constructing an ample monoid from a weak loganathan pair

Semigroup Forum (2012) 84:527–561DOI 10.1007/s00233-012-9391-5

R E S E A R C H A RT I C L E

Constructing an ample monoid from a weakLoganathan pair

Laila Tunsi

Received: 1 December 2010 / Accepted: 14 March 2012 / Published online: 18 April 2012© Springer Science+Business Media, LLC 2012

Abstract This paper is based on a categorical approach to the study of inversemonoids. The main idea is to extend this to the class called ample monoids (typeA monoids). We generalise the notion of a Loganathan category pair to obtain whatwe call a weak Loganathan category pair and take two categories associated with anample monoid and examine their properties. We prove that each of these categoriestogether with its subcategory of idempotents forms a weak Loganathan category pair.Then we construct an ample monoid from them.

Keywords Ample monoids · Type A monoids · Loganathan category pair · WeakLoganathan category pair

The purpose of this paper is to investigate to what extent the theory of Leech [7] canbe extended from inverse monoids to ample monoids. Ample monoids were intro-duced in [3] (under the name type A monoids) as a generalisation of inverse monoids,and their structure has been studied in a number of papers including, for example, [3]and [6].

In [7], Leech showed that inverse monoids can be described in terms of certainsmall categories which he called division categories. Given an inverse monoid S withidentity 1, there is a category D(S) whose objects are the idempotents of S suchthat the pair (D(S),1) is a division category. On the other hand, given a divisioncategory (D, I), one can construct an inverse monoid S[D,I ]. Moreover, given aninverse monoid S, the monoids S and S[D(S),1] are isomorphic, and given a division

Communicated by Norman R. Reilly.

This work is taken from my PhD thesis, Ample Monoids and The Theory of Small Categories [8].

L. Tunsi (�)Department of Mathematics, University of Tripoli, Tripoli, Libyae-mail: [email protected]

mailto:[email protected]

528 L. Tunsi

category (D, I), the division categories (D, I) and (D(T ),1) are equivalent whereT = S[D,I ].

For us, a particularly significant aspect of Leech’s theory is the connection withwhat Leech calls “Loganathan pairs”. A Loganathan pair (D,E) consists of a smallcategory D and a subcategory E with the same set of objects as D satisfying a numberof conditions. In particular, E has an initial object. If S is an inverse monoid, thenfor idempotents e, f with f ≤ e, there is a distinguished morphism from e to f

in the category D(S). The objects of D(S) together with these morphisms form asubcategory E with initial object 1, and the pair (D(S),E) forms a Loganathan pair.Leech also proves that if (D,E) is a Loganathan pair, and I is the initial object of E,then the pair (D, I) is a division category. Thus inverse monoids could equally wellbe described in terms of Loganathan pairs. This observation is the key to our work onample monoids and categories.

Given an ample monoid S, we construct two categories DA(S) and DB(S), andexamine their properties. Although they have many properties in common, these twocategories need not be equivalent. We generalise the notion of a Loganathan cate-gory pair to obtain what we call a weak Loganathan category pair. Each of DA(S)

and DB(S) has a subcategory E such that (DA(S),E) and (DB(S),E) are weak Lo-ganathan category pairs. On the other hand, starting with a weak Loganathan pair(D,E), we show how to construct an ample monoid S[D,E]. Moreover, if S isan ample monoid, then S is isomorphic to the monoid S[DA(S),E]. Thus amplemonoids can indeed be described by means of weak Loganathan category pairs.

1 Introductory concepts

In this section we define some of the basic concepts about monoids, ample monoidsand categories which will be used throughout the paper.

Definition 1.1 A (small) category C consists of a set of objects denoted by ObjCtogether with, for each pair u and v of objC, a disjoint collection of sets Mor(u, v).The elements of the sets Mor(u, v) are called morphisms and the set of all morphismsof C denoted by MorC. For each object u of C, there is a distinguished element 1u ofMor(u,u), called the identity morphism at u. Finally, there is a partial operation onMorC called composition and written as juxtaposition which satisfies the followingconditions:

(1) if f and g are morphisms of C, the composite gf of f and g is defined if and onlyif there exist objects u,v and w of C such that f ∈ Mor(u, v) and g ∈ Mor(v,w);when this is the case we have gf ∈ Mor(u,w);

(2) for any objects u,v and w of C and for any morphisms f ∈ Mor(v,u) and g ∈Mor(u,w), we have 1uf = f and g1u = g;

(3) for all objects u,v,w and x of C and for all morphisms f ∈ Mor(u, v), g ∈Mor(v,w) and h ∈ Mor(w,x) we have (hg)f = h(gf ).

Note that a monoid is an example of a category with one object.

Constructing an ample monoid from a weak Loganathan pair 529

We say that C1 is a subcategory of C if it is a category with

objC1 ⊆ objC, while MorC1(u, v) ⊆ MorC(u, v)

for all u and v in objC1.

Definition 1.2 A morphism f : u −→ v in the category C is an epimorphism if forevery object w and every pair g : v −→ w and h : v −→ w of morphisms in C withgf = hf we must have g = h. The morphism f is a monomorphism if for everyobject w and for every pair g : w −→ u and h : w −→ u of morphisms in C withfg = f h we must have g = h. The morphism f is an isomorphism if there is amorphism g : v −→ u such that fg = 1v and gf = 1u.

Proposition 1.1 In a small category C, an isomorphism f is an epimorphism and amonomorphism.

Definition 1.3 An object t of a category C is called weak terminal if and only if forevery object u of C there is a morphism u −→ t .

An object i of a category C is called weak initial if and only if for every object u

of C there is a morphism i −→ u.

Definition 1.4 Let g1 : w −→ u2 and g2 : w −→ u1 be morphisms in a category witha common domain. We say the commutative diagram

w u2

u1 v

�g1

�g2

�f2

�f1

is a pushout for g1 and g2, if for every pair of morphisms g′1 : u1 −→ w′ and g′

2 :u2 −→ w′ such that g′

2g1 = g′1g2 there exists a unique morphism ψ : v −→ w′ such

that ψf2 = g′2 and ψf1 = g′

1.

Definition 1.5 A division category is a small category D with the following proper-ties:

(1) D has a weak initial object I;(2) every morphism in D is an epimorphism;(3) every pair of morphisms with common domain in D can be completed to a

pushout.

Using Definition 1.4, we can prove:

Proposition 1.2 Suppose that the two small squares in the diagram

u v w

u′ v′ w′

�α

�ρ

�β

�σ

�τ

�γ

�δ

are pushout squares. Then the outer square is also a pushout square.

530 L. Tunsi

Proposition 1.3 If the following square

. .

. .

�p

�q

�s

�t

is a pushout square and u is an isomorphism with cod(p) = dom(u), then the outerdiagram in

. . .

. .

�p

�q

�u

�s

��

�� su−1

�t

is a pushout square.

Using this result we can prove

Corollary 1.1 Suppose that we have a morphism y = ue where u is an isomorphism,and that

. .

. .

�x

�y

�s

�t

is a pushout square. Then

. .

. .

�x

�e

�s

�tu

is also a pushout square.

Lemma 1.1 Let p and q be morphisms in a category C with common domain. If thesquares

. .

. .

�p

�q

�r

�s

and

. .

. .

�p

�q

�r ′

�s′

are both pushouts, then there is an isomorphism t in C such that r ′ = tr and s′ = ts.

Definition 1.6 The relations L∗ and R∗ are defined on a monoid S by the rule thataL∗b (aR∗b) if and only if the elements a and b of S are related by Green’s relationL(R) in some overmonoid of S.

Note that the following lemma is an immediate consequence of the definition of L∗and R∗.


Lemma 1.2 For any monoid S, we have L ⊆ L∗ and R ⊆ R∗.

The proof of the following lemma is in [4].

Lemma 1.3 Let a and b be elements in a monoid S. The following are equivalent:

(1) (a, b) ∈ L∗.(2) for all x, y ∈ S, ax = ay if and only if bx = by.

There is a corresponding dual result for the relation R∗.

Definition 1.7 A monoid S is called left (right) abundant if each R∗(L∗)-class con-tains an idempotent and abundant if it is both left and right abundant [2].

A left (right) abundant monoid is left (right) adequate if its idempotents commutewith each other.

Lemma 1.4 Let S be an adequate monoid and let a be an element of S. Then eachL∗(R∗)-class contains a unique idempotent, which is denoted by a∗(a+).

Note that the natural order on the idempotents of any monoid S is defined by: fore, f in E(S) we have

e ≤ f if and only if e = ef = f e

We have the following result from [5].

Proposition 1.4 Let a and b be elements of an adequate monoid S. Then a ≤r b ifand only if a+ ≤ b+ and a = a+b. The dual result hold for ≤l .

Proposition 1.5 Let S be an adequate monoid, with semilattice of idempotents E.Then

(1) for all a, b ∈ S,

a R∗ b if and only if a+ = b+, and a L∗b if and only if a∗ = b∗;(2) for all a, b ∈ S, (ab)∗ = (a∗b)∗ and (ab)+ = (ab+)+;(3) for all a, b ∈ S, (ab)∗ ≤ b∗ and (ab)+ ≤ a+;(4) if e ∈ E, then

(ae)∗ = a∗e and (ea)+ = ea+.

