Ordered groupoid quotients and congruences on inverse semigroups

0.3 MB·

by  Nouf AlYamani

ORDERED GROUPOID QUOTIENTS AND CONGRUENCES ON
INVERSE SEMIGROUPS
6
1
0 NOUFALYAMANI
2

n
a
J KingdomofSaudiArabia,MinistryofHigherEducation,

9
UniversityofDammam,
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P.O.Box1982,Dammam31441,SaudiArabia.

] [email protected]
R
G
.
h
t N.D.GILBERT
a
m
[

1 SchoolofMathematicalandComputerSciences
v andtheMaxwellInstitutefortheMathematicalSciences,
4
Heriot-WattUniversity,
9
1 EdinburghEH144AS,U.K
8
[email protected]
0
.
1
0
6 ABSTRACT. WeintroduceapreorderonaninversesemigroupSassoci-
1 atedtoanynormalinversesubsemigroupN,thatliesbetweenthenatural
: partialorderandGreen’sJ–relation.Thecorrespondingequivalencere-
v
lation(cid:39) isnotnecessarilyacongruenceonS,butthequotientsetdoes
i N
X inheritanaturalorderedgroupoidstructure. Weshowthatthisconstruc-
r tionpermitsthefactorisationofanyinversesemigrouphomomorphism
a
into a composition of a quotient map and a star-injective functor, and
thatthisdecompositionimpliesaclassificationofcongruencesonS.We
give an application to the congruence and certain normal inverse sub-
semigroupsassociatetoaninversemonoidpresentation.
2010MathematicsSubjectClassification. Primary: 20L05;Secondary18D15,18B40.
Keywordsandphrases. inversesemigroup,congruence,normalsubsemigroup.
TheauthorsgratefullyacknowledgesthesupportofaresearchgrantfromtheUniversity
ofDammam.
1
INTRODUCTION
LetS beaninversesemigroupwithsemilatticeofidempotentsE(S). Recall
thatthenaturalpartialorder onS isdefinedby
s (cid:54) t ⇐⇒ thereexists e ∈ E(S) with s = et.
The natural partial order may be characterized in a number of alternative
ways,including:
• thereexistsf ∈ E(S)withs = tf,
• s = ss−1t,
• s = ts−1s.
(see [6, Proposition 5.2.1]). In this paper, we shall generalize the natural
partial order by introducing a preorder (cid:54) on S for any normal inverse
N
subsemigroup N: the natural partial order then corresponds to the minimal
normal inverse subsemigroup E(S), and at the other extreme, the preorder
associated to S itself is the J–preorder. Symmetrizing the preorder (cid:54)
N
yields an equivalence relation (cid:39) (the identity when N = E(S) and the
N
J–relation when N = S). However, this relation need not be a congru-
ence, and so the set of equivalence classes S/ (cid:39) need not be an inverse
N
semigroup.
However, we may investigate (cid:39) further by exploiting the relationship be-
N
tweeninversesemigroupsandorderedgroupoids. Anorderedgroupoidisa
small category in which every morphism is invertible, equipped with a par-
tial order on morphisms. (The definition is recalled in detail in section 1.)
An inverse semigroup can be considered as an ordered groupoid in which
the identities form a semilattice, and from any such ordered groupoid a
corresponding inverse semigroup can be constructed. In the study of in-
verse semigroups, it is often fruitful to extend the point of view to ordered
groupoids, and this is a major theme of [8]. We show that the quotient set
S/ (cid:39) always inherits a natural ordered groupoid structure. Moreover, to
N
any homomorphism φ : S → Σ of inverse semigroups, we associate its
kernelK = {s ∈ S : sφ ∈ E(Σ)},andφthenfactorisesas
S −→ S/ (cid:39) −→ Σ
K
withthemapS/ (cid:39) → Σastar-injectivefunctorfromtheorderedgroupoid
K
S/ (cid:39) toΣ(consideredasanorderedgroupoid).
K
Now any congruence ρ on an inverse semigroup determines a normal in-
verse subsemigroup K, its kernel, which consists of all elements of S that
are ρ–equivalent to idempotents. We compare the structures of the ordered
groupoidS/ (cid:39) andthequotientinversesemigroupS/ρ,andweshowthat
K
2
if(cid:39) isacongruence,thenitistheminimalcongruencewithkernelK. We
K
showhowtoclassifycongruencesbythefactorisation
S −→ S/ (cid:39) −→ S/ρ
K
andlookatcertaincongruencesandtheirkernelsassociatedwithaninverse
monoidpresentation,andtherelationshipsbetweenthem.
