aa r X i v : . [ m a t h . C T ] F e b ON EHRESMANN SEMIGROUPS
MARK V. LAWSON
This paper is dedicated to the memory of Peter M. Neumann
Abstract.
We describe an alternative approach to describing Ehresmann semi-groups by categories in which a class of ´etale actions plays an important rˆole.We also characterize the Ehresmann semigroups that arise as the set of all sub-sets of a finite category. As an application, we prove that every birestrictionsemigroup can be suitably embedded into a birestriction semigroup constructedfrom a category. As a corollary, we determine when a birestriction semigroupcan be suitably embedded into an inverse semigroup. Introduction
Ehresmann semigroups were introduced in [13] as generalizations of inverse semi-groups. We recall their definition here. An
Ehresmann semigroup is a semigroup S with a distinguished subset U ⊆ E ( S ) of the set of all idempotents, called the setof projections , equipped with two functions λ, ρ : S → U satisfying the followingfour axioms:(ES1): U is a commutative subsemigroup.(ES2): The maps λ and ρ are the identity on U .(ES3): aλ ( a ) = a and ρ ( a ) a = a for all a ∈ S .(ES4): λ ( λ ( a ) b ) = λ ( ab ) and ρ ( aρ ( b )) = ρ ( ab ) for all a, b ∈ S .Observe that λ ( a ) is the smallest projection e such that ae = a ; a dual resultholds for ρ ( a ). Ehresmann semigroups will be denoted by ( S, U ) to make the set ofprojections clear or by S alone if the set of projections is already clear. A morphism of Ehresmann semigroups ( S, U ) and (
T, V ) is a semigroup homomorphism θ : S → T such that θ ( U ) ⊆ V and λ ( θ ( a )) = θ ( λ ( a )) and ρ ( θ ( a )) = θ ( ρ ( a )) for all a ∈ S . We say that such a morphism is an isomorphism if θ is an isomorphism ofsemigroups and induces an isomorphism from U to V . Remark 1.1.
Ehresmann semigroups are equivalently described as algebras oftype (2 , ,
1) [1] which in addition to having a single associative binary operationalso have two unary operations denoted by a a ∗ and a a + satisfying thefollowing axioms:(1) x + x = x and xx ∗ = x . I would like to thank Prof. V. A. R. Gould of the University of York for her comments on anearlier draft of this paper. (2) ( x + y + ) + = x + y + = y + x + and ( x ∗ x ∗ ) ∗ = x ∗ y ∗ = y ∗ x ∗ .(3) ( xy ) + = ( xy + ) + and ( xy ) ∗ = ( x ∗ y ) ∗ .(4) ( x ∗ ) + = x ∗ and ( x + ) ∗ = x + .The fact that Ehesmann semigroups form a variety means that free objects exist.We refer the reader to [11] and [7] for more information; we also recommend thereferences to be found in both these paper. We say more about the origins ofEhresmann semigroups in Remark 3.1.The following are standard results about Ehresmann semigroups but they arealso easy to prove directly. Lemma 1.2.
Let S be an Ehresmann semigroup with set of projections U . (1) λ ( ab ) ≤ λ ( b ) . (2) ρ ( ab ) ≤ ρ ( a ) . (3) If U has a zero then a = 0 if and only if λ ( a ) = 0 . (4) If U has a zero then a = 0 if and only if λ ( a ) = 0 . Ehresmann semigroups come equipped with three partial orders which are alge-braically defined: • x ≤ r y if and only if x = ey for some e ∈ U . Observe ρ ( x ) ≤ ρ ( y ). • x ≤ l y if and only if x = yf for some f ∈ U . Observe that λ ( x ) ≤ λ ( y ). • x ≤ y if and only if x = eyf for some e, f ∈ U .Observe that ≤ = ≤ r ◦ ≤ l = ≤ l ◦ ≤ r . It is easy to check that x ≤ r y if and onlyif x = ρ ( x ) y ; x ≤ l y if and only if x = yλ ( x ); x ≤ y if and only if x = ρ ( x ) yλ ( x ).Although these orders generalize the natural partial order on an inverse semigroup(where they are all the same) they do not share such nice properties. This is anissue we shall have to confront in Section 3.Ehresmann semigroups have emerged as an interesting class [1, 2, 4, 15]. Inparticular, they are closely allied to categories in two ways. First of all, underly-ing every Ehresmann semigroup is a category. The following was proved as [13,Theorem 3.17]. Proposition 1.3.
