Title: Continuum semigroups with midunit
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Title: Continuum semigroups with midunit
Physical Description: vii, 83 leaves. : illus. ; 28 cm.
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Creator: Hu, Hung-Tzaw, 1941-
Publication Date: 1974
Copyright Date: 1974
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CONTINUUM SEMIGROUPS WITH MIDUNIT


by

HUNG-TZAW HU











A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE
UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY










UNIVERSITY OF FLORIDA

1 ,'4































to

My Deai Parent5

with Love













ACKNOWLEDGEMEN PS


The author wishes to express his indebtedness and sincere

gratitude to Professor Alexander D. Wallace. chairman of the

supervisory committee prior to his retirement in June 1973,

for sharing his wisdom, great experience and philosophy as

well as his professional guidance throughout the primary pre-

paration of this work. He would like to express his apprecia-

tion to Professor Kermit N. Sigmon for agreeing to chair his

supervisory committee since September 1973 after his return

from Germany. Words can hardly express the depth and sin-

cerity of the author's thanks to Professor Sigmon not only for

his kind guidance and i '-rest in developing the writing

ability of the author but also for his critical comments,

valuable suggestion, and new ideas which made this work to be

done in a possible frank and perfect form. The author also

takes this opportunity to extend hic sincere gratitude to

Professor P. Bacon and David Stadtlander for their introduction

of the theory of cohomology and the theory of compact semi-

groups, which are essential for this work. He also wishes to

thank Professors Thomas Bowman, J. A. Draper, David Stadtlander,

E. E. Shult, and Mark Yang for serving on his supervisory commmit-

tee. A more profound acknowledgement is due to Professor

Thomas Bowman for his useful suggestiorsand encouragement








during the preparation of this work. Finally he extends his

thanks to Professor A. R. Bodnarok. Chairman of the Department

of Mathematics, for his financial support during the past

years.















TABLE OF CONTENTS

Page

ACKNOWLEDGEMENTS. ........................................ iii

ABSTRACT...................................................vi

INTRODUCTION.................................................. 1

CHAPTERS

I PRELIMINARIES.....................................5

II BASIC PROPERTIES OF MIDUNITS......... ..........10

III THE p-CLASS OF MIDUNITS IN A COMPACT SEMIGROUP..15

IV FINITE SEMIGROUPS WITH MIDUNIT..................20

V THE CODIMENSION OF THE y-CLASS OF MIDUNITS IN
A CONTINUUM SEMIGROUP...............................28

VI THE COMPLEMENT OF THE p-CLASS OF MIDUNITS AND
MINIMAL CONTINUUM SEMIGROUPS WITH MIDUNIT........34

VII A METHOD OF CONSTRUCTING SEMIGROUPS WITH
MIDUNIT................ ........ .............39

VIII SEMIGROUPS ON D-SPACES..........................44

IX EXAMPLES OF SEMGROUPS ADMITTING NO PROPER
MIDUNIT............. ... ........................ 54

X SEMIGROUPS WITH MIDUNIT ON THE TWO-CELL......... 64

XI MARGINALITY OF THE MIDUNITS IN A CONTINUUM
SEMIGROUP........................................ 71

XII ACYCLICITY OF CONTINUT SEMIGROUPS WITH
MIDUNIT. .........................................74

REFERENCES .................... ............................81

BIOGRAPHICAL SKETCH..........................................83







Abstract of Dissertation Presented to the
Gilaiduite Council of the University of Florida in
Pi1 (ial Fulfillment of the Requirements for the
Degree of Doctor of Philosophy

CONTINUUM SEMIGROUPS WITH MIDUNIT

By

Hung-tzaw Hu

August, 1974

Chairman: K. N. Sigmon
Major Department: Department of Mathematics

A continuum semigroup with midunit v is defined as a

compact connected semigroup satisfying avb = ab for all a and

b. It is the purpose of this dissertation to investigate

continuum semigroupswith midunit. If S is a compact semigroup

with midunit satisfying S = S and J is the p-class of the

midunits, then J is the unique maximal p-class in S and is a

completely simple subsemigroup whose idempotents are the set

of all midunits and form a usual rectangular band. There

exists an example of a semigroup with five elements which
2
satisfies the condition S = S and has a unique maximal

p-class, but S does not have a midunit. The cohomology used

is that of Alexander-Wallace-Spanier with coefficient group

arbitrary unless otherwise specified. It is shown that if S

is a continuum semigroup of finite codimension, then

cd(H ) < cd(S) for all a in S such that Ha 4 K. It follows

that the p-class of the midunits in a continuum semigroup of
2
codimension one satisfying S = S / K is totally disconnected.

There is a method of constructing a new semigroup with mid-

unit from a given known semigroup, which is used to prove

that if S is a compact semigroup, v is a p-maximal idempotent,








J is the 5-class of v and vJv c J, then J is homeomorphic to

a cartesian product of a rectangular hand with H the maxi-

mal subgroup which contains v. A midunit is called proper

if it is neither a left nor right identity. A certain class

of spaces is given which, due to their topological structure,

do not admit a semigroup with proper midunit under the con-
2
edition S = S / K. Let S be a semigroup on the two-cell with
2
midunit and zero satisfying S = S, M be the set of all mid-

units, and H be the maximal subgroup containing the midunit

v. If either H (H ) 4 0 or H1(M) 0, then all midunits lie
v
on the boundary of the two-cell and are either left or right

identities for S. If S is a continuum semigroup with midunit

v satifying S2 = S / K and H "nd [(Sv U vS)\vSv]* are dis-

joint then each point of H is marginal in S. An example is

given of a continuum semigroup with zero and midunit satisfy-

ing S2 = S which is not acyclic. It is shown, however, that

if S is a continuously factorable continuum semigroup with

zero and midunit, then S is acyclic. By a continuously

factorable semigroup is meant a semigroup (S;m) for which

there exists a map r from S ont- S x S such that mor = IS.













INTRODUCTION


The purpose of this chapter is to introduce the motiva-

tion for this work and to give a summary and key results

obtained in this investigation as well as some unsolved prob-

lems. The author was attracted to the idea of midunit when

he presented a report on an example concerning midunit from

the dissertation of McCharen [11] in the seminar of Professor

A. D. Wallace at the University of Florida two years ago.

The underlying space of this example consists of countable

many tangent circles converging to a point. He proved that

if it admits the structure of a semigroup S satisfying
2
S = S, then S = K. His proof used a method which revealed

the wonder and attraction of the notion of midunit both in

the algebraic and topological sense. The idea of midunit

was first introduced by Yamada in his paper "A note on middle

unitary semigroups" nineteen years ago. Recently Ault has

published a paper in this area in which she found that every

(regular) semigroup can be embedded in a single (regular)

semigroup with midunits which are neither left nor right

identities [1].

After giving topological and algebraic preliminaries in

Chapter I, we introduce the main notion of this work, which

is midunit, and discuss some basic properties of it in

Chapter II. Among these basic properties of midunits we







describe the structure of the set of idempotent midunits

and the union of all mnyiimcti groups which contain a midunit

as identity, as well as the interrelation between them.

Chapter III is mainly an investigation of the structure

and interrelation of J and M in a compact semigroup S with
2
midunit satisfying S = S. where J is the p-class of the

midunits and M is the set of all midunits. In this direc-

tion we prove that J is the unique maximal ,-class in S,

which is a completely simple semigroup whose idempotents

are midunits and form a rectangular band. Professor Sigmon

raised the question of whether a compact semigroup S satisfying
2
S = S, with a unique maximal p-class forming a completely

simple semigroup must contain a midunit. The answer to this

question is given in Chapter IV. More precisely, we prove

that if S contains at most four elements and satisfies
2
S = S, then any idempotent v satisfying SvS = S must be a

midunit. On the other hand we give an example of a semigroup
2
with five elements which satisfies the condition S = S and

has a unique maximal g-class which is a completely simple

subsemigroup, but S does not have a midunit. As a continua-

tion of Chapter III, we prove in Chapter V that the p-class

J of the midunits in a continuum semigroup of codimension
2
one satisfying S = S / K is totally disconnected. In the

process of proving this, we present a rather short proof of

an interesting result; namely, that if S is a continuum

semigroup with finite codimension, then cd(Ha) < cd(S) for

all a in S such that H / K.
a







The structure of the complement of the p-class of

midunits in a continuum semigroup satisfying S = S I K

will be studied in Chapter VI. Also as a generalization of

the notion of irreducible clan, we introduce and investigate

the structure of minimal continuum semigroups with midunits

in the latter part of Chapter VI. In Chapter VII we study

in detail a method of constructing new semigroups with mid-

units from given known semigroups. The importance of this

method is that it could transfer problems in general semi-

groups into problems of semigroups with midunits and vice-

versa. By using this method and results of preceding chapters

it is not difficult to prove that if S is'a compact semi-

group, v is a p-maximal idempotent, J is the p-class of v,

and vJv C J, then J is horeomorphic to a cartesian product

of a rectangular band with H the maximal subgroup which

contains v. Furthermore, if cdS = 1 and S / K, then J is

totally disconnected.

