollary 3.12

Let (I, Λ, E) be an admissible triple. Then (I, Λ, E) is isomorphic to the admissible triple (IC, ΛC, EC) of C = C(I,Λ,E).

Proof
By Corollary 3.11, we can define the following surjective mappings
φ:IIC,e(θe,τe),ψ:ΛΛC,f(ηf,ξf).
On the other hand, if e, gI and (θe, τe) = (θg, τg), then τe = τg and so e = 1τe = 1τg = g. Thus φ is also injective. Finally, let e, gI and xI1. Then we have
x(θeθg)=g(ex)=gex=(eg)x=(eg)x=θeg
by Lemma 3.4. This shows that θeθg = θeg. Dually, τgτe = τeg. Thus
(eg)φ=(θeg,τeg)=(θeθg,τgτe)=(θe,τe)(θg,τg)=(eφ)(gφ).

This implies that φ is a morphism. Dually, ψ is also a morphism. Furthermore, by Lemma 3.7 and Corollary 3.11, we can see that φ |E = ψ|E and Eφ = EC. □

Corollary 3.13

Let (I, Λ, E) be an admissible triple and D a (2,1,1,1)-subalgebra of C = C(I,Λ,E) containing EC. Then D is fundamental. In particular, C itself is fundamental.

Proof
Let D be a (2,1,1,1)–subalgebra of C and (α, β), (γ, δ) ∈ D such that (α, β) is μD-related to (γ, δ). Then (α, β)+ = (γ, δ)+ and (α, β)* = (γ, δ)* whence α+ = γ+, β+ = δ+ and α* = γ*, β* = δ*. By α* = γ*, we get 1α = 1α* = 1γ* = 1γ. Now let xI. Then xE. By Corollary 3.11, (θx, τ_x) ∈ ECD. Since μD is a semigroup congruence on D, it follows that
(θxα,βτx)=(θx,τx)(α,β)μD(θx,τx)(γ,δ)=(θxγ,δτx)
whence (θxα)* = (θxγ)*. This implies that
xα=(x)α=(1θx)α=1(θxα)=1(θxα)=1(θxγ)=1(θxγ)=(1θx)γ=(x)γ=xγ.

On the other hand, we have = xα and xy = xγ by the fact (α, β), (γ, δ) ∈ C. This implies that = xy. Thus α = γ. Dually, β = δ. Therefore μD is the identity relation on D. That is, D is fundamental. □

4 A representation of generalized Ehresmann semigroups

In this section, we always assume that (S, ., +, *, ) is a generalized Erhesmann semigroup. Then we have the admissible triple (IS, ΛS, ES) of S and the semigroup C(IS,ΛS,ES) by Remark 3.3 and Theorem 3.9. The aim of this section is to show that there exists a (2,1,1,1)-homomorphism Φ : SC(IS,ΛS,ES) whose kernel is μS. To accommodate with the notations of Section 3, we use the notations from Section 3 for the admissible triple (IS, ΛS, ES) throughout this section.

We first consider some properties of the admissible triple (IS, ΛS, ES) of S and the semigroup C(IS,ΛS,ES). Denote C(IS,ΛS,ES) by C for convenience. In view of Remark 3.3, in the admissible triple (IS, ΛS, ES), for all iIS and λ ∈ ΛS, we have
i=i and λ=λ+.
For aS, there are functions
αa:IS1ΛS,βa:ΛS1IS
given by
xαa=(xa),xβa=(ax)+.
Lemma 4.1

With above notation, we have the following results:

  1. For all aS, αa ∈ 𝓞(IS1 → ΛS) and βa ∈ 𝓞(ΛS1IS).
  2. For all xIS1 and yΛS1, a = xαa and yβa = yβa.
  3. For all aS and xIS1 and yΛS1, σaβaρxαa and ρaαaσyβa.
  4. For all a, bS and xIS1 and yΛS1, (x αa)αb = ab and (b)βa = ab.