Let S be an Ehresmann semigroup with set of projections U . Onthe set S define the restricted product a · b = ab when λ ( a ) = ρ ( b ) and undefinedotherwise. Then ( S, · ) is a category in which λ ( a ) = d ( a ) and ρ ( b ) = r ( b ) . Inaddition, for any x, y ∈ S we have that xy = ( xe ) · ( ey ) where e = λ ( x ) ρ ( y ) . Second of all, examples of Ehresmann semigroups directly arise from categories;the following is a special case of [12]
Example 1.4.
Let C be a small category with set of identities C o . We denotethe domain and codomain maps on C by d and r , respectively. The product xy isdefined in the category if and only if d ( x ) = r ( y ); in this case, we shall also write ∃ xy . Let S = P ( C ) be the set of all subsets of C equipped with the multiplicationof subsets. This is a semigroup. Put U = P ( C o ), the set of all subsets of the set N EHRESMANN SEMIGROUPS 3 of identities. For A ⊆ C define λ ( A ) = { d ( a ) : a ∈ C } and ρ ( A ) = { r ( a ) : a ∈ C } .We now show that with these definitions P ( C ) is an Ehresmann monoid. Let E, F ⊆ C o . Then EF = E ∩ F . It follows that U is a commutative subsemigroupof the set of all idempotents of S so that (ES1) holds. It is immediate from thedefinitions that (ES2) holds. It is evident that (ES3) holds. It remains to showthat (ES4) holds. By symmetry, it is enough to show that λ ( λ ( A ) B ) = λ ( AB )holds; observe that in a category if ab is defined then d ( a ) = r ( b ). Thus ab isdefined if and only if d ( a ) b is defined and d ( ab ) = d ( d ( a ) b ). A special case ofthis construction shows that the monoid of all binary relations on a set X , whichwe denote by B ( X ), is also an Ehresmann monoid; the category in this specialcase is the set X × X where d ( x, y ) = ( y, y ) and r ( x, y ) = ( x, x ) and product( x, y )( y, z ) = ( x, z ).In [13], we showed what additional structure a category needed to be equippedwith in order that it arise from an Ehresmann semigroup. In Section 2 of thispaper, we shall prove Theorem 2.7, which describes a different (though, obviously,equivalent) way, in which this can be accomplished. It arose from reading [14, Page184]. In Section 3 of this paper, we shall characterize the Ehresmann monoids ofthe form P ( C ), where C is a finite category, in Theorem 3.13. This result usesideas first described in [12].2. Ehresmann biactions
The goal of this section is to describe a different categorical approach to charac-terizing Ehresmann semigroups from the one described in [13]; a related approachis also discussed in [5]. Our goal is to prove a kind of converse of Proposition 1.3.To do this, we shall need a class of actions called ´etale actions [16] though weprefer the term ‘supported actions’. Let S be an inverse semigroup and X a set.Let p : X → E ( S ) be a function. A supported action is a left action S × X → X ,denoted by ( s, x ) s · x , such that p ( x ) · x = x and p ( s · x ) = sp ( x ) s − . We shallonly be interested in supported actions where the acting inverse semigroup is ameet semilattice where the above properties simplify somewhat.We define an Ehresmann biaction starting from a category C by the followingsix axioms:(E1): The set of identities C o of C is equipped with the structure of a com-mutative, idempotent semigroup.(E2): There are two supported actions: there is a left action C o × C → C denoted by ( e, a ) e · a such that r ( a ) · a = a and r ( e · a ) = e r ( a ); there isa right action C × C o → C denoted by ( a, e ) a · e such that a · d ( a ) = a and d ( a · e ) = d ( a ) e .(E3): The biaction property ( e · a ) · f = e · ( a · f ) holds.(E4): We require that e · a = ea and a · e = ae if a ∈ C o .(E5): d ( e · a ) ≤ d ( a ) and r ( a · e ) ≤ r ( a ). MARK V. LAWSON (E6): When ∃ xy then e · ( xy ) = ( e · x )( d ( e · x ) · y ) and ( xy ) · e = ( x · r ( y · e ))( y · e ) . Remark 2.1.
The essential difference between the approach to Ehresmann semi-groups described here and the one described in [13] is that whereas here the effectof an idempotent on an element is always defined — we have an action — in [13],we only consider certain products of an idempotent on an element.We can refine (E5) as follows.