In Chapter VIII and Chapter IX we discuss those semi-

groups which, due to their topological structure, do not
2
admit a proper midunit under the condition S = S / K, In

this class of spaces there are D-spaces, a circle with a

closed interval issuing from a point on the circle, etc.

Chapter X is devoted to exploring those semigroups S on the
2
two-cell with midunit and zero satisfying S = S. The

cohomology theory used is that of Alexander-Wallace-Spanier

with arbitrary coefficient group unless otherwise specified.

Let S be a semigroup on the two-cell with midunit and zero

satisfying S2 = S, M be the set of all midunits, and Hv be






the maxima] group containing the midunit v. If either

II (H l) J 0 or II (M) V 0. then every midunit must be either

left or right identity for S.

The lasi two chapters deal with cohomological results
2
in continuum semigroups with midunit satisfying S = S K

which are analogous to those holding for continuum semi-

groups with identity. The problem of whether a midunit in

a continuum semigroup satisfying S2 = S 1 K must be marginal

remains open. We prove however, in Chapter XI that if
*
H n [(Sv U vS)\vSv] = E, where H is the maximal group

containing v, then each point of H in marginal in S. In

Chapter XI we are concerned with the acyclicity of a con-

tinuum semigroup with midunit. An example is given of

a continuum semigroup with zero and midunit satisfying

S = S which is not acyclic. We introduce, however, the

notion of a continuously factorable semigroup and prove that

if S is a continuously factorable continuum semigroup with

zero and midunit, then S is acyclic. We end this introduc-

tion with an unsolved problem: Must a continuously factor-

able continuum semigroup with zero be acyclic?














CHAP'r'IT I


PRELIMINARIES


Topological Preliminaries. Throughout this work a space will

always be a Hausdorff topological space. If A and B are

subsets of a space X, then the complement of B in A will be

denoted by A\B. The empty set is denoted by C. If A is a

subset of the space X, then the closure of A in X will be

denoted by A the interior by A, and the boundary (A A )
o
by F(A). A is said to be nowhere dense if (A ) = ]. A

continuum is a compact connected Hausdorff space. The terms

map and continuous function are synonomous. If a map is

one-to-one, onto and open it is a homeomorphism. The term

iff will mean "if and only if".

Let X and Y be spaces. A relation from X to Y is a

subset of the cartesian product Y x Y. A relation on X is a

subset of X x X. If F is a closed subset of X x Y, we call

F a closed relation from X to Y. If r is a relation on X,

then r-1 = {(y,x) (x,y) e T). The diagonal of X x X will

be denoted by AX; that is, AX = ((x,x) x e X). The relation
-l
r on X is reflexive if Ax C r, symmetric if r = anti-

symmetric if r n F 1C A, and transitive if (x,y) e F and

(y,z) e r imply (x,z) e F. A quasi-order is a reflexive

transitive relation. A partial order is an anti-symmetric

quasi-order. An equivalence relation is a symmetric







quasi-order. For an equivalence relation i on a space X,

X denotes the set of equiv ience classes of T. If X is

compact and F is closed. X/ enrlc';nd with the quotient

topology is a space. If A i' a closed subset of the space,

we may define F on X to be the set of all ordered pairs (x,y)

in X x X such that either x = y or x and y are elements of
x x
A. Then the space X/ is written X .

If r is a relation on X and b is a point of X, then

L(b) = (b) U [x (y.b) e F], M(b) = [b] U [y (b,y) e T], and

Lb = L(b) n M(b). If A is a subset of X, then L(A) denotes

the union of all sets L(a) for which a e A. It is easily

verified that if X is compact and F is closed, L(A) is closed

for each closed subset A of X.

The cohomology theory used is that Alexander-Wallace-

Spanier with arbitrary coefficient oroup unless otherwise

specified. Throughout this work we shall use reduced

groups in dimension 0. On occasion we shall require cer-

tain standard results concerning the cohomology of compact

spaces as follows [13].

1-1 Mayer-Vietoris Sequence Theorem. If for each quadruple

(X;A,B;C), where X is fully normal, A and B are closed subsets
*-1
of X with X = A U B, and C C A N B, we set A = j k 5

where 6: HP-(A n B,C) HI(AA n B), k: (A,A n B) C (X,B),

and j: (X,C) C (X,B), then the following sequence
'*
HP(A n B,C) A HP(X,C) t Xt HP(AC) X HP(B,C) i

Hp(A n B,C) is exact, where i, i', t, and t' are inclusion

maps.







In particular, this holds whi,, X is compact, since a

compact space is illyy r1!mnl.

Let X be a space and hr c' (X). If A C X. then hJA

denotes the image of h under the natural homomorphism

H (X)- ip (A).

1-2 Reduction Theorem. Suppose A is closed in the fully

normal space X and h e HP(X). If hIA = 0, then there is an

open set M containing A such that hjM = 0.

Let X be a space and A be a subset of X. Let i: A X

be the inclusion map and h e HP(A). Then h is extendible to

HP(X) if h Im(i ). A closed subset R of X is a roof for h

if h is not extendible to HP(R U A) while R is extendible to

HP(R' U A) for any proper closed subset R' of R.

1-3 Roof Theorem. Suppose X is a compact space, A is a

closed subset of X and h E HP(A).

(1) If h is not extendible to HP(X), then h has a roof.

(2) If R is a roof for h and ho = hR n A, then h / 0,

R is a roof for ho, R = (R\A) and R\A is connected.

Two maps f,g: X Y are weakly homotopic if there is

a continuum T, a map F: X x T Y and a pair of elements t
o
and tl of T such that, for all x of X, f(x) = F(x,t ) and

g(x) = F(x,tl) [14].

1-4 Homotopy Theorem. If X is a compact space and if

f,g: X Y are weakly homotopic maps, then

f g*: H(Y) HP(X) for each p.

Topological Algebraic Preliminaries. Throughout this work

a topological semigroup (S;m) will be a space S endowed

with a continuous, association multiplication m. If no








confusion seems likely, we simply let xy denote m(x.y) and

AB = fxylx E A and y f B1. where A ani1 B are subsets of S.
2
An idempotent is an idempi'ont e such that e = e. The

set of idempotents will be denoted by E. Each idempotent

e belongs to a unique maximal subgroup of S, denoted by He,

and distinct idempotents have disjoint maximal subgroups.

A nonempty set A is a left (right. two-sided) ideal of S

provided SA C A (AS C A, AS U SA C A). It is well known

that a compact semigroup S has a unique minimal two-sided

ideal, denoted by K, which is known to be a retract and is

a union of maximal groups [17]. Let S be a semigroup. An

element v is a left (right) identity if va = a (av = a) for

all a in S; v is an identity if it is both a left and right

identity ofS [2]. A clan is a continuum semigroup with identity.

A clan S with identity v is called irreducible if it does not

contain a proper subcontinuum semigroup T containing v with

T n K / 3. For more information about irreducible clans

the reader is referred to [6] and [15].

The following is a result due to Wallace [16] which

will be of importance in the sequel.

1-5 Swelling Lemma. Let S be a compact semigroup and A a

closed subset of S. If A C Ax for some x e S, then A = Ax.

If S is compact, then maximal elementsexist and each ele-

ment of S is below a maximal element. Also if S is compact

and a is p-maximal, then a is I-maximal as well as R-maximal

[11]. The following theorem, due to McCharen [11], concern-

ing maximal elements in a compact semigroup will be important

in the sequel.







1 'i' m. If S is a compact semigroup satisfying

S S and b is a ,-maximal element of S, then b E ESE.

A point p in a continuum X is a weak cut point between

a and b if a and b are points in X different from p, and any

subcontinuum of X containing a and b also contains p. The

point p is simply a weak cut point if there exists a and b

such that p is a weak cut point between a and b [21].

1-7 Theorem. Let S be a continuum semigroup satisfying

S2 = S. If e is a p-maximal idempotent of S which is a weak

cut point of S, then S = K.

A point p in a continuum X is a local separating point

of X if p separates the component containing p of some closed

neighborhood of p [21].
2
1-8 Theorem. Let S be a continuum semigroup with S = S,

a an f-maximal element of S, and a E SE. If a is a local

separating point of S, then a E K.

It is true that if S is compact semigroup and a is

p-maximal then a is -maximal. It follows from Theorem 1-6.
2
that if S is a continuum semigroup satisfying S = S ; K,

and b is a p-maximal element of S, then b is not a local

separating point of S.

Also under some conditions on S, a p-maximal idempotent

cannot have a two-dimensional Euclidean neighborhood as

follows.

1-9 Theorem. Let S be a continuum semigroup satisfying

S = ESE, and suppose e is a p-maximal idempotent element of S.