Definition 1.8 A right adequate monoid S is called right ample or right type A if

ea = a(ea)∗

for all elements a and idempotents e of S. A left adequate monoid S is called leftample or left type A if

ae = (ae)+a

532 L. Tunsi

for all elements a in S and all idempotents e in S. An adequate monoid is ample ortype A if it is both left and right ample.

Parts of the following proposition are well known, but we give a proof here for thesake of completeness.

Proposition 1.6 Let S be an ample monoid, then we have

(1) the natural partial order ≤ is compatible with the multiplication on S;(2) if e is an idempotent, and a is an element with a ≤ e, then a is an idempotent;(3) if a, b,∈ S, a ≤ b and a L∗b (aR∗b), then a = b;(4) (ef a)∗ = (ea)∗(f a)∗ and (aef )+ = (ae)+(af )+ for all e, f ∈ E and a ∈ S.

Proof

(1) Let a, b ∈ S, such that a ≤ b. Then a = a+b. For c ∈ S we have ca = ca+b, andsince S is an ample monoid, we have ca = (ca+)+cb = (ca)+cb. Then ca ≤ cb,and ≤ is left compatible. It is clear that ≤ is right compatible.

(2) Let e be an idempotent and a in S, with a ≤ e. Then a = a+e is idempotent.(3) Let a, b ∈ S such that a ≤ b and a L∗b. Then there is an idempotent a∗ ∈ L∗

a

such that a = ba∗ = bb∗ = b.(4) Since S is ample, we have ef a = ea(f a)∗ and so (ef a)∗ = (ea(f a)∗)∗ =

(ea)∗(f a)∗. Also since S is ample, we have aef = (ae)+af and so (aef )+ =((ae)+af )+ = (ae)+(af )+. �

2 Two categories associated with an ample monoid

In [7], Leech showed that inverse monoids can he described in terms of certain smallcategories which he called division categories.

We generalise the constructions of Leech’s categories DA(S) and DB(S) frominverse to ample monoid.

Given an ample monoid S, we construct two categories DA(S) and DB(S), andexamine their properties. Although they have many properties in common, these twocategories need not be equivalent.

Define a category DA(S) as follows:DA(S) is a small category whose set of objects is the set of idempotents of S, and

whose morphism sets are defined to be

Mor(e, f ) = {(e, x, f ) : x∗ ≤ e and f = x+}

.

If (e, x, f ) : e −→ f and (f, y, g) : f −→ g, we define the composition as

(f, y, g)(e, x, f ) = (e, yx, g).

The composition is well defined, because we have

x∗ ≤ e, f = x+, y∗ ≤ f and g = y+


since (e, x, f ) and (f, y, g) are morphisms, so that

(yx)+ = (yx+)+ = (yf )+ = y+ = g

by Proposition 1.5(2). Now we have yxx∗ = yx, so that yx ≤ x∗ and (yx)∗ ≤ x∗ ≤ e.We prove now that DA(S) is a category. It is clear that the composition is associa-

tive. Let (e, x, f ) and (f, y, e) be two morphisms in DA(S), so that we have x∗ ≤ e,f = x+, y∗ ≤ f and e = y+. Then (e, e, e) is the identity at the object e. Certainly,(e, e, e) is a morphism, and we have

(e, x, f )(e, e, e) = (e, xe, f ) = (e, x, f )

because x∗ ≤ e and

(e, e, e)(f, y, e) = (f, ey, e) = (f,y+y, e

) = (f, y, e).

Theorem 2.1 Let S be an ample monoid. Then the following hold in DA(S):

(1) the morphism (e, x, f ) : e −→ f is an isomorphism if and only if

x∗ = e and x regular;(2) every morphism in DA(S) is an epimorphism;(3) if (e, x, f ) : e −→ f is such that e = x∗ and f = x+, then (e, x, f ) is a

monomorphism;(4) if e, f, i ∈ E(S) with f ≤ e and x ∈ S are such that x∗ ≤ e and x+ = i, so that

(e, x, i) ∈ Mor(e, i), then the pair

e f

i

�(e,f,f )

�(e,x,i)

is a pushout pair.

Proof

(1) Suppose that the morphisms (e, x, f ) and (f, y, e) are mutually inverse, so thatx∗ ≤ e and (f, y, e)(e, x, f ) = (e, e, e) that is, yx = e. Now we have x = xe =xyx, so that x is regular, and also e = yx = yxx∗ = ex∗ = x∗, since x∗ ≤ e.

Conversely, suppose that (e, x, f ) : e −→ f is a morphism such that e =x∗, x+ = f and x is regular. The idempotents of S commute, and so, sincex is regular, it has a unique inverse, say x−1. Since x ∈ R∗

x+ , it is clear thatxx−1 = x+ = f , and similarly, since x ∈ L∗

x∗ , we have x−1x = x∗ = e. It fol-lows that (f, x−1, e) is a morphism in DA(S), and also that

(e, x, f )(f,x−1, e

) = (f,xx−1, f

) = (f,f,f ),

534 L. Tunsi

and

(f,x−1, e

)(e, x, f ) = (

e, x−1x, e) = (e, e, e).

Hence the morphisms (e, x, f ) and (f, x−1, e) are mutually inverse and the mor-phism (e, x, f ) is an isomorphism.

(2) Let e, f,∈ E(S), x ∈ S and (e, x, f ) : e −→ f be a morphism in DA(S).So we have x∗ ≤ e and x+ = f . Suppose that there are two morphisms(f, y, g) : f −→ g and (f, z, g) : f −→ g where g ∈ E(S), and y, z ∈ S, suchthat (f, y, g)(e, x, f ) = (f, z, g)(e, x, f ). Then (e, yx, g) = (e, zx, g), that is,yx = zx. We want to show that y = z.

Since x+R∗x and yx = zx, we have yx+ = zx+ so that yf = zf andyy∗f = zz∗f . But y∗ ≤ f and z∗ ≤ f , that is, y∗f = y∗ and z∗f = z∗, andso we have yy∗ = zz∗ which implies that y = z. Thus the morphism (e, x, f ) isan epimorphism.

(3) Suppose that (e, x, f ) : e −→ f is a morphism in DA(S) such that e = x∗ andf = x+. Suppose that there are two morphisms (g, y, e) : g −→ e and (g, z, e) :g −→ e where y, z,∈ S and g ∈ E(S) such that

(e, x, f )(g, y, e) = (e, x, f )(g, z, e),

that is, (g, xy,f ) = (g, xz, f ). Then xy = xz. We want to show that z = y. SincexLx∗ and xy = xz, we have ey = ez as e = x∗. But since (g, y, e) and (g, z, e)

are morphisms, we have e = y+ = z+, and so y+y = z+z. Hence y = z, and themorphism (e, x, f ) is a monomorphism.

(4) Let e, f, i ∈ E(S) with f ≤ e and let x ∈ S such that x∗ ≤ e and x+ = i.Then (e, x, i) ∈ Mor(e, i). We want to show the pair (e, f,f ) and (e, x, i) isa pushout pair. Now (e, f,f ) ∈ Mor(e, f ) because f ∗ = f ≤ e and f + = f .Also since (xf )∗ ≤ f ∗ = f , we have (f, xf, (xf )+) ∈ Mor(f, (xf )+), and since((xf )+)∗ = (xf )+ ≤ x+ = i, we have (i, (xf )+, (xf )+) ∈ Mor(i, (xf )+). Nextwe show that the square

e f

i (xf )+

�(e,f,f )

�

(e,x,i)

�

(f,xf,(xf )+

)

�(i,(xf )+,(xf )+

)

commutes. We have(f,xf, (xf )+

)(e, f,f ) = (

e, xf, (xf )+)

and

(i, (xf )+, (xf )+

)(e, x, i) = (

e, xf, (xf )+)


since S is an ample monoid. Hence(f,xf, (xf )+

)(e, f,f ) = (

i, (xf )+, (xf )+)(e, x, i)

and so the square commutes.Now consider the diagram

e f

i (xf )+

h

�(e,f,f )

�

(e,x,i)

�

(f,xf,(xf )+

)��

(f,r,h)�

(i,(xf )+,(xf )+

)

��

(i,s,h)

�

ψ

such that (f, r, h)(e, f,f ) = (i, s, h)(e, x, i) so that (e, rf,h) = (e, sx,h), thatis, rf = sx and r = sx since r∗ ≤ f as (f, r, h) is morphism. Now wewant to show that there is a unique morphism ψ in Mor((xf )+, h) such thatψ(i, (xf )+, (xf )+) = (i, s, h) and ψ(f,xf, (xf )+) = (f, r, h). We put ψ =((xf )+, s, h) and claim that this is the required morphism.