Lemma 2.2.
Assume that C is an Ehresmann biaction. If e ≤ f , where e, f ∈ C o ,then d ( e · a ) ≤ d ( a · f ) , and dually.Proof. By (E5), we have that d (( a · f ) · e ) ≤ d ( a · f ). But ( a · f ) · e = a · ( ef ) = a · e . (cid:3) Given an Ehresmann biaction, we make the following definitions: a ≤ r b if andonly if a = e · b for some e ∈ C o ; a ≤ l b if and only if a = b · e for some e ∈ C o ; x ≤ y if and only if a = e · y · f for some e, f ∈ C o (we can omit brackets becauseof axiom (E3).) Lemma 2.3.
In an Ehresmann biaction, we have that a ≤ r b if and only if a = r ( a ) · b , and dually.Proof. Let a = e · b for some e ∈ C o . Then e · a = a . Thus by axiom (E2), we havethat r ( a ) ≤ e . It follows that r ( a ) · b = ( r ( a ) e ) · b = r ( a ) · ( e · b ) = r ( a ) · a = a. The proof of the converse is immediate. (cid:3)
We shall prove first that an Ehresmann biaction is equivalent to what we calledan Ehresmann category in [13]. This is not strictly needed for our main theorembut is an obvious first step.
Lemma 2.4.
We work in an Ehresmann biaction. (1) ≤ r is a partial order. (2) If x ≤ r y then r ( x ) ≤ r ( y ) and d ( x ) ≤ d ( y )(3) Suppose that x ≤ r y and x ′ ≤ r y ′ where ∃ xx ′ and ∃ yy ′ . Then xx ′ ≤ r yy ′ . (4) Suppose that x ≤ r y and r ( x ) = r ( y ) . Then x = y . (5) Let e ≤ r ( x ) . Then e · x is the unique element satisfying the following twoproperties: e · x ≤ r x and r ( e · x ) = e .Similar results hold for the partial order ≤ l .Proof. (1) This follows by [16, Proposition 3.2].(2) Suppose that x ≤ r y . Then x = e · y . We have that r ( x ) = r ( e · y ) = e r ( y ) ≤ r ( y ). By axiom (E5), we have that d ( x ) = d ( e · y ) ≤ d ( y ).(3) We have that x = r ( x ) · y and x ′ = r ( x ′ ) · y ′ . Now xx ′ = ( r ( x ) · y )( r ( x ′ ) · y ′ ) = ( r ( x ) · y )( d ( x ) · y ′ = r ( x ) · ( yy ′ ) N EHRESMANN SEMIGROUPS 5 using axiom (E6). Thus xx ′ ≤ yy ′ .(4) Suppose that x ≤ r y and r ( x ) = r ( y ). Then x = ( r )( x ) · y . It follows that x = r ( y ) · y = y .(5) Suppose that y ≤ r x is such that r ( y ) = e . Then, as we have seen, y = r ( y ) · x = e · x , as required. (cid:3) Lemma 2.5.
We work in an Ehresmann biaction. Suppose that x ≤ r y . Then x · ( d ( x ) f ) ≤ r y · ( d ( y ) f ) , and dually.Proof. By assumption x = e · y . Thus x · f = ( e · y ) · f . Now use axiom (E3) toget x · f = e · ( y · f ). Thus x · f ≤ r y · f . We now use the fact that x = x · d ( x )and y = y · d ( y ) and that we are dealing with an action. (cid:3) It follows from Lemma 2.4 and Lemma 2.5 that every category which is anEhresmann biaction is also an Ehresmann category in the sense of [13].We now prove that every Ehresmann semigroup gives rise to an Ehresmannbiaction.
Proposition 2.6.
Let ( S, U ) be an Ehresmann semigroup. Put C = S equippedwith the restricted product. Then C o = U , a commutative idempotent semigroup,and C is an Ehresmann biaction.Proof. The left action C o × C → C and the right action C × C o → C are bothdefined by multiplication. All the axioms are easy to check from the axioms andproperties of an Ehresmann semigroup. (cid:3) Proposition 2.6 tells us that Ehresmann semigroups give rise to Ehresmannbiactions. We now prove the first theorem of this paper which shows us that wecan construct Ehresmann semigroups from Ehresmann biactions.
Theorem 2.7.