1. ORDERED GROUPOIDS AND INVERSE SEMIGROUPS
AgroupoidGisasmallcategoryinwhicheverymorphismisinvertible. We
consideragroupoidasanalgebraicstructurefollowing[5]: theelementsare
the morphisms, and composition is an associative partial binary operation.
The set of identities in G is denoted E(G), and an element g ∈ G has
domaingg−1 andrangeg−1g.
Anorderedgroupoid(G,(cid:54))isagroupoidGwithapartialorder(cid:54)satisfy-
ingthefollowingaxioms:
OG1 forallg,h ∈ G,ifg (cid:54) htheng−1 (cid:54) h−1,
OG2 if g (cid:54) g ,h (cid:54) h and if the compositions g h and g h are
1 2 1 2 1 1 2 2
defined,theng h (cid:54) g h ,
1 1 2 2
OG3 if g ∈ G and x is an identity of G with x (cid:54) gd, there exists
a unique element (x|g), called the restriction of g to x, such that
(x|g)(x|g)−1 = xand(x|g) (cid:54) g,
Asaconsequenceof[OG3]wealsohave:
OG3* if g ∈ G and y is an identity of G with y (cid:54) gr, there exists a
unique element (g|y), called the corestriction of g to y, such that
(g|y)−1(g|y) = y and(g|y) (cid:54) g,
sincethecorestrictionofg toy maybedefinedas(y|g−1)−1.
Let G be an ordered groupoid and let a,b ∈ G. If a−1a and bb−1 have a
greatest lower bound (cid:96) ∈ E(G), then we may define the pseudoproduct
of a and b in G as a ⊗ b = (a|(cid:96))((cid:96)|b), where the right-hand side is now a
composition defined in G. As Lawson shows in Lemma 4.1.6 of [8], this is
apartiallydefinedassociativeoperationonG.
If E(G) is a meet semilattice then G is called an inductive groupoid. The
pseudoproduct is then everywhere defined and (G,⊗) is an inverse semi-
group. Ontheotherhand,givenaninversesemigroupS withsemilatticeof
idempotentsE(S),thenS isaposetunderthenaturalpartialorder,andthe
restrictionofitsmultiplicationtothepartialcomposition
s·t = st ∈ S definedwhen s−1s = tt−1
3
gives S the structure of an ordered groupoid, with set of identities E(S).
These constructions give an isomorphism between the categories of in-
verse semigroups and inductive groupoids: this is the Ehresmann-Schein-
Nambooripad Theorem [8, Theorem 4.1.8]. We call a product st ∈ S with
s−1s = tt−1 a trace product. Any product in S can be expressed as a trace
product,attheexpenseofchangingthefactors,sincest = stt−1 ·s−1st.
Let e ∈ E(G). Then the star of e in G is the set star (e) = {g ∈ G :
G
gg−1 = e}. A functor φ : G → H is said to be star-injective if, for each
e ∈ E(G), the restriction φ : star (e) → star (eφ) is injective. A star-
G H
injectivefunctorisalsocalledanimmersion. IfGisinductive,thenstar (e)
G
isjusttheGreenR–classofeintheinversesemigroup(G,⊗).
2. NORMAL INVERSE SUBSEMIGROUPS AND QUOTIENTS
An inverse subsemigroup N of an inverse semigroup S is normal [13] if it
isfull–thatis,ifE(N) = E(S)–andif,foralls ∈ S andn ∈ N,wehave
s−1ns ∈ N. A normal inverse subsemigroup N of S determines a relation
(cid:54) onS,definedusingthenaturalpartialorder(cid:54)onS,asfollows:
N
(2.1) s (cid:54) t ⇐⇒ thereexist a,b ∈ N suchthat a·s·b (cid:54) t.
N
Note the requirement that trace products occur here. We define the relation
(cid:39) bysymmetrizing(cid:54) :
N N
(2.2)
s (cid:39) t ⇐⇒ thereexist a,b,c,d ∈ N suchthat a·s·b (cid:54) t and c·t·d (cid:54) s
N
Lemma2.1.
(a) Therelation(cid:54) isthenaturalpartialorder(cid:54)onS.
E(S)
(b) Therelation(cid:54) istheJ–preorder(cid:22) onS.
S J
(c) Ifs (cid:54) tinthenaturalpartialorderonS thens (cid:54) tforanynormal
N
inversesubsemigroupN ofS.