First, we show that ψ is a morphism. We have s+ = h, since (i, s, h) is amorphism. Also r∗ ≤ f , since (f, r, h) is a morphism, so that we have xr∗ ≤ xf

and so (xr∗)+ ≤ (xf )+. Now (xr∗)+ = (x(sx)∗)+ as r = sx, so that (xr∗)+ =(x(sx)∗)+ = (x(s∗x)∗)+ = (s∗x)+ = s∗x+ as S is an ample monoid. Then(xr∗)+ = s∗i as x+ = i, because (e, x, i) is a morphism. Hence (xr∗)+ = s∗since s∗ ≤ i as (i, s, h) is a morphism. Then s∗ = (xr∗)+ ≤ (xf )+, so thats∗ ≤ (xf )+. Also s+ = h since (i, s, h) is a morphism, and so ((xf )+, s, h) isa morphism. Now we have

((xf )+, s, h

)(i, (xf )+, (xf )+

) = (i, s(xf )+, h

) = (i, s, h),

and((xf )+, s, h

)(f,xf, (xf )+

) = (f, sxf,h) = (f, rf,h) = (f, r, h).

Now suppose that there is a morphism ((xf )+, s′, h) in Mor((xf )+, h), whichmakes the diagram commute. Then

((xf )+, s′, h

)(f,xf, (xf )+

) = (f, s′xf,h

) = (f, r, h)

= ((xf )+, s, h

)(f,xf, (xf )+

).

Since, in DA(S), every morphism is an epimorphism, we have ((xf )+, s′, h) =((xf )+, s, h). Hence the morphism ((xf )+, s, h) is unique.

536 L. Tunsi

Let S be an ample monoid and define DB(S) as follows: DB(S) is a small categorywhose set of objects is the set E(S) of idempotents of S, and whose morphisms aredefined by

Mor(e, f ) = {(e, x, f ) : x+ ≤ e and x∗ = f

}.

The composition of two morphisms is defined by

(f, y, g)(e, x, f ) = (e, xy, g).

It is easy to check that the composition is well defined and DB(S) is a category. �

Theorem 2.2 Let S be an ample monoid. Then the following hold in DB(S):

(1) the morphism (e, x, f ) : e −→ f is an isomorphism if and only if

x+ = e and x is regular;(2) every morphism in DB(S) is an epimorphism;(3) if (e, x, f ) : e −→ f is such that e = x+, then (e, x, f ) is a monomorphism;(4) if e, f, i ∈ E(S) with f ≤ e and x ∈ S are such that x+ ≤ e and x∗ = i, so that

(e, x, i) ∈ Mor(e, i), then the pair

e f

i

�(e,f,f )

�(e,x,i)

is a pushout pair.

Proof The proof is similar to that for DA(S). �

Although the categories DA(S) and DB(S) enjoy similar properties, for a generalample monoid, the categories are not isomorphic. In fact, they need not be equivalent,as shown by Example 5.3 in [8] using a six element monoid.

3 Loganathan category

In this section we define the term “Loganathan category pair” and give an account oftheir properties, expanding that of Leech [7]. Loganathan category pairs are closelyrelated to division categories and inverse monoids.

Definition 3.1 Let D be a category and E be a subcategory of D with objE = objD.We say that the pair (D,E) is a Loganathan category pair if:

(1) objE is a semilattice with greatest element 1;(2) for a, b ∈ objE, the set of morphisms MorE(a, b) is non-empty if and only if

b ≤ a, and if MorE(a, b) is non-empty, then it contains just one element;


(3) every morphism x in D can be written uniquely as x = ue where e is a morphismin E and u is an isomorphism.

We call ue the canonical factorisation of x. We write |x| for the element e in thecanonical factorisation of x.

Lemma 3.1 Let (D,E) be a Loganathan category pair. Let x, y ∈ MorD. Then

(1) if x has canonical factorisation x = ue, and if x = ve for any v ∈ MorD, thenv = u;

(2) if the composition ye is defined for some e ∈ MorE, then |ye| = |y|e.

Proof

(1) By assumption, ue = x = ve. Let v have canonical factorisation v = wf wherew is an isomorphism and f ∈ MorE. Then f e ∈ MorE because E is a category,and so ue = w(f e) where u,w are isomorphisms, and e, f e ∈ MorE. By theuniqueness of canonical factorisations, u = w and e = f e. Let a be the codomainof e. Then f ∈ MorE(a, a), and since |MorE(a, a)| = 1, we have f = 1a . Thusv = wf = uf = u1a = u as claimed.

(2) Let y = v|y| and ye = u|ye| where v and u are isomorphisms. Then u|ye| =ye = v|y|e, and |y|e ∈ MorE because E is category. Hence by the uniqueness ofcanonical factorisations, |ye| = |y|e. �

Definition 3.2 Let D be a category with objD a partially ordered set. For a subcate-gory E of D with objE = objD, we say that the pair (D,E) is a semilattice categorypair if

(1) objD is a semilattice with greatest element 1;(2) for a, b,∈ objD, we have |MorE(a, b)| ≤ 1 and MorE(a, b) is non-empty if and

only fi b ≤ a.

We observe that the object 1 is a weak initial object in D and an initial object in E.

Lemma 3.2 If (D,E) is a Loganathan category pair, then (D,E) is a semilatticecategory pair.

Proof By Definition 3.1(1), objD is a semilattice with greatest element 1. By Defini-tion 3.1(2), we have that if MorE(a, b) is non-empty, then it contains just one element,so that |MorE(a, b)| ≤ 1. Moreover, |MorE(a, b)| = 1 if and only if b ≤ a. �

Definition 3.3 Let (D,E) be a semilattice category pair. We say that a morphism w

of D with domain a is a minimal cancellable morphism if it is cancellable (that is,is a monomorphism and an epimorphism), and if, for any i ∈ MorE and cancellablev ∈ MorD,

vi = w implies i = 1a and w = v.

Lemma 3.3 Let (D,E) be a Loganathan category pair and let u ∈ MorD. Then u

is an isomorphism if and only if u is minimal cancellable.

538 L. Tunsi

Proof First, suppose that u is an isomorphism in D. Then u is certainly a monomor-phism and an epimorphism by Proposition 1.1, that is, u is cancellable. Let a be thedomain of u and suppose that u = vi where i ∈ MorE and v is cancellable. Since(D,E) is a Loganathan category pair, v has a canonical factorisation v = wf forsome isomorphism w and f ∈ MorE. Then u1a = u = vi = wf i, and f i ∈ MorEsince E is a category. As 1a, f i ∈ MorE and u,w are isomorphisms, the unique-ness of canonical factorisations gives u = w and 1a = f i. Now the domain of i

is a (since i, u have the same domain), so that if i has codomain b, then b ≤ a.But then f ∈ Mor(b, a) and since f ∈ MorE we have a ≤ b. Thus a = b so thati, f ∈ MorE(a, a) and hence i = f = 1a . We now have u = vi = v1a = v so that u

is a minimal cancellable morphism.Conversely, suppose that u is a minimal cancellable morphism in D. Since (D,E)

is a Loganathan category pair, u has a canonical factorisation u = ve for some iso-morphism v and morphism e in E. Now v is certainly cancellable, and so by thedefinition of minimal cancellable, u = v and e = 1a where a is the domain of u. Butv is an isomorphism, so u is an isomorphism. �

4 Weak Loganathan category pair

In this section we introduce the notion of a weak Loganathan category pair. This isa generalisation of a Loganathan category pair and as we show, is closely related toample monoids.

Definition 4.1 Let (D,E) be a semilattice category pair. We say that (D,E) is aweak Loganathan category pair if

(1) every morphism x in D has a unique factorisation (called the canonical factorisa-tion of x) as x = ue where u is a minimal cancellable morphism, and e ∈ MorE;we write x for u and |x| for e;

(2) the composition of two minimal cancellable morphisms is minimal cancellable;(3) for e ∈ MorE and v ∈ MorD with dom(e) = dom(v), and v minimal cancellable,

there is a morphism p in MorE such that pv is defined and has canonical fac-torisation ue for some minimal cancellable morphism u. The morphism u can bechosen to be an isomorphism if v is an isomorphism;

(4) for a minimal cancellable morphism v of D and morphisms p,q in E such thatpv and qv are defined,

cod(|qv|) ≤ cod

(|pv|) implies cod(qv) ≤ cod(pv).

Lemma 4.1 Let (D,E) be a weak Loganathan category pair. Let x ∈ MorD havecanonical factorisation x = ue, where e ∈ MorE(a, b). If x = vf where v is can-cellable and f ∈ MorE(a, c), then b ≤ c.


Proof Consider the diagram

a b

c d

�e

�f

�u

�v

Now v has canonical factorisation v = wh say, where h ∈ MorE and w is a minimalcancellable morphism. Now hf ∈ MorE, since E is a category, so that ue = vf =w(hf ). By uniqueness, u = w and e = hf . Since e ∈ MorE(a, b), we have hf ∈MorE(a, b). Since f ∈ MorE(a, c); h ∈ MorE(c, b), so that MorE(c, b) �= ∅. Hence,since (D,E) is semilattice category pair, we have b ≤ c. �

Lemma 4.2 Let (D,E) be a weak Loganathan category pair. Let y ∈ MorD ande ∈ MorE such that the composite ye is defined. Then |ye| = |y|e.