Let C be an Ehresmann biaction. Given x, y ∈ C , put e = d ( x ) r ( y ) and define x • y = ( x · e )( e · y ) , called the pseudoproduct . Then ( C, • ) is anEhresmann semigroup with set of projections C o .Proof. The fact that the pseudoproduct is a binary operation follows from axioms(E1) and (E2). To prove that the pseudoproduct is associative, check that ( x • y ) • z and x • ( y • z ) are both equal to( x · r ( y · r ( z )))( d ( x ) · y · r ( z ))( d ( d ( x ) · y ) · z )and so are equal to each other. Define λ ( a ) = d ( a ) and ρ ( a ) = r ( a ). Then a • λ ( a ) = a • d ( a ) = a · d ( a ) = a by (E2). We calculate λ ( λ ( a ) • b ) = d ( d ( a ) • b ) = d ( e · b ) where e = d ( a ) r ( b ). Thus λ ( λ ( a ) • b ) = d ( d ( a ) · b ). On the other hand λ ( a • b ) = d ( a • b ) = d (( a · e )( e · b )) where e = d ( a ) r ( b ). Thus λ ( a • b ) = d ( e · b ) = d ( d ( a ) · b ). The dual results are proved by symmetry. Thus ( C, • ) is an Ehresmannsemigroup with set of projections C o , as claimed. (cid:3) MARK V. LAWSON A class of finite Ehresmann monoids
In this section, we shall characterize the finite Ehresmann monoids which areisomorphic to those arising from finite categories as in Example 1.4. We havefound [10, Chapter 1] a very useful reference in helping us to clarify our thoughts.We have also used ideas from [12] in an essential way.
Remark 3.1.
Ehresmann semigroups arose within the York School of semigrouptheory led by J. B. Fountain [9]. The paper [6] was particularly influential; inparticular, the conditions below which we refer to below as ‘deterministic’ and‘codeterministic’ were first introduced in this paper. The theme of this Schoolbecame that of determining which properties of (von Neumann) regular semigroupscould be generalized to a non-regular setting. There was a particular emphasis onthe non-regular generalizations of inverse semigroups with the focus of attentionbeing on the abstract relationship between an element a and its idempotents a − a and aa − . Subsequently, these ideas were then developed not only for semigroupsbut also for categories [3]. My own paper [13] was written within the York Schoolframework with the goal being to describe the most general class of non-regularsemigroups that could be regarded as natural generalizations of inverse semigroups.This was achieved by developing ideas due to Charles Ehresmann and his students.The key stumbling block became that of order: in an inverse semigroup, the naturalpartial order encodes algebraically the order induced by subset-inclusion. For moregeneral classes of Ehresmann semigroups, this neat relationship between algebraand order does not hold. Trying to deal with this problem led to the notions of‘bideterministic element’ and ‘partial isometry’ which play a major rˆole in thissection.We shall need to define two classes of elements within an Ehresmann semigroup.The first requires only the algebraic structure of an Ehresmann semigroup. Weuse the terminology from [3]. An element a ∈ S of an Ehresmann semigroup issaid to be deterministic if ea = aλ ( ea ) for all e ∈ U ; it is said to be codeterministic if ae = ρ ( ae ) a for all e ∈ U . An element is said to be bideterministic if it isboth deterministic and codeterministic. An Ehresmann semigroup is said to be a birestriction semigroup if every element is bideterministic. We shall say more aboutbirestriction semigroups in Section 4. We now characterize the bideterministicelements of Ehresmann monoids of the form P ( C ) where C is a category. Lemma 3.2.
Let C be a finite category with P ( C ) being its associated Ehresmannmonoid. (1) A non-empty subset A ⊆ C is deterministic if and only if whenever a, b ∈ A and d ( a ) = d ( b ) then r ( a ) = r ( b ) . (2) A non-empty subset A ⊆ C is codeterministic if and only if whenever a, b ∈ A and r ( a ) = r ( b ) then d ( a ) = d ( b ) . N EHRESMANN SEMIGROUPS 7
Proof.