(d) s (cid:54) eforsomee ∈ E(S)ifandonlyifs ∈ N.
N
(e) Ifs (cid:54) s2 thens ∈ N.
N
(f) Ifs (cid:54) tthenst−1 ∈ N.
N
(g) Therelation(cid:54) isapreorderonS andhence(cid:39) isanequivalence
N N
relationonS.
Proof. (a) This is clear, since if e,f ∈ E(S), a trace product e·s·f is
equaltos.
(b) Ifa·s·b (cid:54) tthenaa−1 (cid:54) tt−1 anda−1a = ss−1. Hences (cid:22) t. On
J
the other hand, if s (cid:22) t then there exists p ∈ S with pp−1 (cid:54) tt−1
J
4
andp−1p = ss−1. Then
p·s·(s−1 ·p−1 ·pp−1t) (cid:54) t
andsos (cid:54) t.
S
(c) SinceN isfull,ss−1,s−1s ∈ N ands = ss−1 ·s·s−1s (cid:54) t.
(d) Supposethata,b ∈ N witha·s·b (cid:54) e. Thereforea·s·b = f (cid:54) ewith
f ∈ E(S),andthens = a−1a·s·bb−1 = a−1fb−1 ∈ N. Conversely,
ifn ∈ N thennn−1 ·n·n−1 = nn−1 andson (cid:54) nn−1.
N
(e) Wehavea,b ∈ N witha·s·b (cid:54) s2 andsos (cid:54) a−1s2b−1. Therefore
s = ss−1a−1s2b−1 anditfollowsthat
s−1 = s−1ss−1 = s−1ss−1a−1s2b−1s−1 = (s−1a−1s)(sb−1s−1) ∈ N
andsos ∈ N.
(f) If a · s · b (cid:54) t then s = a−1a · s · bb−1 (cid:54) a−1tb−1 and so st−1 (cid:54)
a−1(tb−1t−1) ∈ N. SinceN isfull,wededucethatst−1 ∈ N.
(g) It is clear that (cid:54) is reflexive. Suppose that s,t,u ∈ S and that
s (cid:54) t (cid:54) u. There exist a,b,p,q ∈ N such that a·s·b (cid:54) t and
N N
p·t·q (cid:54) u. Then(pa)s(bq) (cid:54) u,and(pa)s(bq)isthetraceproduct
(pa)·s·(bq)since
(pa)−1(pa) = a−1p−1pa = a−1tt−1a = a−1a = ss−1,
and aa−1 (cid:54) tt−1. Similarly s−1s = (bq)(bq)−1. Therefore (pa)·s·
(bq) (cid:54) uands (cid:54) u.
N
(cid:3)
Corollary 2.2. The normal inverse subsemigroup N is determined by the
preorder (cid:54) , and we obtain an order-preserving embedding of the poset
N
of normal inverse subsemigroups of S into the poset of preorders on S that
containthenaturalpartialorder.
Proof. Part(d)ofLemma2.1showsthat
N = {s ∈ S : thereexists e ∈ E(S) with s (cid:54) e}.
N
(cid:3)
Remark 2.3. Not every preorder containing the natural partial order arises
fromanormalinversesubsemigroup. Considerthesymmetricinversemonoid
I and define a preorder (cid:22) by α (cid:22) β ⇐⇒ d(α) ⊆ dβ where d(γ) is the
n
domain of γ ∈ I . Then α (cid:22) id for all α ∈ I , and so the normal inverse
n n
subsemigroup associated to (cid:22) is I itself, but (cid:22) is not the J–preorder on
n
I andsoisnotequalto(cid:54) .
n In
Wedenotethe(cid:39)–classofs ∈ S by[s] .
N
5
Proposition2.4.
(a) Ifn ∈ N thennn−1 (cid:39) n (cid:39) n−1 (cid:39) n−1n,
N N N
(b) Theequivalencerelation(cid:39) saturatesN,
N
(c) Theequivalencerelation(cid:39) determinesN as
N
(cid:91)
N = [e] .
N
e∈E(S)
(d) Ifs (cid:39) tthenss−1 (cid:39) tt−1,s−1s (cid:39) t−1t,ands−1 (cid:39) t−1.
N N N N
(e) The restriction of (cid:39) to E(S) coincides with Green’s J–relation
N
J inducedonE(S) = E(N),
N
(f) In the case N = S the relation (cid:39) coincides with Green’s J–
S
relationonS,
(g) InthecaseN = E(S)therelation(cid:39) isthetrivialrelationonS.