If S is a subset of the complex plane C, S f K, then e lies

on the boundary of S in E.













CHAPTER II


BASIC PROPERTIES OF MIDUNITS


It is our purpose in this chapter to introduce the main

notion of this work, which is midunit, and discuss some

basic properties of it. In particular, we study the structure

of the set of all idempotent midunits and the union of all

maximal subgroups which contain an idempotent midunit as an

identity. A point v of a semigroup S is called a midunit

if avb = ab for any pair of elements a and b of S. It is

evident that a one-sided identity is a midunit, while a

midunit may not even be an idempotent. A midunit v of a

semigroup S is proper if it is neither a left identity

nor a right identity of S. We begin with some examples of

semigroups in which a midunit could exist without being a

left identity, right identity, or even an idempotent.

2-1 Example. Let S be topologically the closed unit interval

[0,1]. We define two multiplications on S as follows:

(1) Define ab = 0 for any pair of elements a and b of

S. Then every point of S is a midunit and 0 is

the only idempotent midunit of S.

(2) Define ab = minimum (-,a,b) for any pair of elements

a and b of S. Then every point of [-,1] is a mid-
1unit of S but onl is an idempotent midunit.
unit of S but only I is an idempotent midunit.







Besides showing the existence of proper midunits these

xiil'l, s also show than a midunit can occur more pathologi-

c.'lly thin a one-side1 ide'it.ity, rot only in an algebraic

but clso in a topological sense.

T'li following is a lemma concerning idempotent midunits

which is also known to Janet E. Ault [1].

2-2 Lemma. Let S be a semigroup with midunit. If a and b

are midunits of S, then ab is an idempotent midunit.

2-3 Lemma. Let S be a semigroup with midunit b. If for

some point d, bd is a midunit of S, then d is a midunit and

bd is an idempotent.

Proof. Let x and y be a pair of elements of S. Then

xdy = xb(dy) = x(bd)y = xy, i.e., d is a midunit of S. By

Lemma 2-2 we know that bd is an idempotent.

Let S and S be two semigroups. A mapping of S into

S is called a homomorphism if *,(ab) = )(a) (b) for all a,

b in S. A one-to-one homomorphism ( of S onto S is called

an isomorphism of S onto S. If v is a midunit of S and 4 is

a homomorphismfrom S onto S, then 4(v) is a midunit of S.

A semigroup is called a band if all points of S are idempo-

tent. A semigroup S is called a rectangular band if S is

isomorphic to L x R, where L is a semigroup with left zero

multiplication, R is a semigroup with right zero multipli-

cation and L x P is a semigroup with coordinatewise

multiplication.

2-4 Proposition. Let S be a semigroup with midunit and M

be the set of all idempotent midunits of S. If v is an ele-

ment of M then






(1) Mv is a subsemigroup with left zero multiplication

and vM is a subsemigroup with right zero multipli-

cation.

(2) Let Mv x vM have coordinatewise multiplication, i.e.,

(u,w)(u,w) = (uu,ww), and r: Mv x vM M be defined

by f((u,w)) = uw. Then < is an isomorphism. Hence,

M is a rectangular band.

Proof. (1) From Lemma 2-2 we know that M is not empty.

Let u and w be two elements of Mv. Then uw = (uv)wv =
2
(uv)v = uv = u, i.e., Mv is a subsemigroup with left zero

multiplication. Similarly we could prove that vM is a sub-

semigroup with right zero multiplication.

(2) To prove that ( is a homomorphism, we let (u,w) and

(u,w) be two elements of My x vM. Then

((u,w)) ((u,w)) )= (uw)(uw) = (uw)w = uw, and

((u,w)(u,w)) = 4((uu,ww)) = t((u,w)) = uw.

Then is a homomorphism. To prove that $ is a one-to-one

mapping, we let (u,w), (u,w) be any two elements of Mv x vM

such that r((u,w)) = 1((u,w)), i.e., uw = uw. Then

uwv = uv = u and uwv = uv = u, but uwv = uwv hence u = u.

Similarly we can prove that w = w. It is easy to see that

' is an onto mapping, since any idempotent u can be written
2
as u = u = uvvu = ((uv,vw)). The proof is complete.

Recall that if S is a semigroup and b an element of S,

the smallest left (right, two-sided) ideal containing b is

denoted by L(b), (R(b), J(b)), and the M-class which contains

b is denoted by Hb. If b is an idempotent then Hb is the

maximal subgroup of S which contains b.






2-5 Lcin ma. L,' S be a s *miroup) with units. Tf u and v

are two id impotcl nt nidir its of S. then 11i II '

Proof. Lct' x I I ad \ H wo wait tC 1c p Cve I'
'I V

uv
xv IIU V. By the dc finittion of 11-class. w- .'.Te

xS = uS. Sx = Su, yS = 'e :nd Iy Sv.

Then L(xy) = xy Sxy = uxy U Sxy = Sxv = Suy = Sv = Sv = Suv

= L(uv), and similarly R(xy) = R(uv).

This implies that xy e II

If S is a semigroup without zero, then S is simple if

S does not properly contain an i'kal. Further S is

completelysiLple if S is simple and contains a primitive

idempotent. An idrepotcnt e is called Tprimi tive if
2
ef = fe = f, f = f implies e f 21.

2-6 Proposition. Let S be a semigrour with ridunits and M

be the set of all iclrrm tent midunits of S. If N = U H ,
veM
then N is a completely simp1- ''ibsemigroup, in which

M is a subsemigroup. More precisely, let v be a midunit and

consider Mv x Hv < vM with coordinatewise multiplication.

If *: Mv x H X vM N is defined by *(u,x,w) = uxw, then 4

is an isomorphism under which *[Mv x (v] x vM] = M.

Proof. From Lemnra 2-5, we know that N is a subsemi-

group. If P is an ideal of N, then it is evident that

P n M / ]. Let v e P n M, and z be an element in N. Then

z e H for some w e M, i.e., z = wz. Since v is a midunit
w
we have wvz = wz = z, thus z e P. This proves that P = N.

In order to prove that the midunit v is primitive, we let

u E M such that uv = vu = u. Then v = (vu)v = (uv)v = uv = u,







which implies that v is primitive. Hence N is a completely

simple subsemigroup. The proof of 9 is an isomorphism and

([Mv x (v1 x vMJ = M are esscutially the same as what we

did in Proposition 2-4.













CHAPTER III


THE 9-CLASS OF MIDUNITS IN A COMPACT SEMIGROUP


In this chapter we investigate the structure and inter-

relation of J and M in a compact semigroup S with midunit

satisfying S2 = S, where J is the g-class of the midunits

and M is the set of all midunits. We note here that under

the condition S = S all midunits belong to the same g-class

as described in Proposition 3-1. From Theorem 1-6 we have

that if S is a compact semigroup satisfying S = S and z is

a 9-maximal element of S then there exists a pair of idempo-

tents e and f such that z = ezf.

3-1 Proposition. Let S be a compact semigroup with midunit

satisfying S = S. Then

(1) There is cnly one maximal 2-class in S and this

9-class contains all midunits.

(2) All midunits are idempotent.

Proof. Let z be a midunit in S. Then
2
J(z) = z U Sz U zS U SzS 3 SzS = S = S; thus J(z) = S.

This implies that all midunits are in the same 9-class,

which is the only maximal 9-class in S. Let e and f be two

idempotents such that z = ezf. Then z = ez so that by Lemma

2-3 we know that z is an idempotent.





16

3-2 Remark. Without the algebraic condition S2 = S, none of

the results in Proposition 3-1 remain true. Recall in Example
2 1 1
2-1 part (2) that S = [0, ] / S and every point of [ ,1]
1
is a midunit of S while only is an idempotent midunit. On

the other hand, J( ) = [0, ] and J(1) = [O, ] U (1]. Hence
1
midunits 1 and are not even in the same y-class.

The following lemma is due to Koch and Wallace. For

its importance to this work and convenience to the reader,

we repeat its proof [20].

3-3 Lemma. Let S be a compact semigroup. If x, y are two

points in S and L(x) C L(y) C J(x), then L(x) = L(y).

Proof. To avoid the triviality of its proof we may

assume that y / x. y E J(x) implies that y = xtl, y = t2x or

y = t3xt4 for some elements tl,t2,t3 and t4 of S. In case

y = t2x, then L(y) C L(x), by assumption L(x) C L(y), we

have the equality L(x) = L(y). In case y = t3xt4, then

L(y) = t3xt4 U St3xt4; thus L(x) C L(y) C L(x) t4 so that

L(x) = L(y) by the Swelling Lemma 1-5. If y = xtl, the

proof is similar.

The following hypothesis is held fixed throughout.the

remainder of this chapter:
2
S is a compact semigroup with midunit satisfying S = S.

We let J be the y-class of the midunits, M be the set of all

midunits, and H be the H-class of midunit v.