We prove (1) since (2) follows by symmetry. Let A be a non-empty subsetof C . Suppose that a, b ∈ A are such that d ( a ) = d ( b ) = e but r ( a ) = i isdifferent from r ( b ) = j . We prove that A is non-deterministic. The singleton set { i } is an element of U . The product { i } A is an element of P ( S ) which contains a but which does not contain b . On the other hand, λ ( { i } A ) contains e . Itfollows that Aλ ( { i } A ) contains both a and b . Thus A cannot be deterministicsince { i } A = Aλ ( { i } A ). Suppose now that A is such that whenever a, b ∈ A and d ( a ) = d ( b ) then r ( a ) = r ( b ). We prove that A is deterministic. Let E be anyprojection. Let x ∈ Aλ ( EA ). Then x = ae where a ∈ A and e ∈ λ ( EA ). Thus e = d ( e ′ a ′ ) where e ′ ∈ E and a ′ ∈ A . It follows that x, a ′ ∈ A and d ( x ) = d ( a ′ ).By assumption, r ( x ) = r ( a ′ ). But then x = e ′ x where e ′ ∈ E . (cid:3) Ehresmann monoids of the form P ( C ) come equipped with subset inclusion asan order. We shall formalize some of its properties below. Let S be an Ehresmannsemigroup with set of projections U . We shall say that S is ordered if S is equippedwith a partial order ⊆ such that the following properties hold:(OE1): If a ∪ b exists then c ( a ∪ b ) = ca ∪ cb , and dually.(OE2): If e ∈ U and a ⊆ e then a ∈ U .(OE3): If e, f ∈ U then e ≤ f if and only if e ⊆ f .(OE4): If a = eb then a ⊆ b and if a = be then a ⊆ b where e ∈ U .(OE5): If a ∪ b exists then λ ( a ∪ b ) = λ ( a ) ∪ λ ( b ), and dually.Observe that a ∪ b denotes the lub of a and b with respect to the partial order ⊆ .The motivation for the above definition comes, of course, from Example 1.4 asthe following lemma shows. The proof is by routine verification. Lemma 3.3.
Let C be a (finite) category. Then P ( C ) is a partially ordered Ehres-mann monoid. The proof of the following lemma is immediate from axiom (OE1) and (OE5).
Lemma 3.4.
Let ( S, U ) be an ordered Ehresmann semigroup. Then the followingtwo properties hold: (1) If a ⊆ b and c ⊆ d then ac ⊆ bd . (2) If a ⊆ b then λ ( a ) ⊆ λ ( b ) and ρ ( a ) ⊆ ρ ( b ) . The following notion first arose in [14] but was then extended in [12]. In orderedEhresmann semigroups, we shall need a stronger notion than that of a bidetermin-istic element. Let S be an ordered Ehresmann semigroup. An element a ∈ S issaid to be a partial isometry if whenever b ⊆ a then b ≤ a . The following wasproved as [12, Lemma 2.26]. Lemma 3.5.
In an ordered Ehresmann semigroup, the set of partial isometries isan order-ideal.Proof.
Let a be a partial isometry and let b ⊆ a . Since a is a partial isometry,we have that b ≤ a . Thus b = ρ ( b ) aλ ( a ). We prove that b is a partial isometry. MARK V. LAWSON
Let c ⊆ b . Then c ⊆ a . It follows that c ≤ a . Thus c = ρ ( c ) aλ ( c ). Now, ρ ( c ) bλ ( c ) = ρ ( c ) ρ ( b ) aλ ( b ) λ ( c ). By axiom (OE5) applied to c ⊆ b we deduce that ρ ( c ) ≤ ρ ( b ) and λ ( c ) ≤ λ ( b ). Thus ρ ( c ) bλ ( c ) = ρ ( c ) aλ ( c ) = c . It follows that c ≤ b , as required. (cid:3) The proof of the following is straightforward.
Lemma 3.6.
Every partial isometry is bideterministic.
We next describe the partial isometries in the ordered Ehresmann monoids P ( C ). Lemma 3.7.
Let C be a category. Then a non-empty subset A ⊆ C is a partialisometry in P ( C ) if and only if the following two conditions hold: (1) If a, b ∈ A and d ( a ) = d ( b ) then a = b . (2) If a, b ∈ A and r ( a ) = r ( b ) then a = b .Proof. Let A be a subset that satisfies both conditions and suppose that B ⊆ A .Then, clearly, B ⊆ Aλ ( B ). We prove that B ≤ A . We shall show that B = Aλ ( B ).Let x ∈ Aλ ( B ). Then x = a d ( b ) where a ∈ A and b ∈ B . Thus d ( x ) = d ( b ).Now, x, b ∈ A . By the assumption that A satisfies the condition (1), we musthave that x = b . We have therefore proved that B = Aλ ( B ). The fact that B = ρ ( B ) A follows by symmetry. We now prove the converse. Suppose that A isa partial isometry. We prove that condition (1) holds; the fact that condition (2)holds follows by symmetry. Suppose that a, b ∈ A be distinct elements such that d ( a ) = d ( b ) = e . Clearly, { a } , { b } ⊆ A , but we do not have that { a } ≤ A . Thereason is that λ ( { a } ) = { e } and Aλ ( { a } ) = A { e } contains both a and b . Thusthis set cannot be equal to { a } . (cid:3) Remark 3.8.