E(S)
Proof.
(a) If n ∈ N then n−1 ·n·n−1 = n−1 and n·n−1 ·n = n, and hence
n (cid:39) n−1. Similarlynn−1·n·n−1 = nn−1 andnn−1·nn−1·n = n,
N
whencen (cid:39) nn−1.
N
(b) Suppose that s ∈ S and that for some n ∈ N we have s (cid:39) n. By
N
part (a) we may assume that n ∈ E(S): then there exist p,q ∈ N
suchthatp·s·q (cid:54) n. Henceforsomee ∈ E(S)wehavep·s·q = e
andsos = p−1 ·e·q−1 = p−1q−1 ∈ N.
(c) Thisfollowsfromparts(a)and(b).
(d) Suppose that s (cid:39) t, with a,b,c,d ∈ N as in (2.2). Then bb−1 =
N
s−1s, b−1b (cid:54) t−1t,dd−1 = t−1t,d−1d (cid:54) s−1s and so s−1s (cid:39) t−1t.
N
Similarlyss−1 (cid:39) tt−1. Sinceb−1·s−1·a−1 (cid:54) t−1andd−1·t−1·c−1 (cid:54)
s−1,wealsohaves−1 (cid:39) t−1.
(e) Ife (cid:39) f thenthereexista,b,p,q ∈ N witha·e·b (cid:54) f andp·f·q (cid:54)
N
e. Therefore we have a−1a = e and aa−1 (cid:54) f, and so e (cid:54) f. By
JN
symmetry f (cid:54) e and so e J f. Conversely, if e J f there
JN N N
exist m,n ∈ N with mm−1 (cid:54) f and m−1m = e,nn−1 (cid:54) e and
n−1n = f. Thenm·e·m−1 (cid:54) f andn·f ·n−1 (cid:54) e,andsoe (cid:39) f.
N
(f) Thisfollowsfrompart(b)ofLemma2.1.
(g) By Lemma 2.1(a), (cid:54) is the natural partial order , which is of
E(S)
courseanti-symmetric.
(cid:3)
However,(cid:39) neednotbeacongruenceonS.
N
Example2.5.
6
(a) InthesymmetricinversemonoidI ,letf : {1} → {2},letS bethe
4
inversesubsemigroup
S = {id ,id ,id ,f,f−1,0}
{1,3} {1} {2}
of I , and let N = S. Then id = ff−1 (cid:39) f−1f = id . But
4 {1} S {2}
id id = id is not (cid:39) –related to id id = 0. In this
{1,3} {1} {1} S {1,3} {2}
example,theposetofJ–classesisjustathree-elementchainandso
isasemilattice.
(b) Nowletg : {3} → {4}inI andletT betheinversesubsemigroup
4
ofI generatedby{id ,id ,f,g}. HeretheJ–classesdonot
4 {1,3} {2,4}
form a semilattice, and so (cid:39) is not a congruence, and the quotient
T
T/ (cid:39) isnotaninversesemigroup: itistheposet
T
Following the notation in [1], we shall denote the quotient S/ (cid:39) by S//N
N
and let π : S → S//N be the quotient map. Our next result sets out the
ordered groupoid structure on S//N: it is a special case of [1, Theorem
3.14],butthedescriptionissimplerforquotientsofinversesemigroupsand
seemsworthstatingindetail.
Theorem 2.6. For any inverse semigroup S and normal inverse subsemi-
group N, the quotient set S//N is an ordered groupoid, with the following
structure:
(a) the identities are the classes [e] where e ∈ E(S), and a class [s]
N N
hasdomain[ss−1] ,range[s−1s] ,andinverse[s−1] .
N N N
(b) asaposet,E(S//N)isisomorphictoN/J .
N
(c) If s,t ∈ S and s−1s (cid:39) tt−1, then there exists a ∈ N with aa−1 (cid:54)
N
s−1s and a−1a = tt−1: the composition of [s] and [t] is then
N N
definedas[sat] .
N
(d) Theordering(cid:54) of(cid:39) –classesisgivenby
N N
[s] (cid:54) [t] ⇐⇒ thereexist a,b ∈ N suchthat a·s·b (cid:54) t.
N N N
Proof. Itfollowsfrompart(d)ofProposition2.4thatthedomainandrange
of[s] anditsinverse[s]−1 arewell-defined.