3-4 Proposition. x E J iff SxS = S.

Proof. (=) If x is an element of J, then by Theorem 1-6

there exist a pair of idempotents e and f such that x = exf.







Since x U xS U Sx = cxf U exS U Sxf C SxS, we have

J(x) = x UL xS U Sx i SxS = SxS. Also since x is in the

5-class of midunits, hence J(x) = S, or SxS = S.

(c) If SxS = S, the J(x) = S i.e., x e J, because

there is only one maximal >-class in S, as mentioned in

Proposition 3-1.

3-5 Proposition. If x c J, then L(xv) = L(v), R(vx) = R(v)

and vxv e H for any v e M.

Proof. From Proposition 3-4 we have that SxS = S.

Then SxS = SxvS since v is a midunit; thus SxvS = S, which

implies that xv e J by Proposition 3-4. Hence we have

L(xv) C L(v) C J(xv), v'* ",h implies that L(xv) = L(v) by

Lemma 3-3. Similarly we can prove R(vx) = R(v). Hence it

follows that vxv H .
v
3-6 Proposition. xy s J iff x e J and y e J.

Proof. (e) If x and y are a pair of elements of J and

v is a midunit, then vvv and vyv belong to H by proposition

3-5. From Proposition 3-1 we know that v is an idempotent

so that H is a group. Since H is a group we have that

vxyv = vxvvyv is an element of H On the other hand we have

SxyS = SvxyvS = SvS = S; then xy E J by Proposition 3-4.

(=) It follows easily from Proposition 3-4.

3-7 Proposition. M = J n E.

Proof. Let v be a midunit and u an element of J n E.

It follows from Proposition 3-5 that vuv eH ,I and also
2 2
we have (vuv) = vuvvuv = vuvuv = vu v = vuv, i.e. vuv is an

idempotent in the group H Since there is only one idempo-







tent in a group we have vuv = v. From Lemma 2-3 we know

that vu is a midunit. But since (vu)u = vu, one has that u

is a midunit. This proves that J n E C M. On the other

hand, M C J N E follows from Proposition 3-1.

3-8 Proposition. J is a completely simple subsemigroup.

Proof. Since J is a 9-class in a compact semigroup

hence J is a closed set in S. From Proposition 3-6 we have

that J is a compact subsemigroup. If K(J) is its minimal

ideal and v is an idempotent in K(J), then v is a midunit

from Proposition 3-7. By assumption S = S so for each

element z in J we can write z as z = xy for some elements

x and y in S. Thus

z = xy = xvy E K(J), which implies K(J) = J.

To prove the midunit v is primitive, we let v E J n E,

such that vu = uv = u. Then v = (vu)v = (uv)v = uv = u,

which implies that v is primitive. Hence J is a completely

simple subsemigrouF [5].

We now collect the preceding results of this chapter

into the main theorem of this chapter.

3-9 Theorem. Let S be a compact semigroup with midunit

satisfying S = S. Let J be the g-class of the midunits and

M be the set of all midunits of S. Then

(1) J is the unique maximal 9-class in S and is a

completely simple subsemigroup, whose idempotents

form a rectangular band.

(2) M = J n E.








(3) More precisely, let v be a midunit. Then Mv and

vM are subsemigroups with left zero and right

zero multipliu ticns, respectively, and if

i: Mv H x vM J iq defined by i((u,x,w)) = uxw,

then i is an isomorphism under which the image of

Mv x (v} x vM is M.

Proof. (1) and (2) follow from Proposition 2-4,

Proposition 3-1, and Proposition 3-7. Analogous to the

proof of Proposition 2-4, we can prove that Mv and vM are

subsemigroups with left zero and right zero multiplication,

respectively. From (1) and (2) we have J = U [H lv e M),

so that the proof that is an isomorphism is the same as

that of Proposition 2-6.













CHAPTER IV


FINITE SEMIGROUPS WITH MIDUNIT


In this chapter we study finite semigroups which satisfy
2
S = S with a unique maximal p-class which is a completely

simple subsemigroup. Recall that we proved in Theorem 3-9

that in a compact semigroup S with midunit satisfying S = S,

the p-class of the midunits i' maximal and is a completely

simple subsemigroup. The converse of this'is not true in

general, even on a semigroup containing only five elements.

That is, we will give an example of a semigroup S with five

elements satisfying S2 = S which has a unique completely

simple maximal p-class, but does not have a midunit. Also

we will give an example of a semigroup S with four elements

satisfying S = S and containing a proper midunit. We begin

with a generalized definition of midunit.

4-1 Definition. Let S be a semigroup and A and B be two

nonempty subsets of S. A point v of S is called a midunit

relative to [A,B] if avb = ab for all ordered pairs (a,b) of

A X B.

A midunit, as previously defined, is then a midunit

relative to [S,S].
2
4-2 Proposition. Let S be a semigroup satisfying S = S and

v be an idempotent of S such that SvS = S. Then v is a

midunit of S iff v is a midunit relative to (vS\vSv, Sv\vSv].







Proof. (-) The proof is trivial.

(c) If a and b are two elements of S. then there

exists al.a2, bl and b2 in S such that a = a va and b = blvb2

In case va2 e Sv, i.e., va2v = va2, then

avb = alva2vb = al(va2v)b = al(va2)b = ab.

In case blv e vS, then avb = ab, similarly. Suppose that

neither of these cases occur, i.e., va2 E vS\vSv and

blv e Sv\vSv. Then by assumption that v is a midunit relative

to [vS\vSv, Sv vSv], we have (va2)v(blv) = va2b v; thus

avb = al(va2vb v)b2 = al(va2blv)b2 = ab.

Hence the proof is complete.

4-3 Proposition. Let S be a semigroup satisfying S2 = S and

v be an idempotent of S such that SvS = S. Then the follow-

ing statements hold.

(1) vS\vSv = C iff v is a right identity for S,

(2) Sv\vSv = ] iff v is a left identity for S.

Proof. (1) (c) If v is a right identity, then

vS\vSv = vS\v(Sv) = vS\vS = ]. Similarly if v is a left

identity for S, then Sv\vSv = 0.

(-) If vS\vSv = [, then vS = vSv. Therefore

S = S(vS) = SvSv = Sv; i.e., v is a right identity for S.

The proof of (2) is similar.

If A is a set, then its cardinality will be denoted by

n(A).

4-4 Proposition. Let S be a semigroup satisfying S = S and

v is an idempotent of S such that SvS = S. If S is finite







and v is neither a left nor a right identity of S, then

n(vSv) < n(S) 1.

Proof. Suppose the conclusion is not true and let S

be a finite semigroup with cardinality n(S) = m, and v be an

idempotent such that SvS = S and n(vSv) n(S) 1 = m-1. If

v is neither a left identity nor a right identity we know

that vS\vSv i and Sv\vSv / n by Proposition 4-3. It is

evident that [vS\vSv] n [Sv\vSv] = n. Hence we have

n[vS U Sv] = n(vS\vSv) + n(Sv\vSv) + n(vSv)

1 + 1 + (m-l) = m + 1.

That is a contradiction since vS U Sv is a subset of S.

Therefore

n(vSv) < n(S) 1.

4-5 Proposition. Let S be a finite semigroup satisfying
2
S = S, and v be an idempotent of S such that SvS = S. Then

the following statements hold.

(1) If n(S) = 1,2, or 3, then v is either a left or a

right identity for S.

(2) If n(S) = 4, then v is a midunit for S.

Proof. (1) In the case n(S) = 1, it is evident that

v is a identity for S. In the cases n(S) = 2 and n(S) = 3,

we prove it as follows. If S = K, then there are only three

possible multiplications on S; the group multiplication,

left zero multiplication, or right zero multiplication. In

case of left zero multiplication, every point is a right

identity, in particular, v is a right identity. Similarly

in the case of right zero multiplication, v is a left identity.




23

If the multiplication is the group multiplication, then v is

the identity for S. Suppose S / K and v is neither a left

identity nor a right identity for S. According to Proposi-

tion 4-4 we have n(vSv) = 0 when n(S) = 2 and n(vSv) = 1 when

n(S) = 3. But n(vSv) = 0 is impossible, since vSv is always

a nonempty set. This implies that v is either a left or a

right identity for S when n(S) = 2. For the case n(S) = 3

we obtain a contradiction as follows. Since vSv n K i O

and v e vSv. one has v c K, which implies that SvS = S = K.

That is again a contradiction since we assume S / K. Hence

v must be a left or a right identity for S.

(2) The proof is immediate if S = K, so we assume that

S / K. Then v J K, since otherwise S = SvS C K, which is a

contradiction to the assumption S / K. If v is a right or a left

identity for S, then v is a midunit. Hence we may assume

that v is neither a left identity nor a right identity for S.