Subsets of groupoids satisfying both the conditions of Lemma 3.7are called local bisections . The set of all local bisections of a groupoid forms aninverse semigroup. See [14, page 164].
Lemma 3.9.
Let S be an ordered Ehresmann semigroup. Then any atom is apartial isometry.Proof. Let a be an atom. If b ⊆ a then either b = a or b = 0. In both cases, b ≤ a .Thus a is a partial isometry. (cid:3) The following example shows that the concepts we have introduced are distinct.
Example 3.10.
Consider the following category C : e f a h h b v v The monoid P ( C ) has 16 elements. The element { a } is a partial isometry; theelements { a, b } is bideterministic but not a partial isometry (since { a } ⊆ { a, b } but { a } (cid:2) { a, b } ); the element { a, e } is not bideterministic. N EHRESMANN SEMIGROUPS 9
Lemma 3.11.
Let C be a category. (1) The product of partial isometries in P ( C ) is also a partial isometry. (2) If C is finite, then every element of P ( C ) is a finite join of partial isome-tries.Proof. (1) We use our characterization of partial isometries from Lemma 3.7. Let A and B be non-empty partial isometries. We prove that AB is a partial isometry.Let x, y ∈ AB and suppose that d ( x ) = d ( y ). Then x = ab and y = a ′ b ′ where ∃ ab and ∃ a ′ b ′ in the category C and a, a ′ ∈ A and b, b ′ ∈ B . We have that d ( b ) = d ( b ′ ).But B is a partial isometry and so b = b ′ . We may similarly show that a = a ′ andso x = y . The other case follows by symmetry.(2) Singleton sets are atoms and these are partial isometries by Lemma 3.9.Every subset is a finite union of singleton sets. (cid:3) Proposition 3.12.
Let C be a finite category. Then P ( C ) is a finite orderedEhresmann monoid with the order being subset-inclusion. In addition, the productof partial isometries is a partial isometry and every element is a join of partialisometries. Both P ( C ) and P ( C o ) are Boolean algebras with respect to the partialorder.Proof. This follows by Lemma 3.3 and Lemma 3.11. (cid:3)
We now prove the second theorem of this paper which characterizes the monoidsarising in Proposition 3.12.
Theorem 3.13.
Let ( S, U ) be a finite ordered Ehresmann monoid such that both S and U are Boolean algebras with respect to the partial order on the Ehresmannmonoid. Suppose, in addition, that within S the product of partial isometries is apartial isometry and that every element is a join of partial isometries. Then thereis a finite category C such that S is isomorphic to P ( C ) as an Ehresmann monoidand in such a way that the order on S is isomorphically mapped to the order on P ( C ) .Proof. Let S be an ordered Ehresmann monoid having the stated properties. Ourproof is in six steps. In the first three steps, we construct the category C . To dothis, we shall use the atoms of S ; these exist (and in sufficient quantities) since S is a finite Boolean algebra.(1) If a is an atom then both ρ ( a ) and λ ( a ) are atoms. Suppose that a is anatom. Let e ≤ ρ ( a ). Then ea ≤ r a and so ea ⊆ a by axiom (OE4). But a is anatom and so either ea = a or ea = 0. In the first case, ρ ( a ) ≤ e . We thereforededuce that e = ρ ( a ). In the second case, eρ ( a ) = 0 by the properties of ρ andLemma 1.2. Thus e = 0. We have therefore proved that ρ ( a ) is also an atom. Bysymmetry, if a is an atom then so too is λ ( a ).(2) If a and b are atoms and ab is a restricted product then ab is an atom. Both a and b are atoms and so by Lemma 3.9 each of a and b is a partial isometry. By assumption, their product ab is a partial isometry. Let c ⊆ ab . Then c ≤ ab since ab is a partial isometry. It follows that c = ρ ( c ) abλ ( c ). Now, bλ ( c ) ≤ l b and so bλ ( c ) ⊆ b . But b is an atom. It follows that bλ ( c ) = 0 or bλ ( c ) = b . Supposefirst that bλ ( c ) = 0. Then λ ( b ) λ ( c ) = 0 by axiom (ES4). But from c ⊆ ab we getthat λ ( c ) ⊆ λ ( ab ) ≤ λ ( b ) where we have used axiom (OE5) and Lemma 1.2. Thus λ ( c ) ≤ λ ( b ) by axiom (OE3). It follows that λ ( c ) = 0 and so c = 0 by Lemma 1.2.Now suppose that bλ ( c ) = b . We have therefore shown that c = ρ ( c ) ab . We nowrepeat the above argument in its dual version. As a result, we deduce that either c = 0 or c = ab and so we have proved that ab is an atom.