Proof Let y = v|y| and ye = u|ye| be the canonical factorisation of y and ye respec-tively, where v and u are minimal cancellable morphisms. Now ye = v|y|e and sov|y|e = ye = u|ye| and |y|e ∈ MorE, since E is a category. Hence by the unique-ness of canonical factorisations, |y|e = |ye|. �

Lemma 4.3 Let (D,E) be a weak Loganathan category pair and let u ∈ MorD.Then u is an isomorphism if and only if u is minimal cancellable.

Proof Let u be an isomorphism in a weak Loganathan category pair (D,E). Bydefinition, u = vi for some minimal cancellable v and i ∈ MorE. We show that u =v, so that u is minimal cancellable. Let u have domain a, so that u1a = vi and hence1a = u−1vi since u is an isomorphism. By Lemma 4.2, |u−1vi| = |u−1v|i and so1a = |1a| = |u−1v|i. Let cod(i) = b. Then b ≤ a since i ∈ E. But dom(|u−1v|) = b

and cod(|u−1v|) = a so that a ≤ b since |u−1v| ∈ E. Hence a = b and we havei ∈ Mor(a, a) so that i = 1a . Thus u = vi = v1a = v as required.

Conversely, suppose that u is a minimal cancellable morphism in D. Since (D,E)

is a Loganathan category pair, u has a canonical factorisation u = ve for some iso-morphism v and e in E. Now v is cancellable, then by definition u = v and e = 1a

where a is the domain of u. But v is an isomorphism, so u is an isomorphism. �

Lemma 4.4 Let (D,E) be a weak Loganathan category pair, and let t be a morphismin D and p,q be morphisms in E such that pt and qt are defined. Then

cod(|qt |) ≤ cod

(|pt |) implies cod(qt) ≤ cod(pt).

Proof Let t have canonical factorisation t = ve. By Lemma 4.2, |qt | = |qve| =|qv|e. Hence cod(|qt |) = cod(|qv|), and similarly, cod(|pt |) = cod(|pv|). Thus, ifcod(|qt |) ≤ cod(|pt |) then cod(|qv|) ≤ cod(|pv|), and so, by condition (4) of thedefinition of weak Loganathan category pair, we have cod(qv) ≤ cod(pv).

However, cod(qv) = cod(qve) = cod(qt), and cod(pv) = cod(pt), so cod(qt) ≤cod(pt) as required. �

540 L. Tunsi

Proposition 4.1 Let (D,E) be Loganathan category pair. Then (D,E) is a weakLoganathan category pair.

Proof By Lemma 3.2, we have that (D,E) is a semilattice category pair, and byLemma 4.3, we know that the minimal cancellable morphisms in D are exactly theisomorphisms of D. Since (D,E) is a Loganathan category pair, every morphism x

in D has a unique factorisation as x = ue where u is isomorphism and e ∈ MorE.We have just noted that u must be a minimal cancellable morphism; moreover, ifx = vf with v a minimal cancellable morphism, then v is an isomorphism, so u = v

and e = f . Thus x has a unique factorisation x = ue with u minimal cancellable ande ∈ MorE.

Now, let u,v be minimal cancellable morphisms in D for which the composite uv

is defined. Then u,v are isomorphisms, and so uv is an isomorphism, and thereforeuv is minimal cancellable. Let v ∈ MorD be minimal cancellable and let e ∈ MorEbe such that dom(e) = dom(v). By Lemma 4.3, v is an isomorphism and so ev−1 is amorphism in D. Since (D,E) is a Loganathan pair, ev−1 has a canonical factorisationev−1 = wp for some isomorphism w and morphism p in E. Put u = w−1 so that u isan isomorphism. Then ue = ue(v−1v) = u(ev−1)v = u(wp)v = (uw)pv = pv. Thusthere is a morphism p in E such that pv is defined and pv has canonical factorisationue for some minimal cancellable morphism u.

Finally, let v be a minimal cancellable morphism in D, that is, an isomorphism andp,q be morphisms in E such that pv and qv are defined and cod(|qv|) ≤ cod(|pv|).Then there is a morphism e in E from cod(|pv|) to cod(|qv|)

. .

. .

m

�v

��

��

��

��

|qv| �|pv|

�p��

q�w

��

��e

�u

Since e|pv| and |qv| are both in E and are coterminal, we have e|pv| = |qv|.Now pv and qv have canonical factorisations pv = w|pv| and qv = u|qv|, for

some isomorphisms u and w.Hence qv = u|qv| = ue|pv| = uew−1w|pv| = uew−1pv, and cancelling v we

get q = uew−1p, so that the above diagram is commutative. Now q is a morphismin E so that q = |q| = |uew−1p|. Hence, by Lemma 3.1(2), q = |uew−1|p. Thus|uew−1| ∈ MorE, dom(|uew−1|) = cod(p) and cod(|uew−1|) = cod(q). Conse-quently, cod(q) ≤ cod(p) and we have cod(qv) = cod(q) ≤ cod(p) = cod(pv).

We have now shown that (D,E) satisfies all the properties in the definition ofweak Loganathan category pair. �

The following lemma is easy to prove.

Lemma 4.5 Let (D,E) be a weak Loganathan category pair. Let x ∈ MorD. If x

has canonical factorisation x = ue and if x = ve for any v ∈ MorD, then u = v.


5 (DA(S),E) and (DB(S),E) are weak Loganathan category pairs

Let S be an ample monoid, and consider the category DA(S). Recall that the set ofobjects of DA(S) is the set E(S) of idempotents of S. It is easy to see that E(S)

together with the set of morphisms

MorE = {(e, f,f ) : e, f ∈ E(S) and f ≤ e

}

form a subcategory E of DA(S).

Lemma 5.1 The pair (DA(S),E) is a semilattice category pair.

Proof Certainly objDA(S) = E is a semilattice with greatest element 1. For e, f ∈E(S), it is clear that |MorE(e,f )| ≤ 1. Also MorE(e,f ) �= ∅ if and only if(e, f,f ) ∈ MorE(e,f ), and (e, f,f ) is a morphism if and only if f ≤ e. Thus(DA(S),E) is a semilattice category pair. �

Next we identify the minimal cancellable morphisms of the pair (DA(S),E).

Lemma 5.2 A morphism (e, x, f ) of DA(S) is minimal cancellable if and only ife = x∗.

Proof If (e, x, f ) is a morphism in DA(S), then f = x+. Suppose also that e = x∗.Every morphism in DA(S) is an epimorphism, so that (x∗, x, x+) is an epimorphism.Also we have that the morphism (x∗, x, x+) is monomorphism. Hence (x∗, x, x+)

is a cancellable morphism. Let (e, y, f ) be any cancellable morphism in DA(S),and let (g, e, e) be any morphism in E. Suppose that (e, y, f )(g, e, e) = (x∗, x, x+),so that (g, ye, f ) = (x∗, x, x+), that is, g = x∗, ye = x and f = x+. Now ye = y

as (e, y, f ) is a morphism, so that y = x. Now, (e, y, f ) = (x∗, x, x+). Moreover,g = x∗ = e, so that (g, e, e) = (e, e, e). Hence (x∗, x, x+) is a minimal cancellablemorphism.

Conversely, suppose that the morphism (e, x, f ) is minimal cancellable. Nowx∗ ≤ e and we know that (x∗, x, x+) is cancellable, and we have

(e, x, f ) = (e, x, x+) = (

x∗, x, x+)(e, x∗, x∗).

Hence, since (e, x, f ) is minimal cancellable, we have (e, x∗, x∗) = 1e = (e, e, e), sothat e = x∗ as required. �

Theorem 5.1 The pair (DA(S),E) is a weak Loganathan category pair.

Proof By Lemma 5.1, (DA,E) is a semilattice category pair. Let (e, x, f ) be a mor-phism in DA(S). Then f = x+ and e ≤ x∗ so that we can write (e, x, f ) as

(e, x, f ) = (x∗, x, x+)(

e, x∗, x∗)

542 L. Tunsi

where (e, x∗, x∗) ∈ MorE and, by Lemma 5.2, (x∗, x, x+) is minimal cancellable.If (e, x, f ) = (h, y, k)(g,h,h) where (g,h,h) ∈ MorE and (h, y, k) is minimal can-cellable, then

(e, x, f ) = (g, yh, k)

so that e = g, x = yh and k = f = x+. Moreover, by Lemma 5.2, (h, y, k) =(y∗, y, y+) so that h = y∗ and so x = yh = yy∗ = y. Thus (h, y, k) = (x∗, x, x+)

and (g,h,h) = (e, x∗, x∗) so that the expression for (e, x, f ) as a product of a mini-mal cancellable morphism and a morphism of E is unique.

Next let (e, x, f ) and (g, y, e) be two minimal cancellable morphisms. Then byLemma 5.2, e = x∗ and g = y∗. Moreover, f = x+ and e = y+ so that

(xy)+ = (xy+)+ = (xe)+ = (

xx∗)+ = x+.