Then by Proposition 4-3 we have vS\vSv / ] and Sv\vSv / 7, i.e.,

n(vS\vSv) 2 1 and n(Sv\vSv) 1. It is claimed that

n(vSv) = 2. For we know from Proposition 4-4 that n(vSv) 2

and if n(vSv) = 1, then it implies that v e K. It is easy

to see that vSv, vS\vSv and Sv\vSv are mutually disjoint

sets. Hence we have the following inequality:

4 = n(S) n(vS U SV) = n(vSv) + n(vS\vSv) + n(Sv\vSv) 4.

It follows that n(vS\vSv) = 1 and n(Sv\vSv) = 1. Since

v E vSv, vSv n K / 0, and n(vSv) = 2, we may write vSv = (v.k)

where k = vkv e K. Also we let vS\vSv = (al and Sv\vSv = (b).

To prove v is a midunit for S, it is sufficient to prove that







v is a midunit relative to [vS\vSv, Sv\vSv] by Proposition

4-2, i.e., to prove that avb = ab.

We prove first that either av = k or vb = k. Otherwise,

one has that av = v and vb = v, since (av,vb) C vSv and

vSv = fv,k). Then we have b(av) = by = b. In order to arrive

at a contradiction we consider the element ba. There are

only four possibilities that is ba = v, ba = a, ba = b, or

ba = k. In the case ba = v, we have b = by = b(av) = (ba)v = v,

which is a contradiction since b i v. In the case ba = k,
2
we have v = v = (vb)(av) = v(ba)v = vkv = k, which is a

contradiction since v g k. Hence we may assume ba = a.

Then (ba)v = av = v and b(av) = by = b, which implies that

b = v. That is impossible since b / v. Similarly if ba = b,

then v(ba) = vb = v and (vb)a = va = a, which is again a

contradiction since a / v. Hence it is true that either

av = k or vb = k. Therefore, we have avb = ak or avb = kb.

Then avb = k since either ak = v or kb = v will imply that

v e K, and hence S = K.

In order to prove ab = k, let us assume that ab = v,

since ab e vSv and vSv = (v,kj. It follows that (a,b) C J,

where J is the p-class of the p-maximal element v in S. Thus

both a and b are p-maximal elements of S. By Theorem 1-6,

there is an element x of S such that ax = a and an element y

of S such that yb = b. In the case x = v, then av = a implies

that a E vSv. It is a contradiction since a vSv. In the

case x = b, then ab = a,contradicting our assumption ab = v.

Further, if x = k, then ak = a, contradicting the assumption







that v = ab. since v = ab = akb = k. Hence the only possible

case will be x = a which implies that a = a. This implies

that ava = a. where av v vSv. It follows that av = v, since

vSv = Iv.k) and if av = k, then ]h = av = a(ab) = a2b = ab = v.

Similarly, one can check that the only possibility for y is

y = b so that b = b and hence vb = v. But by the preceding

argument we know it is impossible that av = v and vb = v.

Therefore, we proved that ab = k. The proof is complete.

4-6 Example. We give two examples of semigroups of four

2
elements with midunit satisfying S = S. The first one is

an example of a semigroup S with zero and a proper midunit

v satisfying SvS = S. Let S = [v,a,b,0) with multiplication

as follows.
Table 4-1


v a b 0

v v a 0 0

a 0 0 0 0

b b 0 0 0

0 0 0 0 0






It is easy to see that vSv = (v,0), vS = (v,a,0],

Sv = (v,b,03, and SvS = S. It is evident that b is neither

a left identity nor a right identity for S, since vS and Sv

both are proper subsets of S. To check that v is a midunit

for S it is sufficient to check that v is a midunit relative

to [vS\vSv, Sv\vSv]. We see that vS\vSv = [a] and








Sv\vSv = [b]. While ab = 0 = avb according to the multipli-

cation table. Hence v is a proper midunit for S.

The second example is of a semigroup satisfying S = K

in which every point is a proper midunit. We construct this

example as follows. Let S1 = (a,b) with left zero multipli-

cation and S2 = [c,d] with right zero multiplication, and

we define S = S1 x S2 with coordinatewise multiplication.

Then it is not difficult to check that S = K and each point

in S is a proper midunit for S.

We end this chapter with an example of a semigroup with

five elements which satisfies the condition SvS = S for an

idempotent v and has a unique maximal p-class which is a

completely simple subsemigroup, but S does not have a mid-

unit.

4-7 Example. Let S = (v,a,b,c,0] with multiplication as in

the following table.
Table 4-2

v a b c 0

v v a b a 0

a a 0 0 0 0

b a 0 0 a 0

c c 0 0 0 0

0 0 0 0 0 0





Then it is easy to see that S is a semigroup, v is an

idempotent, J = Iv) is the unique maximal 9-class in S which

is a completely simple subsemigroup. Also we have that





27

SvS = S. for vS = fv.a.b.01 and Sv = (v,a,c,0]. But v is

not a midunit because bc = a while bvc = 0.













CHAPTER V


THE CODIMENSION OF TIE P-CLASS OF
MIDUNITS IN A CONTINUUM SEMIGROUP

This chapter is devoted to proving that the p-class J

of the midunits in a continuum semigroup of codimension one

with midunit satisfying S = S / K is totally disconnected.

Recall that a compact space X has codimension (abbreviated

"cd") less than or equal to a nonnegative integer n if

Hn(X) Hn(A) is epimorphism for each closed subset A of X,

where i is the inclusion map i: A X. This is equivalent

to the statement that Hn(B) k* Hn(A) is epimorphism for

each pair of closed subsets A and B of X with A C B, where

k is the inclusion map k: A B. We define cd] = -1, and

cd(X) = inf(nlcd(X) ; n for a nonempty compact space X [3]. Recall

that a floor for a nonzero element h c H (X) is a closed

subset F of X such that h|F # 0 while hIF' = 0 for any closed

proper subset F' of F. If X is compact Hausdorff and

0 h EHP(X), then h has a floor. Moreover, any floor is

connected [13]. We proceed with a theorem due to Wallace [20].

5-1 Theorem. Let S be a continuum semigroup with cd(S) n

and A be a closed subset of S which is a floor for some

0 f h E H(A). It tlA = A (AtI = A) for some tl E S, then

tA = A (At = A) for each t E S.







The following lemma is a result concerning 11-classes in

compact semigroups.

5-2 Lemma. Let S be a compact semigroup and b be a point of

S. If 1b is not a singleton set, then there exists an

idcmpotcnt o such that Ib = oiHb.

Proof. If d is an element in Hb which is different from

b. then we have b = xd and d = yb for some x and y in S.

This implies that b = xyb; thus b = eb for some idempotent e.

from the Swelling Lemma 1-5. If z is an element of Hb other

than b, then there is a point u in S such that z = bu, then

ez = ebu = bu = z, i.e., Hb = eHb'

It is well known that if S is a compact semigroup then

each M-class is homeomorphic to a topological group [6; p.35].

Also if X is a compact topological group with codimension n

and Z is the additive group of integers then Hn(X;Z) / 0

[6;p.54]. Consequently we have that if S is a compact semi-

group and Ha is an h-class of a in S with cd(Ha) = n, then

Hn(Ha;Z) / 0.

5-3 Theorem.If S is a continuum semigroup with finite co-

dimension n, then cd(H ) < cd(S) for all a in S such that

Ha / K.
a
Proof. We prove first that cd(Ha) < cd(S) for each

a e S\K. Suppose on the contrary that there exists an element

a e S\K such that cd(Ha) = n. If Ha is a singleton set, then

cd(H ) = 0, which implies that cd(S) = 0; i.e., S is a totally

disconnected space. But this is impossible since S is a con-

tinuum with more than one element. Lemma 5-2 guarantees the

existence of an idempotent e in S such that Ha = eHa. Let
a







h be a nonzero element of Hn(Ha;7) and let A be a floor for h.

Since A C Ha we have A = eA. Therefore tA = A for each t c S

by Theorem 5-1. Since a c S\K, we have that H n K = -,
a
which implies that A A K = n. If we let t0 be a point of K.

then we have A = t0A C K, since K is an ideal in S. It is

a contradiction since we know that A n K = 0.

We next prove that cd(Ha) < cd(S) for each a E K with

Ha / K. Suppose on the contrary that cd(H ) = n for some

a E K with H K. Then K consists of at least two H-classes.
a
We know that the h-class in K are mutually homeomorphic groups,

so that we have cd(Hu) = n for each u E K n E. In order to

obtain a contradiction, we let e and f be distinct idempotents

in K. This is possible since K contains at least two

h-classes. As before, we let h be a nonzero element of

H (H ;Z) and let A be a floor for h. Since A C H is a
e e
group with identity e, it follows that A = eA. By Theorem

5-1 we know that tA = A for each t E S. This implies that

e e A since A is a subset of the group Ha while S contains

all possible inverses in H of A. Therefore fe E H since

tA = A for each t e S. Similarly we could prove that fe e Hf.