Proof

  1. Let x, yIS1 with xy. Then xy = yx = x. It follows that a = (xa)* = (xya)* ≤ (ya)* = a by Lemma 2.7 (b), whence αa ∈ 𝓞(IS1 → ΛS). Dually, βa ∈ 𝓞(ΛS1IS).
  2. Observe that x = x* for all xIS, it follows that
    xαa=(xa)=(xa)=xαa=xαa
    by the identity (4)′ in Table 1. Moreover, since 1 = 1, we have 1αa = 1αa. Thus a = xαa for all xIS1. Dually, a = yβa for all yΛS1.
  3. Let uΛS1. Then
    uσxαa=u(xαa)=u(xa)ΛS,u(βaρxαa)=(x(au)+a)ΛS.
    By Lemma 2.7 (a),
    u(xa)a=u(xa),(x(au)+a)a=(x(au)+a).
    Since ΛS is a right normal band, we have
    u(xa)(x(au)+a)=u(xa)(x(au)+a)a=(x(au)+a)u(xa)a=(x(au)+a)u(xa).
    On the other hand, since u(xa)* ∈ ΛS, we obtain that (u(xa)*)* = u(xa)* by Lemma 2.6 (a). Using the identity (3)′, (4)′, (1), (4)′ and the fact that ΛS is a right normal band successively, we get
    (x(au)+a)u(xa)=(x(au)+a)(u(xa))=((x(au)+a)(u(xa)))=(x(au)+au(xa))=(xau(xa))=((xa)u(xa))=(u(xa))=u(xa).
    Thus
    uσxαa=u(xa)(x(au)+a)=u(βaρxαa)
    for all uΛS1. That is, σaβaρxαa. Dually, ρaαaσyβa.
  4. For xIS1, we have ∈ ΛS and (x α) = (x αa)+ = ((xa)*)+. Using Lemma 2.7 (a), the identities (6), (6)′, Lemma 2.6 (a), (b), the identity (4)′ and Lemma 2.7 (a) in that order, we have
    (xαa)αb=(((xa))+)αb=(((xa))+b)=(((xa))+¯b¯)b=((xa)¯+b¯)b=((xa¯)+b¯)b=(xa¯b¯)b=(xa¯b¯)b=(xab)=xαab.
    Dually, (b)βa = ab. □

Now we can state our main result in this section.

Theorem 4.2

Define Φ : SC, a ↦ (αa, βa). Then Φ is a(2,1,1,1)-homomorphism whose kernel is μS. Moreover,

  1. Φ|IS is a(2,1,1,1)-isomorphism from IS to IC.
  2. Φ|ΛS is a(2,1,1,1)-isomorphism from ΛS to ΛC.
  3. Φ|ES is a(2,1,1,1)-isomorphism from EStoEC.

Proof
By (a), (b) and (c) of Lemma 4.1, (αa, βa) ∈ C for all aS. Observe that (αa, βa)(αb, βb) = (αaαb, βbβa) in C and
x(αaαb)=(xαa)αb=xαab,y(βbαa)=(yβb)βa=yβab
for all xIS1 and yΛS1 by Lemma 4.1 (d). It follows that
(aΦ)(bΦ)=(αa,βa)(αb,βb)=(αab,βab)=(ab)Φ.

Thus Φ preserves the binary operation.

Let xIS1. Then
xαa+=(xa+)=(xa+)=x(a+)=(a+)x=(a+)x=(1βa)x=xαa+
by Lemma 3.4. This shows that αa+=αa+. Dually, βa+=βa+. So
a+Φ=(αa+,βa+)=(αa+,βa+)=(αa,βa)+.

This shows that Φ preserves “+″. Dually, Φ preserves “*″.