N N
Supposethats,t ∈ S ands−1s (cid:39) tt−1 buthatwechooseanotherelement
N
z ∈ N with zz−1 (cid:54) s−1s and z−1z = tt−1. Then az−1z = att−1 =
7
aa−1a = aands−1sz = (s−1s)(zz−1)z = zz−1z = z. Hence
sat = saz−1zt = saz−1s−1szt = (saz−1s−1)·szt (cid:39) szt
N
andsoforfixeds,tthe(cid:39) –classoftheelementsatdoesnotdependonthe
N
choice of a. We denote this class by s (cid:71) t. Suppose that s (cid:39) s and that
N 1
we choose a ∈ N to form s (cid:71) t = [s a t] . There exist u,v ∈ N with
1 1 1 1 N
u·s ·v (cid:54) s,andsovv−1 = s−1s (cid:62) a a−1. Then
1 1 1 1 1
(v−1a )(v−1a )−1 = v−1a a−1v (cid:54) v−1v (cid:54) s−1s
1 1 1 1
and
(v−1a )−1(v−1a ) = a−1vv−1a = a−1a = tt−1.
1 1 1 1 1 1
Thereforewecanusetheelementv−1a toformtheclasss (cid:71) t = [sv−1a t] .
1 1 N
Now sv−1s−1 = (sv−1v)(v−1s−1 = (us v)(v−1s−1 = u and since s−1s (cid:62)
1 1 1 1
v−1v,wehave
sv−1a t = sv−1vv−1a t = sv−1s−1s a t = u·(s at) (cid:39) s a t
1 1 1 1 1 1 N 1 1
and so s (cid:71) t (cid:39) s (cid:71) t. Similarly, s (cid:71) t does not depend on the choice of
N 1
the element t within its (cid:39) – class, and the product [s] ·[t] = [sat] is
N N N N
well-defined.
Nowtherelations−1s (cid:39) tt−1 furnishesnotonlya ∈ N withaa−1 (cid:54) s−1s
N
and a−1a = tt−1 but also p ∈ N with pp−1 (cid:54) s−1s and pp−1 = tt−1.
Considersaps−1 ∈ N: wehave
(saps−1)(saps−1)−1 = saps−1sp−1a−1s−1 (cid:54) sapp−1a−1s−1 = (sat)(sat)−1
and
(saps−1)−1(saps−1) = sp−1a−1s−1saps−1
= sp−1a−1aps−1
= sp−1ps−1 = ss−1
and, since (sat)(sat)−1 (cid:54) ss−1, we have ss−1 (cid:39) (sat)(sat)−1. There-
N
fore, [sat] has domain [ss−1] and range [t−1t] , and the composition
N N N
[s] ·[t] = [sat] doesgiveagroupoidstructureonS//N.
N N N
NowbyLemma2.1,(cid:54) inducesthegivenpartialorderonthe(cid:39) –classes,
N N
and it remains to show that this partial order makes S//N into an ordered
groupoid.
Now if a·s·b (cid:54) t then b−1 ·s−1 ·a−1 (cid:54) t−1 and so [s] (cid:54) [t] implies
N N N
that [s]−1 (cid:54) [t]−1. Now suppose that [s ] (cid:54) [s] ,[t ] (cid:54) [t] and that
N n N 1 N N 1 N N
the compositions [s] ·[t] and [s ]·[t ] exist. There exist m,n,u,v ∈ N
N N 1 1
8
such that m·s ·m (cid:54) s and u·t ·v (cid:54) t: since [m·s ·n] = [s ] and
1 1 1 N 1 N
[u·t ·v] = [t ] wemayaswellassumethats (cid:54) sandt (cid:54) t.
1 N 1 N 1 1
We now have a,p ∈ N with aa−1 (cid:54) s−1s,a−1a = tt−1 and b,q ∈ N with
bb−1 (cid:54) s−1s ,b−1b = t t−1,qq−1 (cid:54) t t−1 andq−1q = s−1s . Now
1 1 1 1 1 1 1 1
s bt (cid:39) sab−1s−1 ·s bt = sab−1s−1s bt (cid:54) sat (cid:54) sat
1 1 N 1 1 1 1 1 1
andso[s ] ·[t ] = [s bt ] (cid:54) [sat] = [s] ·[b] .