Also

(xy)∗ = (x∗y

)∗ = (ey)∗ = (y+y

)∗ = y∗.

Hence

(e, x, f )(g, y, e) = (g, xy,f ) = (y∗, xy, x+) = (

(xy)∗, xy, (xy)+)

and also by Lemma 5.2, the product (e, x, f )(g, y, e) is a minimal cancellable mor-phism. Let (e, g, g) ∈ MorE and v = (e, x, f ) be a minimal cancellable morphismin DA(S). Then f = x+, and by Lemma 5.2, e = x∗.

Note that (xg)∗ = x∗g = g so that (g, xg, (xg)+) = ((xg)∗, xg, (xg)+) is minimalcancellable by Lemma 5.2 also (xg)+ ≤ x+ and so (x+, (xg)+, (xg)+) is a morphismof E. Now the square

x∗x+

g (xg)+

�(x∗,x,x+)

�

(x∗,g,g)

�

(x+,(xg)+,(xg)+)

�(g,xg,(xg)+)

commutes because,

(g,xg, (xg)+

)(x∗, g, g

) = (x∗, xg, (xg)+

) = (x∗, (xg)+x, (xg)+

)

= (x+, (xg)+, (xg)+

)(x∗, x, x+)

since S is ample. Putting p = (x+, (xg)+, (xg)+) we have pv = (g, xg, (xg)+)(x∗,g, g) and since (g, xg, (xg)+) is minimal cancellable, we see that |pv| = (x∗, g, g).

Now suppose that v is an isomorphism. We know that x is a regular ele-ment of S. Also, p = (x+, (xg)+, (xg)+) and if we put u = (g, xg, (xg)+), thenpv = u(x∗, g, g) and |pv| = (x∗, g, g). Since the regular elements of S form a


subsemigroup of S, the element xg is regular, and so u is an isomorphism as re-quired.

Finally, let (x∗, x, x+) be a minimal cancellable morphism of DA(S) and(x+, e, e), (x+, f, f ) be morphisms of E. Now

(x+, e, e

)(x∗, x, x+) = (

x∗, ex, e) = (

x∗, ex, (ex)+),

and since it is a morphism in DA(S), it has canonical factorisation

(x∗, ex, e

) = ((ex)∗, ex, (ex)+

)(x∗, (ex)∗, (ex)∗

).

Hence cod(|(x∗, e, e)(x∗, x, x+)|) = (ex)∗. Similarly, |(x∗, f, f )(x∗, x, x+)| has co-domain (f x)∗. Now if (f x)∗ ≤ (ex)∗, then since the natural order is compatible,x(f x)∗ ≤ x(ex)∗. But S is ample, so f x = x(f x)∗ and ex = x(ex)∗, so f x ≤ ex.Thus

f = f x+ = (f x)+ ≤ (ex)+ = ex+ = e.

Hence cod((x+, f, f )(x∗, x, x+)) ≤ cod((x+, e, e)(x∗, x, x+)), and condition (4) inthe definition of weak Loganathan category pair is satisfied. �

With similar way we can prove the following theorem

Theorem 5.2 The pair (DB(S),E) is a weak Loganathan category pair.

6 Properties of weak Loganathan category pairs

Leech in [7] proves that if D is a Loganathan category with distinguished subcate-gory E, then D is a division category. We will prove next that if (D,E) is a weakLoganathan category pair, it has properties similar to those of DA(S) and DB(S).

Proposition 6.1 If (D,E) is a weak Loganathan category pair, then every morphismin D is an epimorphism.

Proof For a, b, c ∈ objD, let x ∈ Mor(a, b). We want to show that x is an epimor-phism. Suppose that y, z ∈ Mor(b, c) such that yx = zx. Let x, y, z have canonicalfactorisations

x = ue, y = vf and z = wg,

where u,v and w are minimal cancellable and e, f and g are morphisms in E. Thenvf ue = wgue. Let f u and gu have canonical factorisations f u = sh and gu = tk

where s, t are minimal cancellable and h, k are morphisms in E. Then vshe = wtke.Now, by the definition of weak Loganathan pair, vs and wt are minimal cancellablemorphisms. Hence vshe and wtke are both canonical factorisations, so that vs = wt

and ke = he. Thus k,h have the same domains and codomains, and since they are

544 L. Tunsi

in E, we have h = k. Now, using f u = sh, vs = wt , and th = tk = gu, we havevf u = vsh = wth = wgu. Since u is cancellable, this gives vf = wg, that is, y = z.Thus x is an epimorphism. �

Lemma 6.1 If (D,E) is a weak Loganathan category pair, then every pair of mor-phisms in E with common domain is a pushout pair in D.

Proof Let e ∈ MorE(a, b) and f ∈ MorE(a, c) be morphisms in E with com-mon domain a. Since objE is a semilattice, the product bc is a member of objEwith bc ≤ b and bc ≤ c. Hence there are morphisms h ∈ MorE(b, bc) and k ∈MorE(c, bc). Since kf and he are coterminal morphisms in E, they are equal, sothe square

a b

c bc

�e

�f

�h

�k

is commutative. Now let m be an object of D and suppose that there are morphismsr ∈ MorD(b,m) and s ∈ MorD(c,m) such that the diagram

a b

c bc

m

�e

�f

�h��

r�k��

s

is commutative. Let r and s have canonical factorisations r = u|r| and s = v|s| re-spectively, where u and v are minimal cancellable morphism. Then

u|r|e = re = sf = v|s|f.

By Lemma 4.2, |r|e = |re| and |s|f = |sf |, so that u|re| and v|sf | are both canon-ical factorisations of re = sf . Thus u = v and |re| = |sf |, that is, |r|e = |s|f . Thus|r| and |s| have a common co-domain, say n and so we have a commutative dia-gram

a b

c bc

n

�e

�f

�h��

|r|�k��

|s|


in E. Since |r| and |s| are both morphisms in E, we have n ≤ b and n ≤ c. Hencen ≤ bc since objE is a semilattice. Consequently, there is a morphism t : bc −→ n inE. Since tk and |s| are both morphisms in E from c to n, we have tk = |s|. Similarly,th = |r|, and the diagram

a b

c bc

n

�e

�f

�h��

|r|�k��

|s|�

t

is commutative. Since s = u|s| an r = u|r|, it follows that u ∈ MorD(n,m) and thatthe diagram is commutative. Thus given morphisms r and s in D with common

co-domain m and such that re = sf , there is a morphism ut in MorD(bc,m) withr = (ut)h and s = (ut)k. Since all morphisms in D are epimorphisms, it follows thatut is the unique such morphism and so

a b

c bc

�e

�f

�h

�k

is a pushout square in D. �

Lemma 6.2 Let (D,E) be a weak Loganathan category pair; t be a morphism inD; e, f be morphisms in E; u be a minimal cancellable morphism in D and supposethat the square

546 L. Tunsi

a b

c d

�t

�e

�f

�u

is commutative. Then the square is a pushout square.

Proof First, we note that since the square is commutative, we have f t = ue and, asu minimal cancellable, and e is in E, we have |f t | = e. Let r and s be morphisms inD such that rt = se, that is, the diagram

a b

c d

m

�t

�e

�f��

r�u��

s

is commutative. Let r and s have canonical factorisations r = r|r| and s = s|s| re-spectively. By Lemma 4.2, |se| = |s|e. We claim that ||r|t | = |s|e. To prove the claim,let ||r|t | = h. Then |r|t = wh for some minimal cancellable morphism w. Hencert = r|r|t = rwh. Now, by condition (2) of the definition of weak Loganathan cate-gory pair, rw is minimal cancellable because both r and w are minimal cancellable.As rt = se = s|s|e, we have, by uniqueness of canonical factorisation, h = |s|e ands = rw. It follows that the diagram

a b

c d

. .

m

�t

�e

�f��

|r|�u

�|s|

�w��

s�r

is commutative. Now, we have noted that |f t | = e, and so

cod(|f t |) = cod(e) ≥ cod

(|s|e) = cod(h) = cod(||r|t |).


Hence, cod(f t) ≥ cod(|r|t), and so, there is a morphism q in MorE from d tocod(|r|) = cod(|r|t). We claim that the diagram

a b

c d

. .

m

�t

�e

�f��

|r|�u

�|s|

�

q

�w��

s�r

is commutative. We have to show that |r| = qf and qu = w|s|.First, we observe that |r| and qf are coterminal morphisms in E, so we certainly

have |r| = gf .Next, note that rt = r|r|t = rqf t = rque, and that, se = s|s|e = rw|s|e so that

rque = rt = se = rw|s|e. Now r is cancellable, and every morphism in D is anepimorphism, so qu = w|s|. Thus we have a morphism rq from d to m, and r =r|r| = rqf , and

s = s|s| = rw|s| = rqu.