(3) Let a be an atom. Define d ( a ) = λ ( a ) and r ( a ) = ρ ( a ). By (Step 1) abovethese are both atoms. Put C equal to the set of all atoms of S . By (Step 2), theset C is a category under the restricted product. We have therefore proved that C is a category whose set of identities is the set of atoms in U .We may accordingly construct the Ehresmann monoid P ( C ) whose set of pro-jections is P ( C o ). We prove that S is isomorphic to P ( C ) as semigroups and thatthis induces an isomorphism between U and P ( C o ) as semigroups.(4) Since we are in a Boolean algebra, every non-zero element is a join of theatoms below it. If a is an element of S define φ (0) = ∅ and if a = 0 define φ ( a )to be the join of all atoms below it. The map φ determines a bijection from S to P ( C ). Observe that by axiom (OE2), the elements below a projection are allprojections. Thus each projection is a finite join of atoms which are themselvesprojections. It follows that φ also determines a bijection from U to P ( C o ). Thesebijections are actually isomorphisms of Boolean algebras.(5) By (Step 2) above, we have that φ ( a ) φ ( b ) ⊆ φ ( ab ) . We prove the reverseinclusion. Let x be an atom such that x ⊆ ab . By assumption, each of a and b can be written as unions of partial isometries. Let a = S mi =1 a i and b = S nj =1 b j where a i and b j are partial isometries. Then ab = S ≤ i ≤ m, ≤ j ≤ n a i b j . Since theproduct of partial isometries is a partial isometry, we know that a i b j is a partialisometry. We now use the distributivity property of Boolean algebras to deducethat x = S i,j ( x ∧ a i b j ). Now x is an atom and so x ∧ a i b j = 0 or x ⊆ a i b j . Supposethat x ⊆ a i b j for some i and some j . Then x ≤ a i b j since a i b j is a partial isometry.By definition x = ( ρ ( x ) a i )( b j λ ( x )). Put e = λ ( ρ ( x ) a i ) and f = ρ ( b j λ ( x )). Then x = uv where u = ρ ( x ) a i e and v = eb j λ ( x ). This is a restricted product byProposition 1.3. Observe that u ⊆ a i and v ⊆ b j by axiom (OE4). Thus both u and v are partial isometries by Lemma 3.5. We shall prove that u and v areatoms. In fact, we prove explicitly that v is an atom; the fact that u is an atomthen follows by symmetry. Since x = uv we have that λ ( x ) = λ ( v ). Suppose that w ⊆ v . Since v is a partial isometry we have that w ≤ v . Thus w = vλ ( w ) Now λ ( w ) ≤ λ ( v ) = λ ( x ). But x is an atom implies that λ ( x ) is an atom. Thus either λ ( w ) = 0 in which case w = 0 or λ ( w ) = λ ( x ) in which case w = v . We havetherefore shown that v is an atom. N EHRESMANN SEMIGROUPS 11
We have proved that φ is an isomorphism of semigroups between S and P ( C )and between U and P ( C o ).(6) φ is an isomorphism of Ehresmann semigroups. Let a ∈ S be a non-zeroelement. Then a is the join of all the atoms below it. Thus by axiom (OE5), wehave that λ ( a ) is equal to the join of all the atoms of the form λ ( b ) where b ⊆ a .We now prove that every atom e below λ ( a ) is of the form λ ( b ) where b is an atomand b ⊆ a . Let e ≤ λ ( a ) where e is an atom. Then ae ≤ l a and so ae ⊆ a byaxiom (OE4). Observe that λ ( ae ) = e . If ae is an atom then we are done. If ae isnot an atom then x ≤ ae where x is an atom since we are working in a Booleanalgebra. But λ ( x ) ⊆ λ ( ae ) = e . But e is an atom. Thus either λ ( x ) = 0 whichimplies that x = 0, which is ruled out since x is an atom, or λ ( x ) = e . We havetherefore found an atom x ≤ a such that λ ( x ) = e . We have therefore proved that θ ( λ ( a )) = λ ( θ ( a )). A dual result holds for ρ . (cid:3) Birestriction semigroups
In this section, we shall, in effect, combine results we found in the previous twosections. The following results are all well-known. We include them for the sakeof completeness.