1 N 1 N 1 1 N N N N
Finally we suppose that [n] (cid:54) [ss−1] for some n ∈ N and s ∈ S. By
N N N
part (a) of Proposition 2.4 we may replace n bye = nn−1. Then there exist
a,b ∈ N with a · e · b (cid:54) ss−1 and so aa−1 (cid:54) ss−1 and a−1a = e. Then
the class [aa−1s] has domain [aa−1] = [e] and [aa−1s] (cid:54) [s] . We
N N N N N N
wishtoshowthat[aa−1s] istheunique(cid:39) –classwiththeseproperties.
N N
Suppose that [k] (cid:54) [s] and that [kk−1] = [e] . There exist u,v ∈ N
N N N N N
with u · k · v (cid:54) s and b,q ∈ N with bb−1 (cid:54) kk−1,b−1b = e,qq−1 (cid:54) e,
and q−1q = kk−1. We have a ∈ N as in the previous paragraph. Then
kvs−1 = u−1 andu−1uq−1 = q−1: hence
k (cid:39) k ·vs−1uq−1a−1s = q−1a−1s = q−1a−1aa−1s
N
= q−1a−1 ·aqq−1a−1s (cid:39) = aqq−1a−1s (cid:54) aa−1s
N
andso[k] (cid:54) [aa−1s] . Bysymmetry,theyareequal. (cid:3)
N N N
Corollary 2.7. [1, Theorem 4.15] Given a homomorphism φ : S → Σ of
inverse semigroups, let K = {s ∈ S : xφ ∈ E(Σ)}. Then K is a normal
inversesubsemigroupofS,andφfactorisesasacomposition
π κ
S −→ S//K −→ Σ
whereκ,definedby[s] κ = sφ,isastar-injectivefunctor.
K
Proof. The map κ is well-defined, since if [s] = [s(cid:48)] then, for some
K K
a,b ∈ K we have a · s · b (cid:54) s(cid:48). Now aφ = (a−1a)φ = (ss−1)φ and
similarly, bφ = (s−1s)φ). It follows that (a · s · b)φ = sφ (cid:54) s(cid:48)φ. By
symmetry,s(cid:48)φ (cid:54) sφ.
To show that κ is a functor, suppose that [s] and [t] are composable in
K K
S//K. Thenthereexista,p ∈ K with
aa−1 (cid:54) s−1s,a−1a = tt−1,pp−1 (cid:54) tt−1,p−1p = s−1s
9
and the composition of [s] and [t] is defined, as in Theorem 2.6, by
K K
[s] ·[t] = [sat] . Then
K K K
([sat] )κ = (sat)φ = (sφ)(aφ)(tφ)
K
= (sφ)(a−1a)φ(tφ) (since a ∈ K)
= (sφ)(tt−1)φ(tφ) = (sφ)(tφ)
Now(sφ)(tφ)isatraceproduct(sφ)·(tφ)since
(sφ)−1(sφ) = (s−1s)φ = (p−1p)φ
= (pp−1)φ (since p ∈ K)
(cid:54) (tt−1)φ = (tφ)(tφ)−1
andsimilarly
(tφ)(tφ)−1 = (tt−1)φ = (a−1a)φ
= (aa−1)φ (since a ∈ K)
(cid:54) (s−1s)φ = (sφ)−1(sφ).
Therefore (sφ)(tφ) is a trace product defined in the inductive groupoid
(Σ,·)andκisafunctor.
To show that κ is star-injective, suppose that for some u,v ∈ S we have
[uu−1] = [vv−1] and uφ = vφ. We claim that u (cid:39) v. By symmetry, it
K K K
issufficienttoshowthatu (cid:54) v. Nowbypart(e)Proposition2.4thereexist
K
a,b ∈ K with aa−1 (cid:54) uu−1,a−1a = vv−1,bb−1 (cid:54) vv−1 and b−1b = uu−1.
Then
b·u·u−1b−1v = (buu−1b−1)v (cid:54) v
and
(u−1b−1v)φ = (u−1)φ(b−1b)φvφ (sinceb ∈ K)
= (u−1b−1b)φvφ
= (u−1φ)vφ = (uφ)−1vφ ∈ E(Σ) sinceuφ = vφ,
andsou−1b−1v ∈ K andu (cid:54) v asrequired. (cid:3)
k
Corollary 2.8. The factorization of φ : S → Σ is unique, in the sense that
ν
if φ also factorizes as S → S//N → Σ with ν a star-injective functor, then
N = K (andhenceν = κ.)
Proof. If n ∈ N then by part (a) of Proposition 2.4, we have n (cid:39) nn−1
N
andsonφ ∈ E(Σ). HenceN ⊆ K.
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