All morphisms in D are epimorphisms, and so the square

a b

c d

�t

�e

�f

�u

is a pushout square as claimed. �

Corollary 6.1 Let (D,E) be a weak Loganathan category pair. If e ∈ MorE andv ∈ MorD is minimal cancellable, and if e, v have common domain, then they are apushout pair. Moreover, there is a pushout square

a b

c d

�v

�e

�p

�u

with p in E and u is minimal cancellable. If v is an isomorphism, then u is also anisomorphism.

548 L. Tunsi

Proof By condition (3) in the definition of weak Loganathan category pair, thereis a morphism p in MorE such that pv is defined and has canonical factori-sation ue for some minimal cancellable morphism u. The same condition al-lows us to choose u to be an isomorphism if v is an isomorphism. Now thesquare

a b

c d

�v

�e

�p

�u

is commutative. Since e,p are in E and u is minimal cancellable, it follows from thelast lemma that the square is a pushout square. �

Theorem 6.1 Let (D,E) be a weak Loganathan category pair, and let a, c, d ∈objD, x ∈ MorD(a, c) and e ∈ MorE(a, d). Then the morphisms x and e form apushout pair.

Proof Let x have canonical factorisation x = vf for a minimal cancellable morphismv and f ∈ MorE(a, b). In the following diagram

a b c

d

�f

�e

�v

the pair f, e is a pushout pair by Lemma 6.1, and so there are morphisms h ∈Mor(b, bd) and k ∈ Mor(d, bd) such that the square in the diagram

a b c

d bd

�f

�e

�v

�h

�k

is a pushout square. Moreover, from the proof of Lemma 6.1, we know that h and k

are both morphism in E. By Corollary 6.1, the pair v,h is a pushout pair and so, forsome object m there are morphisms p ∈ Mor(c,m) and u ∈ Mor(bd,m) such that theright hand square in the diagram

a b c

d bd m

�f

�e

�v

�h

�p

�k

�u


is pushout square. Thus both the left and right hand squares are pushouts and henceby Proposition 1.2, the outer rectangle is a pushout. �

7 Constructing an ample monoid from a weak Loganathan pair

Most of this section is devoted to describing a construction which gives an amplemonoid from a weak Loganathan category pair. We conclude the section by showingthat if S is an ample monoid, and if we apply our construction to the weak Loganathancategory pair (DA(S),E), then we obtain a monoid isomorphic to S.

Let (D,E) be a weak Loganathan category pair. Let x be a morphism of D. ByDefinition 4.1(1), x can be written uniquely as x = ue (canonical factorisation of x)where u is minimal cancellable and e ∈ MorE. We put |x| = e and u = x, so thatx = x|x|. We remind the reader that the objects of D form a semilattice with greatestelement 1.

Let U consist of all pairs of morphisms (x1, x2) where x1, x2 ∈ MorD with com-mon domain 1 and a common codomain, and such that x1 is an isomorphism. Sincecod(x1) = cod(x2), we have x2 = x1x1

−1x2. Define a relation ∼ on the set of pairs asfollows:

(x1, x2) ∼ (y1, y2)

if and only if there is an isomorphism u in D such that

x1 = uy1 and x2 = uy2.

It is easy to prove that the relation is an equivalence relation on U . The equivalenceclass of (x1, x2) is denoted by [x1, x2], and the set of classes U/ ∼ is denoted byS[D,E].

Now let [x1, x2] and [y1, y2] ∈ S[D,E]. Since (x1, x2), (y1, y2) ∈ U we have thatx1 and y1 are isomorphisms. Putting p = x1

−1x2 and q = y1−1y2 we have x2 = x1p

and y2 = y1q .We intend to define a product on S[D,E] by the rule:

[x1, x2][y1, y2] = [r1x1, s1q] where

1 .

. .

�x2

�|y1|

�r1

�s1

is pushout square. For this to make sense we require the pair (r1x1, s1q) to be amember of U .

The following lemma allows us to choose one particular pushout square and showthat, for this particular choice of r1 and s1, we have (r1x1, s1q) in U .

550 L. Tunsi

Lemma 7.1 Suppose that

1 .

. .

�x2

�|y1|

�r1

�s1

and

1 .

. .

�x2

�|y1|

�r ′1

�s′1

are both pushout squares. Then (r1x1, s1q) ∈ U if and only if (r ′1x1, s

′1q) ∈ U .

Proof Suppose that (r1x1, s1q) ∈ U . By Lemma 1.1, there is an isomorphism u suchthat r ′

1 = ur1 and s′1 = us1. Clearly, dom(ur1x1) = dom(us1q) = 1 and cod(ur1x1) =

cod(us1q). We also have a canonical factorisation

r ′1x1

∣∣r ′1x1

∣∣ = r ′1x1 = ur1x1 = ur1x1|r1x1|.

Since |r ′1x1|, |r1x1| ∈ E, we have by the uniqueness, r ′

1x1 = ur1x1. Now (r1x1, s1q)

is in U , so that r1x1 is an isomorphism and so ur1x1 is an isomorphism. It followsthat r ′

1x1 is an isomorphism and so (r ′1x1, s

′1q) ∈ U . If (r ′

1x1, s′1q) ∈ U , then a similar

argument shows that (r1x1, s1q) ∈ U . �

The proof of the next lemma is in [8].

Lemma 7.2 There exist morphisms r1, s1 such that

1 .

. .

�x2

�|y1|

�r1

�s1

is a pushout square and (r1x1, s1q) ∈ U .

We prove now that our multiplication is well defined. Suppose that [x1, x2] =[x′

1, x′2] and [y1, y2] = [y′

1, y′2]; then

(x1, x2) ∼ (x′

1, x′2

)and (y1, y2) ∼ (

y′1, y

′2

),

so that there are isomorphisms u,u′ such that

x1 = ux′1, x2 = ux′

2, y1 = u′y′1 and y2 = u′y′

2.

Let x′1|x′

1|, x′2|x2|, y′

1|y1| and y′2|y′

2| be the canonical factorisations of x′1, x

′2, y

′1

and y′2, respectively. Then x1 = ux′

1, x2 = ux′2, y1 = u′y′

1 and y2 = u′y′2 by Defi-

nition 4.1(1).First, we note that |y1| = |y′

1|. This follows from

y1|y1| = y1 = u′y′1 = u′y′

1

∣∣y′1

∣∣ = y1∣∣y′

1

∣∣


and the uniqueness of canonical forms.Let q = y1

−1y2 and note that

q = y1−1y2 = (

y′1

)−1(u′)−1

y2 = (y′

1

)−1y′

2 = q ′.

Using these observations and the definition of product, we have [x1, x2][y1, y2] =[r1x1, s1q] and [x′

1, x′2][y′

1, y′2] = [r ′

1x′1, s

′1q] where the squares

1 .

. .

�x2

�|y1|

�r1

�s1

and

1 .

. .

�x′2

�|y1|

�r ′1

�s′1

are pushouts.Since x2 = ux′

2, the square

1 .

. .

�x2

�|y1|

�r ′1u

−1

�s′1

is a pushout by Proposition 1.3. Hence by Lemma 1.1, there is an isomorphism t suchthat r ′

1u−1 = tr1 and s′

1 = ts1. Hence r ′1x

′1 = r ′

1u−1x1 = tr1x1 and s′

1q = ts1q so that(r1x1, s1q) ∼ (r ′

1x′1, s

′1q), that is, [r1x1, s1q] = [r ′

1x′1, s

′1q]. Thus the multiplication is

well defined.Note that, Leech defined a relation ∼L on the set of pairs of morphisms by

(x, x′) ∼L (y, y′) if and only if there exists an isomorphism w such that

y = wx and y′ = wx′.

We now compare our definitions with those of Leech. Suppose that (D,E) is aLoganathan category pair. Leech’s set of pairs is

UL = {(x1, x2) : x1, x2 ∈ MorD with dom(x1) = dom(x2) = 1 and

cod(x1) = cod(x2)},

which is exactly the same as our set U .Our equivalence relation is the same as that of Leech. We now want to consider

the two definitions of multiplication. Let [x1, x2], [y1, y2] ∈ S[D,E]. For our multi-plication, we put

[x1, x2][y1, y2] = [r1x1, s1q]where q = y1

−1y2

1 .

. .

�x2

�|y1|

�r1

�s1

552 L. Tunsi

is a pushout square. Since y1 is an isomorphism and y1|y1| = y1, we see that

1 .

. .

�x2

�y1

�r1

�s1y1

−1

is a pushout square. Leech’s definition of multiplication is given by

[x1, x2][y1, y2] = [r1x1, s1y1

−1y2].

Since q = y1−1y2 we see that the two multiplications are the same, and thus our

construction does generalise that of Leech.Now we prove that if (D,E) is a weak Loganathan pair, then the set of equivalence

classes S[D,E] with the multiplication defined above is an ample monoid. We splitthe proof into a sequence of subsidiary results. First we have

Proposition 7.1 If (D,E) is a weak Loganathan category pair, then S[D,E] ismonoid.

Proof First, we prove that S[D,E] is a semigroup. Let [x1, x2], [y1, y2] and [z1, z2]be elements in S[D,E]. Then we have [x1, x2][y1, y2] = [r1x1, s1q1] where s1 and r1are in a pushout of x2 and |y1| in the following square

1 .