Let xIS1. If x = 1, then
xαa¯=1αa¯=a¯=(a)+=(1αa)+=(1αa)=1αa¯=xαa¯
by the identity (5)′ in Table 1. If xIS, using the identity (4)′, Lemma 2.6 (a), Lemma 2.7 (c), the identity (5)′ in that order, we have
xαa¯=(xa¯)=(xa¯)=(x¯a¯)=xa¯=((xa))+=((xa))=(xαa)=xαa¯.
This shows that αa = αa. Dually, βa = βa. So
a¯Φ=(αa¯,βa¯)=(αa¯,βa¯)=(αa,βa)¯=aΦ¯.
Thus Φ preserves “−″. We have proved that Φ is a (2,1,1,1)-homomorphism and so
kerΦ={(a,b)S×S|αa=αb,βb=βa}
is a (2,1,1,1)-congruence on S. If (a, b) ∈ ker Φ, then we have a* = 1αa = 1αb = b*. Dually, a+ = b+. This shows that
kerΦ{(a,b)S×S|a+=b+,a=b}.
Let σ be a semigroup congruence on S and aσb such that
σ{(a,b)S×S|a+=b+,a=b}.
Then for all xIS1, we have xaσxb, whence a = (xa)* = (xb)* = b. This shows that αa = αb. Dually, βa = βb. Thus (a, b) ∈ ker Φ. We have shown that ker Φ is the largest semigroup congruence contained in
{(a,b)S×S|a+=b+,a=b}.

That is, μS = ker Φ.

  1. Since IS = {a+|aS} and
    a+Φ=(αa,βa)+IC={(α,β)+|(α,β)C}
    by (9), it follows that IS Φ ⊆ IC. Now, let (α, β)+ = (α+, β+) ∈ IC where (α, β) ∈ C. Then 1βIS. For xIS1, we have
    xα1β=(x(1β))=(x(1β))=x(1β)=(1β)x=(1β)x=xα+
    by Lemma 3.4, which implies that α1β = α+. Dually, β1β = β+. Thus
    (1β)Φ=(α1β,β1β)=(α+,β+)=(α,β)+.
    This gives IS Φ = IC and so Φ|IS is surjective. If i, jIS and i Φ = j Φ, then (i, j) ∈ ker Φ, this implies i = i+ = j+ = j by Lemma 2.6 (a) and the fact that
    kerΦ{(a,b)S×S|a+=b+,a=b}.
    it follows that Φ|IS is also injective. In view of Lemma 2.6, IS and IC are (2,1,1,1)-subalgebras of S and C, respectively. By the first part of the theorem, Φ|IS is a (2,1,1,1)-isomorphism from IS to IC.
  2. This is the dual of (a).
  3. This follows from items (a) and (b). □

A (2,1,1,1)-subalgebra D of a generalized Ehresmann semigroup (S, ., +, *, ) is called quasi-full if IS ∪ ΛSD. Combining Corollaries 3.12 and 3.13 and Theorem 4.2, we obtain the main result of this paper.

Theorem 4.3

Let (I, Λ, E) be a given admissible triple. Then (S, ., +, *, ) is a fundamental generalized Ehresmann semigroup whose admissible triple is isomorphic to (I, Λ, E) if and only if it is (2,1,1,1)-isomorphic to a quasi-full (2,1,1,1)-subalgebra of C(I,Λ,E).

Considering the case that I = Λ = E is a semilattice, by Remark 3.10 we have the following corollary which is Theorem 3.2 in [6] substantially.

Corollary 4.4

Let E be a given semilattice. Then (S, ., *, +) is a fundamental Ehresmann semigroup whose distinguished semilattice is isomorphic to E if and only if it is (2,1,1)-isomorphic to a (2,1,1)-subalgebra of CE containing the distinguished semilattice of CE.

Acknowledgement

The author expresses his profound gratitude to the referee for the valuable comments and suggestions. Thanks also go to the editor for the timely communications. This paper is supported by the National Natural Science Foundations of China (11661082,11301470).

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If the inline PDF is not rendering correctly, you can download the PDF file here.

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