Lemma 4.1.
Let ( S, U ) be a birestriction semigroup. (1) The partial orders ≤ l , ≤ r and ≤ are all the same. (2) The semigroup S is partially ordered with respect to ≤ . (3) If a ≤ bc then there exist b ′ ≤ b and c ′ ≤ c such that b ′ c ′ is a restrictedproduct and a = b ′ c ′ . (4) The set U is an order-ideal of S . (5) If a, b ≤ c and λ ( a ) = λ ( b ) (respectively, ρ ( a ) = ρ ( b )) then a = b . (6) If a ≤ b then λ ( a ) ≤ λ ( b ) and ρ ( a ) ≤ ρ ( b ) .Proof. (1) Suppose that a ≤ b . We prove that a ≤ l b . The fact that a ≤ r b followsby symmetry. We are given that a = ebf where e, f ∈ U . But, by assumption, ea = aλ ( ea ) and so a = bλ ( ea ) f . This gives the result since U is a subsemigroup.(2) This follows by (1) above because ≤ l is left compatible with the multiplicationand ≤ r is right compatible. (3) Suppose that a ≤ bc . Then a = ebcf for some e, f ∈ U . Thus a = ( ebρ ( cf ))( λ ( eb ) cf ). Put b ′ = ebρ ( cf ) and c ′ = λ ( eb ) cf . Then a = b ′ c ′ , the product b ′ c ′ is a restricted product and b ′ ≤ b and c ′ ≤ c . (4) Thisis immediate. (5) The proof of this is straightforward. (6) This follows by theproperties of the partial orders. (cid:3) Let C be a category. Denote the set of partial isometries in P ( C ) by PI ( C ).This is a birestriction monoid by Lemma 3.3, Lemma 3.11, and Lemma 3.6. Theorem 4.2.
Let ( S, C ) be a birestriction semigroup. Then there is a category C and an injective morphism α : S → PI ( C ) . Proof.
For the category C , we take the set S equipped with the restricted productaccording to Proposition 1.3. Then PI ( C ) is a birestriction monoid. Define α : S → PI ( C ) by putting α ( a ) equal to the set of all elements less than or equal to a . Thisis well-defined by part (5) of Lemma 4.1 and injective. By part (4) of Lemma 4.1elements of U are mapped to elements of P ( C o ). It is a homomorphism by part(3) of Lemma 4.1. It remains to check that it is a morphism. By symmetry, it isenough to check the morphism property for λ . If b ≤ a then λ ( b ) ≤ λ ( a ) by part (6)of Lemma 4.1. On the other hand, if e ≤ λ ( a ) then ae ≤ a and λ ( ae ) = λ ( a ). (cid:3) The following is related to questions discussed in [8], although our approach isquite different. A natural question is the following. Given a birestriction semigroup S under what circumstances can S be embedded into an inverse semigroup T insuch a way that λ ( a ) = a − a and ρ ( a ) = aa − — let’s call this a ‘nice embedding’.This question is completely answered by the following theorem. Theorem 4.3.
Let S be a birestriction semigroup. Denote the set S equipped withthe restricted product by C . Then S admits a nice embedding if and only if thecategory C can be embedded into a groupoid G .Proof. Suppose first that the category C can be embedded into a groupoid G . ByTheorem 4.2, there is an injective morphism α : S → PI ( C ). Let C ⊆ G . Then PI ( C ) → PI ( G ) is an injective morphism which is actually an embedding. Butthe elements of PI ( G ) are simply the local bisections of G and so form an inversesemigroup; see Remark 3.8. It follows that S admits a nice embedding into theinverse semigroup PI ( G ). To prove the converse, suppose that S can be nicelyembedded into the inverse semigroup T . Let G be the set T equipped with therestricted product. Then C embeds into G as a subcategory. (cid:3) References [1] M. J. J. Branco, G. M. S. Gomes, V. Gould, Ehresmann monoids,
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