. .

�x2

�|y1|

�r1

�s1

with r1 in E and q1 = y1−1y2. In particular, r1x2 = s1|y1|. Now [y1, y2][z1, z2] =

[r2y1, s2q2] where s2 and r2 are in a pushout of y2 and |z1| in the following square

1 .

. .

�y2

�|z1|

�r2

�s2

with r2 in E and q2 = z1−1z2. In particular, r2y2 = s2|z1|. Now since the first square

is pushout and y1 is an isomorphism with dom(y1) = cod(|y1|), we have that, byProposition 1.2, the rectangle in the following diagram

1 .

.

. .

�x2

�|y1|

�

r1

�y1

�

s1

�s1y1−1


is a pushout. Now r2 is in E and so by Theorem 6.1, there are morphisms m,n withm in E such that the following square

1 .

. .

�s1y1−1

�r2

�m

�n

is a pushout. Now the last two diagrams together give us that the outer rectangle inthe following diagram

1 .

.

. .

. .

�x2

�|y1|

�

r1

�y1

�

s1

�s1y1−1

�r2

�m

�n

is a pushout. We also note that the square

1 .

1 .

�q1

�1

�y1

�y2

is a pushout, because first, we have y1q1 = y1y1−1y2 = y2 so that the square com-

mutes. Next, if h, k are morphisms such that hq = k, then there is a morphism hy1−1

such that (hy1−1)y2 = h(y1

−1y2) = hq = k, and (hy1−1)y1 = h(y1

−1y1) = h. Nowwe have that the diagram

1 .

1 .

. .

�q1

�1

�y1

�y2

�|z1|

�r2

�s2

554 L. Tunsi

is a pushout. The diagrams together give that the following diagram of pushoutsquares

1 .

1 .

. . .

. . .

�x2

�|y1|

�

r1�q

�1

�

y1

�

s1

�y2

�

|z1|

�s1y1−1

�

r2

�

m

�s2

�n

is commutative. Thus we have

mr1x2 = nr2y1 = n˜(r2y1)|r2y1|,so that the following square

1 .

. .

�x2

�r2y1

�mr1

�n

is a pushout because it is obtained by combining two pushout squares. Hence usingthe fact that r2y1 is an isomorphism, we have by Corollary 1.1, that the square

1 .

. .

�x2

�|r2y1|

�mr1

�n(r2y1)

is a pushout. Then

[x1, x2]([y1, y2][z1, z2]

) = [x1, x2][r2y1, s2q2] = [mr1x1, n(r2y1)q3

]

where s2q2 = (r2y1)q3. Hence

[x1, x2]([y1, y2][z1, z2]

) = [mr1x1, ns2q2].Now we have ms1q1 = ns2|z1|, so that by combining pushout squares, the square

1 .

. .

�s1q

�|z1|

�m

�ns2


is pushout. Then ([x1, x2][y1, y2])[z1, z2] = [r1x1, s1q1][z1, z2] = [mr1x1, ns2q2].Hence

[x1, x2]([y1, y2][z1, z2]

) = ([x1, x2][y1, y2])[z1, z2]

and S[D,E] is a semigroup.Secondly, we show that [1,1] is the identity element of S[D,E]. For any element

[x1, x2] in S[D,E], we have the square

1 a

1 a

�x2

�11

�1a

�x2

is a pushout. Then [x1, x2][1,1] = [1x1, x21] = [x1, x2]. Consider the followingsquare

1 1

b b

�11

�|x1|

�|x1|

�1b

where b = cod(|x1|) and 1b is the identity arrow at b. Clearly the square is commu-tative and it is easy to see it is a pushout.

Now x1 = x1|x1| and since [x1, x2] is in S[D,E], we have that x1 is an isomor-phism. Putting p = x1

−1x2 we see that (x1, x2) ∼ (|x1|,p) and therefore

[1,1][x1, x2] = [1,1][|x1|,p] = [|x1|11,1bp

] = [|x1|,p] = [

x1, x2].

Hence S[D,E] is a monoid. �

We now locate the idempotents of S[D,E].

Proposition 7.2 The set of idempotents of S[D,E] is

E(S[D,E]) = {[x, x] : x is a morphism of D with domain 1

},

and E(S[D,E]) is a commutative submonoid of S[D,E].

Proof Suppose that [x1, x2] is an idempotent. Then [x1, x2][x1, x2] = [sx1, rp] =[x1, x2], where s and r are in the following pushout square of x2 and |x1|, and p =x1

−1x2

1 a

. .

�x2

�|x1|

�s

�r

556 L. Tunsi

In particular, sx2 = r|x1|. Now we have (sx1, rp) ∼ (x1, x2), so that there is an iso-morphism u such that sx1 = ux1 and rp = ux2. Since x1 is an epimorphism, we haves = u and since p = x1

−1x2 so that rx1−1x2 = ux2, and since x2 is an epimorphism

we have rx1−1 = u. But s is in E and the only isomorphisms in E are the identity

arrows at the objects. Hence u = s = 1a , so that rx1−1 = 1a , then r = x1. Then

x1 = x1|x1| = r|x1| = sx2 = 1ax2 = x2.

Conversely, suppose that x1 = x2, and let a = cod(x1). Consider the followingsquare

1 a

. a

�x1

�|x1|

�1a

�x1

Since x1 = x1|x1|, clearly the square is commutative. Hence the square is pushout byLemma 6.2. Therefore [x1, x1][x1, x1] = [1x1, x1p] = [x1, x1x1

−1x2] = [x1, x2] =[x1, x1], where x2 = x1p. Hence [x1, x1] is an idempotent.

Now we show that the idempotents commute. Let [x, x] and [y, y] be idempotentsin S[D,E]. Then [x, x][y, y] = [s1x, r1q], where s1, r1 are in a pushout square of x

and |y|1 .

. .

�x

�|y|

�s1

�r1

and q = y−1y = y−1y|y| = |y|. Since s1x = r1|y|, we have [x, x][y, y] =[s1x, r1q] = [s1x, r1|y|] = [r1|y|, s1x] = [|y|, |y|][x, x] since the square

1 .

. .

�|y|

�|x|

�r1

�s1x

is a pushout by Corollary 1.1. Now (y, y) ∈ U , so that y is an isomorphism and hence(y, y) ∼ (|y|, |y|). Hence [|y|, |y|][x, x] = [y, y][x, x], and we have [x, x][y, y] =[y, y][x, x].

It is easy to see that if the idempotents of a monoid commute with each other, thenthey form a submonoid. Hence E(S(D,E)) is a submonoid of S[D,E]. �

Now we prove that S[D,E] is adequate, that is, we prove that each L∗-class andeach R∗-class contains an idempotent.

Proposition 7.3 The monoid S[D,E] is adequate.


Proof Let [x, y] ∈ S[D,E]. Then x is an isomorphism and putting p = x−1y wehave y|y| = y = xp|p|. Consider the idempotent [|y|, |y|] = [|p|, |p|]. We will showthat [x, y]L∗[|p|, |p|]. Since x is an isomorphism, x = x|x| and y = xp, we have[x, y] = [|x|,p]. Consider the product [|x|,p][|p|, |p|]. The following square is apushout

1 a

. .

�p

�|p|

�1a

�p

Hence [|x|,p][|p|, |p|] = [1a|x|, p|p|] = [|x|,p], that is, [x, y][|p|, |p|] = [x, y].Now suppose [z1, z2] and [k1, k2] are elements in S[D,E] such that [x, y][z1, z2] =[x, y][k1, k2]. Then [|x|,p][z1, z2] = [|x|,p][k1, k2]. We put q1 = z−1z2 and q2 =k−1k2. We shall construct pushout squares

1 .

. .

�p

�|z1|

�s1

�r1

and

1 .

. .

�p

�|k1|

�s2

�r2

as follows. First, by Lemma 6.1 we can form pushout squares

1 .

. .

�|p|

�|z1|

�j1

�i1

1 .

. .

�|p|

�|k1|

�j2

�i2

where i1, i2, j1, j2 are in E. Now by Theorem 6.1, we can adjoin pushout squares asfollows:

1 . .

. . .

�|p|

�|z1|

�p

�j1

�s1

�i1

�t1

1 . .

. . .

�|p|

�|k1|

�p

�j2

�s2

�i2

�t2

where s1, s2 are in E and t1, t2 are minimal cancellable. By Proposition 1.1, the outerrectangles in the above diagrams are pushouts and so

[s1|x|, t1i1q1

] = [|x|,p][z1, z2] = [|x|,p][k1, k2] = [s2|x|, t2i2q2

].