But this is a contradiction since He and Hf are mutually

disjoint. Hence the proof is complete.

5-4 Proposition. Let X and Y be two compact Hausdorff spaces.

If cd(X) is finite and Y is totally disconnected, then

cd(X x Y) = cd(X).

Proof. Suppose cd(X x Y) = n and let A be a closed

subset of X x Y such that the homomorphism induced by the







inclusion map from A into X x Y, i.e., i*: Hn- (XxY) Hn- (A)
is not epimorphic. Hence there exists a nonzero element

h Hn-(A) which is not extendible to X x Y. According
to Theorem 1-3, we let R be a roof for h and consider

h = hlR n A f 0. Then R is also a roof for h0 and R is
connected. Let 7i and T2 be the two projection maps from
X x Y to X and Y, respectively, i.e., rr((x,y)) = x and

T2((x,y)) = y. Let j be the inclusion map from R into
X x Y, and let n be defined by n = 7loj: R X. We want to
prove that 7 is injective. Suppose not, and let (a,b) and

(a,c) be two points in R such that b / c. Then 2(R) is a
nondegenerate connected set in Y, since R is connected, T2
is a map and 7r2((a,b)) / n2((a,c)). This contradicts the
assumption that Y is totally disconnected. Then n is an
injective map. Consider the following diagram:


R 1 [R]


kI I ROA I1
R n A I n[R n A]


where k is the inclusion map from R n A into R, and L is the
inclusion map from p[R n A] into n[R]. Also n and RoR N A
are homeomorphisms since n is an injective map.







Then the following diagram is commutative, i.e.,

r*ok* = (nIR n A)* o 4*.



Hn-l(R) Hn-l([R])

k* *

Hn-1 ( ) (RA)* Hn-1 ([RnA])



We know that both 9* and (rIRNA)* are isomorphisms,and

k* is not an epimorphism, so t* is not an epimorphism.

This implies that cd(X) > n-l. But we know that

cd(X)cd(X x Y) = n, therefore cd(X) = cd(X x Y). Hence the

proof is complete.

5-5 Corollary. Let S be a compact semigroup with midunit of
2
finite codimension satisfying S = S T K. Let J be the

y-class of the midunits, M be the set of all midunits and H

be the M-class of midunit v. If H is totally disconnected

for some midunit, then cdJ = cdM.

Proof. It is not difficult to prove from Theorem 3-9

that J is homeomorphic to the space M x H The conclusion

follows from Proposition 5-4.

It has been proved by Cohen and Koch [4] that if A is a

closed subset of a compact Hausdorff space X with cd(X) < n,

then cd(X/A) s n [4]. Also they proved in the same paper

that if S is a continuum semigroup satisfying S = ESE with

zero then H (S) = 0.

5-6 Theorem. Let S be a continuum semigroup of codimension one
with midunit satisfying S2 = S K. Then the -class J of
with midunit satisfying S = S $ K. Then the p-class J of


midunits is totally disconnected.







Proof. As before we let M be the set of all midunits

and H be the Il-class of the midunit v. From Thcore.m 5-3

we know that Hv is totally disconnected for each midunit v.

In order to prove that J is totally disconnected it suffices

by Corollary 5-5 to prove that M is totally disconnected.

In case K is not a singleton set we may consider S = S/K,

the Rees quotient of K. Then S is a continuum semigroup with

zero and cd(S) 1. If we let Q: S S/K be the natural

homomorphism, then (v) is a midunit for S and (J), the

g-class of S(v) in S, is homeomorphic to J. Hence without

loss of generality we may assume K = ({0. Consider T = MSM,

then T is a nondegenerated subcontinuum subsemigroup with

zero and E(T)TE(T) = T since M g E(T). Also we know that

cd(MSM) s 1 and H1(MSM) = 0.

Let C be a component in M with more than one point and

u and v be two different points in C. Then vSv U uSu is a

closed subset in MSM and C n [vSv U uSv] = Iv,u). It follows

from the Mayer-Vietoris Sequence Theorem 1-1 that the follow-

ing sequence is exact,

oH(C) x f(vSv U uSu) i*-io* O([u,v]) H(C U[vSvUuSuj).

If G is the nontrivial coefficient group for the cohomology,

then Ro((u,v)) = G while H(C) = 0 and o(C U [vSv U uSu]) = 0,

since both C and CU[vSv U uSu] are connected. Therefore A

is a monomorphism, which implies that H (C U [vSv U uSuJ) / 0.

This is a contradiction since H (MSM) = 0, cd(MSM) s 1 and

C U [vSv U uSu] is a closed subset of MSM. Hence the proof

is complete.













CHAPTER VI


THE COMPLEMENT OF THE y-CLASS OF MIDUNITS AND
MINIMAL CONTINUUM SEMIGROUPS WITH MIDUNIT

The first part of this chapter is devoted to a study of

the structure of the complement of J in S, where S is a con-

tinuum semigroup with midunit satisfying S2 = S K and J is

the p-class of the midunits of S. We prove that the comple-

ment of J is a dense connected maximal ideal in S and no sub-

set of J cuts S. If, furthermore, S is locally connected and

U is an open set containing J, then there exists an open set

V in U containing J such that S\V is a connected ideal. The

second part of this chapter investigates the structure of
2
minimal continuum semigroupswith midunit satisfying S = S K K

(see Definition 6-5) and their relation to irreducible clans.

We proceed with a proposition that is due to Koch and Wallace

[8]. Since our proof is slightly different from theirs we

present it here.

6-1 Proposition. Let S be a continuum semigroup satisfying
2
S = S / K with midunit and J be the p-class of the midunits.

Then S\J is a dense connected maximal ideal of S.

Proof. From Proposition 3-6 and Theorem 3-9 it follows

that S\J is a maximal ideal in S. Since the closure of ideal

is again an ideal hence [S\J]* is an ideal. From the maximality








of S\J it follows that either [S'.Jj* = S or [S\J]* = S J.

If S\J = [S\JJ*, then J is a proper nonempty closed and open

set of S which is impossible because S is connected. Hence

[S\J]* = S, S\J is dense in S. It remains to show that
2
S\J is connected. Since S = S, for each z e S\J there exists

a pair of elements a and b of S such that z = a b From
z z zz
Proposition 3-6 we know that for each z e S\J, a Sb is dis-
z z
joint from J and z e a Sb since S contains a midunit.
z z
Then S\J = U(azSbzlz e S\J) is a union of connected sets.

so since each a Sb meets K, we have that S\J is a connected

set. The proof is complete.

Let A and B be two subsets of a space X. It is known

in general topology that if A is connected and A C B C A*,

then B is connected. Then the following proposition can

be proved easily.

6-2 Proposition. Let S be a continuum semigroup with midunit

satisfying S2 = S / K and J be the y-class of the midunits.

Then no subset of J cuts S.

Proof. Let N be a subset of J, then S\J C S\N C S.

From Proposition 6-1 we know that S\J is connected and

[S\J]* = S. Therefore S\N is connected.

6-3 Proposition. Let S be a continuum semigroup with mid-
2
unit satisfying S = S K and J be the p-class of the mid-

units. If S is locally connected and U is an open set contain-

ing J, then there exists an open set V containing J with

V C U such that S\V is a connected ideal of S.

Proof. Let A = S\U and for each a E A let N(a) be a





36

closed connected neighborhood which is small enough that it

is disjoint from J. Since A is compact, we let N(al )

N(a2),...,N(a ) be a finite cover of A. For each a let

b. and c. be a pair of elements of S such that a. = b.c..
3 3 3 33
Then b.Sc. is a subcontinuum which contains a. and meets K
J3 3
for each j = 1,2,...,m. Let
m m
B = [ .U N(a )] U [ U b.Sc.] U K.

Then B is a subcontinuum which contains A and is disjoint

from J. nom Proposition 3-6 we know that D = BUESUSBUSBS is a

subcontinuum ideal of S and is disjoint from J. Now we

simply let V = S\D. It is then easy to see that J C V C U

and S\V = D. This completes the proof.

6-4 Definition. Let S be a continuum semigroup with midunit

satisfying S = S / K and M be the set of all midunits of S.

Then S is called minimal if it does not contain a proper

subcontinuum subsemigroup T with M C T and T n K i O.

6-5 Theorem. Let S be a continuum semigroup with midunit

satisfying S = S / K and M be the set of all midunits. Then

S contains a continuum subsemigroup T such that M C T,

T n K / 0 and T is minimal.

Proof. This follows immediately through an application

of Zorn's Lemma.

Recall that a clan S with identity v is called irreducible

if it does not contain a proper subcontinuum semigroup T con-

taining v and satisfying T n K / n. In an irreducible clan

with identity v, one has H = (v) [6; p. 156].