Thus (s1|x|, t1i1q1) ∼ (s2|x|, t2i2q2) so that there is an isomorphism u such thats1|x| = us2|x| and t1i1q1 = ut2i2q2. As morphisms in D are epimorphisms, s1 = us2,and since s1, s2 are in E and u is an isomorphism, the uniqueness of canonical formsgives s1 = s2 and u = 1a where a = cod(s2). Hence t1j1 = s1p = s2p = t2j2. Ast1, t2 are minimal cancellable and j1, j2 are in E, uniqueness of canonical formsgives t1 = t2 and j1 = j2. Hence t1i1q1 = ut2i2q2 = 1at1i2q2 = t1i2q2, and since t1 is

558 L. Tunsi

(minimal) cancellable, we obtain i1q1 = i2q2. Now, using the pushout squares above,we have

[|p|, |p|][z1, z2] = [j1|p|, i1q1

] = [j2|p|, i2q2

] = [|p|, |p|][k1, k2].Thus [|p|, |p|]L∗[x, y] so that the L∗-class of [x, y] contains an idempotent.

Now, again we use the fact that [x, y] = [|x|,p] and consider the idempotent[|x|, |x|]. Let cod(|x|) = a. Then

1 a

a a

�|x|

�|x|

�1a

�1a

is a pushout square and so [|x|, |x|][x, y] = [|x|, |x|][|x|,p] = [1a|x|,1ap] =[|x|,p] = [x, y]. Now let [z1, z2] and [k1, k2] be element in S[D,E] such that[z1, z2][x, y] = [k1, k2][x, y], that is, [z1, z2][|x|,p] = [k1, k2][|x|,p]. Then [r1z1,

s1p] = [r2k1, s2p] where the squares

1 .

. .

�z2

�|x|

�r1

�s1

and

1 .

. .

�k2

�|x|

�r2

�s2

are pushout squares. Since these are pushout squares we see that [z1, z2][|x|, |x|] =[r1z1, s1|x|] and [k1, k2][|x|, |x|] = [r2k1, s2|x|]. Since [r1z1, s1p] = [r2k1, s2p],there is an isomorphism u of D such that r1z1 = ur2k1 and s1p = us2p. We cancancel p because morphisms are epimorphisms, so s1 = us2. Thus s1|x| = us2|x|and hence

(r1z1, s1|x|) ∼ (

r2k1, s2|x|),and consequently, [z1, z2][|x|, |x|] = [k1, k2][|x|, |x|]. Therefore, [|x|, |x|]R∗[x, y]and so every R∗-class contains an idempotent. �

We have now proved that S[D,E] is an adequate monoid. Moreover, for an ele-ment [x, y] of S[D,E] we have [x, y]+ = [|x|, |x|] and [x, y]∗ = [|y|, |y|].

Finally, we want to show that S[D,E] is an ample monoid.

Proposition 7.4 The monoid S[D,E] is ample.

Proof Let a = [x1, x2] be element in S[D,E] and let e = [y, y] be any idempotentin S[D,E]. We want to prove (ae)+a = ae and a(ea)∗ = ea.

First, we note that x2 = x1p where p = x−11 x2 and that (x1, x2) ∼ (|x1|,p), that

is, [x1, x2] = [|x1|,p]. Also [y, y] = [|y|, |y|] and so

ae = [x1, x2][y, y] = [|x1|,p][|y|, |y|] = [

s|x1|, r|y|]


where

1 .

. .

�p

�|y|

�s

�r

is a pushout square. In particular, sp = r|y|. Also we can assume that the morphisms is in E, so that |s|x1|| = s|x1|. Thus (ae)+ = [s|x1|, s|x1|] and so

(ae)+a = [s|x1|, s|x1|

][x1, x2] = [s|x1|, s|x1|

][x1|,p].

The square

1 a

. a

�s|x1|

�|x1|

�1a

�s

is a pushout (where a = cod(s)), so that (ae)+a = [1as|x1|, sp] = [s|x1|, r|y|] = ae.Thus (ae)+a = ae.

Now ea = [|y|, |y|][|x1|,p] = [s|y|, rp] where the square

1 .

. .

�|y|

�|x1|

�s

�r

is a pushout. By Lemma 6.1, we can assume that the morphisms r and s are both inE, and in particular, we have s|y| = r|x1|. Certainly the square

1 .

. .

�p

�|rp|

�r

�(rp)

is commutative; r and |rp| are morphisms in E; and (rp) is minimal cancellative.Hence the square is a pushout square. Now (ea)∗ = [|rp|, |rp|] and so

a(ea)∗ = [|x1|,p][|rp|, |rp|] = [

r|x1|, (rp)|rp|] = [s|y|, rp] = ea.

Thus S[D,E] is an ample monoid as required. �

We conclude this paper by showing that if we start with an ample monoid S,and construct the weak Loganathan category pair (DA(S),E), then the monoidS[DA(S),E], as constructed in this section, is isomorphic to S.

560 L. Tunsi

Theorem 7.1 Suppose that S is ample monoid and consider S[DA(S),E]. Defineθ : S −→ S[DA(S),E] by

θ(s) = [(1, s+, s+)

,(1, s, s+)]

.

Then θ is an isomorphism.

Proof First, we point out that [(1, s+, s+), (1, s, s+)] is an element of S[DA(S),E]for any s ∈ S. Putting x = (1, s+, s+) and y = (1, s, s+) we observe that x =(s+, s+, s+) is an isomorphism and that x, y have common domain 1 and com-mon co-domain. Hence (x, y) ∈ U and [x, y] ∈ S[DA(S),E]. Let s, t ∈ S so that[(1, s+, s+), (1, s, s+)] = [(1, t+, t+), (1, t, t+)] are members of S[DA(S),E]. Toform the product, consider the pushouts

1 s∗s+

t+ s∗t+ (st)+

�(1,s∗,s∗)

�

(1,t+,t+)

�(s∗,s,s+)

�

(s∗,s∗t+,s∗t+)

�

(s+,(st)+,(st)+)

�(t+,s∗t+,s∗t+)

�(s∗t+,st+,(st)+)

Then[(

1, s+, s+),(1, s, s+)][(

1, t+, t+),(1, t, t+

)]

= [(s+, (st)+, (st)+

)(1, s+, s+)

,(s∗t+, st+, (st)+

)(t+, s∗t+, s∗t+

)(1, t, t+

)]

= [(1, (st)+s+, (st)+

),(1, st+s∗t+t, (st)+

)]

= [(1, (st)+, (st)+

),(1, st, (st)+

)]

since (st)+ ≤ s+ and st+s∗t+t = ss∗t+t = st . Hence θ(s)θ(t) = θ(st) and θ is ahomomorphism.

Next we show that θ is one-one. Suppose that s, t ∈ S and θ(s) = θ(t). Then[(

1, s+, s+),(1, s, s+)] = [(

1, t+, t+),(1, t, t+

)]

and so there is an isomorphism u in DA(S) such that (1, s+, s+) = u(1, t+, t+) and(1, s, s+) = u(1, t, t+). Let u = (x∗, x, x+). Since (1, s+, s+) = (x∗, x, x+)(1, t+,

t+) = (1, xt+, x+) we must have x+ = s+ and x∗ = t+. Furthermore, s+ = xt+.Thus s+ = xt+ = xx∗ = x so that x is idempotent and x∗ = x = x+. Hence s+ = t+.We also have

(1, s, s+) = (

x∗, x, x+)(1, t, t+

) = (1, xt, x+)

so that s = xt = s+t = t+t = t . Thus θ is one-one.Finally, we show that θ is onto. Let [(1, s, s+), (1, t, t+)] ∈ S[DA(S)]. Then s+ =

t+ and (1, s, s+) has canonical factorisation(1, s, s+) = (

s∗, s, s+)(1, s∗, s∗).


Put (1, u,u+) = (s+, s−1, s∗)(1, t, t+) = (1, s−1t, s∗). Then u+ = s∗ so that we have

(1, s, s+) = (

s∗, s, s+)(1, u+, u+)

.

Thus ((1, s, s+), (1, t, t+)) ∼ ((1, u+, u+), (1, u,u+)) so that

θ(u) = [(1, u+, u+)

,(1, u,u+)] = [(

1, s, s+),(1, t, t+

)]

and hence θ is onto. Therefore θ is an isomorphism as required. �

Acknowledgements I would like to thank my supervisor John Fountain, for his guidance and providingthe necessary points of the paper.

References

1. El-Qallali, A.: Abundant semigroups with a multiplicative type A transversal. Semigroup Forum 47,327–340 (1993)

2. Fountain, J.: Abundant semigroups. Proc. Lond. Math. Soc. 44, 103–129 (1982)3. Fountain, J.: Adequate semigroups. Proc. Edinb. Math. Soc. 22, 113–125 (1979)4. Howie, J.M.: Fundamentals of Semigroup Theory. Oxford University Press, Oxford (1995)5. Lawson, M.V.: The structure of type A semigroups. Q. J. Math. (Oxford) 37, 279–298 (1986)6. Lawson, M.V.: The natural partial order on an abundant semigroup. Proc. Edinb. Math. Soc. 30, 169–

186 (1987)7. Leech, J.: Constructing inverse monoids from small categories. Semigroup Forum 36, 89–116 (1987)8. Tunsi, L.: Ample monoids and the theory of small categories. PhD thesis. The University of York

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