6-6 Theorem. Let S be a continium semigroup satisfying
2
S = S with zero and midunit. M be the set of all midunits,

J be the p-class of midunits and T be minimal continuum

subscmigroup such that M C T. ind 0 e T. Then the following

statements hold.

(1) MTM = T.

(2) vTv is irreducible for each v e M.

(3) T n = M.

(4) Consider the cartesian product Mv x vTv x vM with

coordinatewise multiplication, and define

t: Mv x vTv X vM T by t((u,x,w)) = uxw. Then v

is a surjective homomorphism.

Proof. (1) MTM C T follows from the facts that M C T

and T is a subsemigroup. On the other hand, since MTM is a

continuum subsemigroup such that M C MTM and MTM contains 0,

we have by the minimal condition of T that T C MTM. Hence

T = MTM.

(2) Suppose not, i.e., there exists a point v e M such

that vTv is not irreducible. Hence there exists a subclan

of vTv, say TO, such that TO is a proper subset of vTv; v e TO

and 0 E T0. Then MTOM is a continuum subsemigroup of S such

that M C MTOM, since v E TO and MvM = M = M, and 0 E MTOM.

Since T is minimal we have MTOM = T, which implies that

vMTOMv = vTv. Since vMTOMv = vT0v, we have that vT0v = T =

vTv, which contradicts the assumption that TO is a proper

subset of vTv. This completes the proof.

(3) It is sufficient to prove that T n J C M. Since

by (2) vTv is irreducible for each v e M, we know that




38


vTv n H = (v1, where H is the maximal group containing v.
v v
If z e T i J then z = vzv for some v e M. since by Theorem 3-9

we know that J is the union of all maximal group H with

w E M. This implies that z E H n vTv. Then z = v and the

proof is complete.

(4) Since MTM = T, each z e T can be written as

z = uzw for some midunits u and w. On the other hand since

r((uv, vzv, uw)) = uvvzvvw = uzw = z, then 9 is surjective.

Also since i[(u,x,w) (u,x,w)] = t[(uu,xx,ww)] = uxxw

and ((u,xw)((u,x,w)) ((, = (uxw) (uxw) = uxxw for any pair

of elements (u,x,w) and (u,x,w) of Mv X vTv x vM, we have that

( is a surjective homomorphism.













CHAPTER VII


A METHOD OF CONSTRUCTING SEMIGROUPS WITH MIDUNIT


McCharen has pointed out a method of constructing new

semigroups with midunit from given known semigroups [11]. It

is our purpose in this chapter to introduce this method and

discuss some basic properties of it. In particular, we

investigate those algebraically defined sets which are invariant

under this construction. The importance of this method to

this work is that it could transfer problems in general semi-

groups into problems of semigroups with midunit and vice-

versa. We prove in this way that if S is a compact semigroup,

v is a J-maximal idempotent, J is the 1-class of v, and

vJv C J, then J is homeomorphic to a cartesian product of a

rectangular band with Hv, the maximal subgroup which contains

v. Furthermore, if cd(S) = 1 and S / K, then J is totally

disconnected.

7-1 Proposition. Let S be a semigroup and v be an idempotent

in S. We define a multiplication "*" on S by

a*b = avb, for all a and b of S.

Then the following statements are true.

(1) (S;*) is a semigroup under the multiplication "*"

and v is an idempotent midunit in (S;*).

(2) S*S = S iff SvS = S.

39







Proof. (1) It is routine to check the associativity

of the multiplicati.- "* because S is a given semigroup.

To prove v is a midiil we let x and y be a pair of elements

of (S;*). Then

x*v*y = (x*v)*y = (xvv)*y = (xv)*y =(xvv)y = xvy = x*y,

which implies that v is a midunit of (S;*). Also v*v = vvv = v,

i.e., v is an idempotent in (S;*).

(2) The proof of this part is trivial.

It is not difficult to prove that if N is a right (left,

two-side) ideal of S, then N is a right (left, two-sided)

ideal of (S;*).

7-2 Proposition. The following two statements are equivalent:

(1) N is a minimal right (left, two-sided) ideal in S.

(2) N is a minimal right (left, two-sided) ideal in (S;*).

Proof. We only prove the case when N is a minimal right

ideal. The other cases can be proved analogously.

(1) = (2) It is not difficult to prove that N is a right

ideal in (S;*). In order to prove that N is a minimal right

ideal in (S;*), we let P be a right ideal in (S;*) such that

P C N. Then we have that PvS = P*S C p C N. Since PvS is a

right ideal of S contained in the minimal right ideal N, we

have that PvS = P = N. This shows that N is a minimal right

ideal of (S;*).

(2) = (1) Since N is a right ideal of (S;*), we have

that NvS = N*S C N. Also it is easy to see that NvS is a

right ideal of (S;*) which is contained in the minimal right

ideal N of (S;*). Therefore N = NvS, so that N is a right







ideal of S. In order to prove that N is a minimal right

ideal in S, we let P be a right ideal in S such that P C N.

Then PS C P C N. On the other hand, PS is a right ideal of

(S;*), so that by the minimality of N we have PS = P = N,

which proves that N is a minimal right ideal of S.

As usual, for an element z in S, we let L(z),(R(z).J(z))

be the smallest left (right, two-sided) ideal of S which

contains z, and L(z),(R(z), J(z)) be the smallest left (right,

two-sided) ideal of (S;*) which contains z. Also we let H
v
be the maximal group in S which contains v, and let H be the

maximal group in (S;*) which contains v.

7-3 Proposition. If z is an element of S, then the following

statements hold.

(1) L(z) C L(z), R(z) C R(z), and J(z) C J(z)

(2) L(vz) = L(vz), R(zv) = R(zv), and J(vzv) = J(vzv)

(3) J(v) = J(v), and Hv = H

Proof. (1) Since L(z) = z U S*z = z U Svz C L(z), we

have that L(z) C L(z). We can prove that R(z) C R(z) and

J(z) C J(z).

(2) Since L(vz) = vz U S*(vz) = vz U Svvz = vz U Svz = L(z)

we have that L(vz) = L(z). We can prove analogously that

R(zv) = R(zv) and J(vzv) = J(vzv).

(3) It follows from (2) that J(v) = J(v), L(v) = L(v)

and R(v) = R(v), which implies that H = H .
v v
7-4 Proposition. Let S be a semigroup and v be an idempotent

of S such that SvS = S. Let "*" be the multiplication defined

on S by a*b = avb for all a and b of S. If we let J be the





42

;-class of v in S and J be the 9-class of v in (S;*) and if

vJv C J. then J = J.

Proof. In order to prove that J C J, we let z e J. Thus

we have v e z U z*S U S*z U S*z*S, which implies that

S*v*S C S*z*S. But S*v*S = SvS = S by assumption, which

implies that S*z*S = S,i.e., SvzvS = S. Therefore z e J

since SvzvS C SzS. In order to prove that J C J, we let z E J.

Since vJv C J, we have that vzv e J; that is, SvzvS = S. On

the other hand, S*z*S = SvzvS = S, which implies that z e J.

Therefore the proof is complete.

7-5 Theorem. Let S be a compact semigroup with minimal ideal

K. Let v be a 9-maximal idempotent of S such that its

p-class J satisfies vJv C J and let "*" be the multiplication

defined on S by a*b = avb for all a and b of S. Then the

following statements hold.

(1) J is homeomorphic to a cartesian product of a

rectangular band and Hv, the maximal group which

contains v. More precisely, (J;*) is the cartesian

product of a rectangular band and H with coordi-

natewise multiplication.

(2) If cd(S) = 1 and S 4 K, then J is totally disconnected.

Proof. Let T be defined by T = SvS, then T is a compact

semigroup satisfying TvT = T and contains v as a p-maximal

idempotent in T. Let J be the y-class of v in T. It is

claimed that J = J. In order to prove that J C J, let t be

an element of J. Then we have J(t) = J(v) = TvT = T = SvS.

It follows that J(v) = SvS C J(t), since J(t) C J(t). By the







maximality of v in S we obtain J(t) = J(v); i.e., t e J.

Therefore J C J. In order to prove that J C J, let z be an

element of J. By the assumption vJv C J. it follows that

vzv e J. This implies that J(vzv) = J(v) = SvS. Also it

can easily be proved that J(vzv) = J(vzv) and J(vzv) C J(z).

Therefore, we have that T = SvS = J(vzv) = J(vzv) C j(z) C J

and hence that J(z) = T or z c J. Hence we have proved that

J = J. If we let H be the maximal subgroup in T which

contains v, then it is evident that H = H .

If J is the p-class of v in (T;*), then by Theorem 3-9,

(J;*) is isomorphic to a cartesian product of a rectangular

band and H with coordinatewise multiplication. By Proposi-

tion 7-4 we have J = J, since vjv C J follows directly from

the fact that J = J and the assumption vJv C J. Consequently,

we have proved part (1).

Since cd(S) = 1 and S K imply that cd(T) s 1 and

T / K(T), the conclusion of part (2) follows easily from

Proposition 7-2 and Theorem 5-6.













CHAPTER VIII


SEMIGROUPS ON D-SPACES


In this chapter we are seeking sufficient conditionson a

semigroup S under which S must have either a left or a right

identity and consequently S does not admita proper midunit. We
2
show that the algebraic condition S = S V K and the topologi-

cal condition that S be D-space (to be defined in Definition

8-6) are one such set of conditions. Recall that a midunit

v of a semigroup S is proper if it is neither a left identity

nor a right identity of S. In Example 2-1 we defined two

multiplications on a closed unit interval, in which a midunit

exists without being a left identity, right identity, or even

an idempotent. It seems difficult to study the notion of

midunit without some algebraic restriction. One algebraic

condition, which neither of the two multiplications given
2
in Example 2-1 satisfies, is S2 = S K. It is not difficult

to prove the following proposition.

8-1 Proposition. Let S be topologically the closed unit
2
interval [0,1] and satisfy S = S / K. Then there exists a

point v of S which is either a left or a right identity of

S and is located at one endpoint of [0.1].

This example motivated us to seek a more general topologi-

cal condition under which a semigroup S satisfying S = S / K

must have either a left or right identity.

44







8-2 Definition. Let X be a continuum. A point p of X is a

D-point iff for any two subcontinuua C1 and C2 with p E C nC2'

either C1 C C2 or C2 C C1.

Let X be a continuum with D-point p. For each point a of

X, let 3a be the collection of all subcontinuua of X which

contains both a and p. Then ja is a non-empty collection

of compact connected subsets of X and 3a is totally ordered

by inclusion. If L[a] = 03a, then L[a] is the minimal sub-

continuum containing a and p. It is easy to see that

L[p] = (p), since (p) is itself a subcontinuum which contains

p. We define a relation < on X as a s b iff L[a] C L[b].

Since L[a] and L[b] are always comparable under inclusion for

any pair of elements a and b of X, s is a total quast-order on

X. By the definition of and construction of L[a] we have

L[a] = (blb s a], and call L[a] the lower set of a in X. The

sets U[a] = (bla b} and La = L[a] n U[a] are called the

upper set and level set of a respectively. For convenience,

we write a b iff a b and b a, and a < b iff a s b and

b j a. Recall that a quasi-order on a set X is called

order dense if and only if for any pair of elements a and b of

X satisfying a < b, there exists a point c of X such that

a < c < b.

8-3 Lemma. Let X be a continuum with a D-point p. If U[a]

is closed for each point a of X, then s is order dense.

Proof. Suppose not; i.e., suppose there exists a pair of

elements a and b in X, with a < b and no point c in X satisfies

a < c < b. Since s is a total quasi-order on X we have







L[a] U U[b] = X and L[a] n U[b] = l. Then X is a union of

a pair of disjoint nonempty closed subsets L[a] and U[b].

which contradicts the assumption that X is connected. Hence

the proof is complete.

8-4 Proposition. Let X be a continuum with a D-point p and s

be the induced quasi-order. Then the following two statements

are equivalent.

(1) "s" is a closed relation on X, and

(2) U[a] is closed for each point a of X.

Proof. (2) = (1) Let b and c be a pair of elements of

X such that (b,c) S a. Since 5 is a total quasi-order on X

we have c < b. By Lemma 8-3 there is a point d in X satisfying

c < d < b. If U = X\L[d] and V = X\U[d], then U is an open

set containing b while V is an open set containing c. It is

claimed that (U x V) n : = r3. Suppose not; i.e., suppose

there exists a pair of elements x and y of X such that

(x,y) E (U x V) n s. Then d < x,y < d and x s y, which

implies that d < d. But this is a contradiction. Hence the

proof is complete.

(1) = (2) We omit the proof because it is trivial.

8-5 Remark. There exists a continuum with a D-point whose

induced quasi-order is not closed. Let X be a space defined


1 1
by X = ((x,y)| y = sin ( -)j- < x 1 O) U



{(x,y) y = sin ( ), 0 x < 1] U


[ (x,y) -l y a 1, x = or x =- ,

with the usual topology as seen in the following figure.







"




P


q X
P-- H ~ ---- 7 --- / I ^ --- x



q


Figure 8-1

Then X is a continuum with a D-point p = (- 1). But

the induced quasi-order is not closed, since

U[q] = X\((x,y) x = ,-1 y 1} is not a closed set in
17
X where q = (- -1). By Proposition 8-1 we know that is

not a closed relation.

8-6 Definition. A continuum X with a D-point p is called a

D-space if the induced quasi-order s is a closed relation on

X.

We have seen in Remark 8-5 that X is not locally connect-

ed at p, while the induced quasi-order s is not closed. But

if X is a D-space, then X is locally connected at its D-point

p.

8-7 Proposition. If X is a D-space with D-point p, then X is

locally connected at p.

Proof. Let V be an open proper subset containing the

D-point p. If is the collection of all L[x] n (X\V) for

each x of X\V, then is a nonempty collection of nonempty

closed subsets of X and is totally ordered by inclusion, so

that n / n. If d is a point of n, then d s x for all x of

X\V, since d E L[x] for all x of X\V. This implies that







X\V C U[d]. It is not difficult to verify that X\U[d] =

U(L[c] c < d). Then X\U[d] C V and is an open connected

set containing p. Hence the proof is complete.

Suppose X is a D-space with a D-point p. We let li be

the collection of all U[b] for each point b of X. Then U

is a collection of closed subsets of X which is totally

ordered by inclusion. If we define M = nl, then M is a non-

empty closed subset of X, and is called the maximal level

set of X.

8-8 Theorem. Let X be a D-space with D-point p. Then the

following statements hold.

(1) X is irreducible between its D-point p and any

point m of M.

(2) Each point z e X\(M U [p)) is a weak cut point of X.

Furthermore, if M contains more than one point, then

every point z, except D-point p, is a weak cut point

of X.

(3) Each level set Ld, except L = (p) and the maximal

level set M, cuts X.

(4) If, furthermore, X is locally connected, then X is

an arc.

Proof. (1) Let A be a subcontinuum in X which contains

p and a point m of M. Since m E M we have L[m] = X. But

L[m] is the minimal continuum which contains p and m so that

L[m] C A and hence A = X.

(2) From (1) we know that each point z e X\(MU(p)) is a

weak cut point between p and m a point of M. In the case







that M contains more than one point, it is sufficient to prove

that each point m of M is also a weak cut point. Let n be a

point of M which is different from m. Then it can be easily

verified that m is a weak cut point between p and n, since

L[n] = X is the minimal continuum which contains p and n.

(3) If we let P = L[d]\Ld and Q = U[d]\Ld, then

X\Ld = PUQ and both P and Q are nonempty sets since M C Q and

p e P. Also P* n Q = n, since P* n Q C L[d] n Q = r. Simi-

larly it is true that P n Q* = O. Thus X\Ld is a disconnected

set, which completes the proof.

(4) We prove this part by steps.

(i) We show that for each a / p, there exists some

point a e La such that if V is an open neighborhood of ao,

then there exists a point b E V with b < a0. Suppose not

and let a be a point such that for each L e La there exists

an open neighborhood V- of t for which t s b for all b e V

Then U[a] = (X\L[a]) U ( U V ) is both open and closed in
aGEL a
X. Since a / p, U[a] is a proper subset of X which is both

open and closed in X, which is impossible because X is con-

nected.

(ii) We prove that each level set La is a singleton set.

Recall that L = (p), since L[p] = (p). So we assume that

a / p and La contains more than one point. By (i) there

exists a point a0 E La which satisfies the statement mentioned

in (i). Since L contains more than one point, we let a1 be
a 1
a point in L which is different from a0 and let Va be a

connected neighborhood of a0 which is small enough that







al Va*. By (i) there is a point b in V such that b < a0'
0 a0
Then L[b] U V is a closed connected set containing p and
a0
aO. Since p is a D-point, we have L[a0] C L[b] U Va*
0
which implies that al e L[b]. But this is a contradiction

since b < a0 and a0 al"

(iii) From (3) and (ii) we know that each point

b e X\MU(p) is a cut point of X. From (ii) we know that M

is a singleton set, hence let M = [m). In order to prove

that X is an arc, it is sufficient to prove that p and m are

noncut points. So suppose p is a cut point, i.e., X\(p) =

P U Q, where P and Q are disjoint nonempty open sets, and

P* n Q* = [p). Since P* n Q* = [p) and P* U Q* = X is a

connected set it is not difficult to prove that both P* and

Q* are connected. Then both P* and Q* contain the D-point

p, but neither of them contains the other as a subset, which

contradicts the fact that p is a D-point. On the other hand

we have X\[m) = U(L[d] d e X and d / m] is a connected set,

which implies that m is not a cut point. Hence the proof

is complete.

8-9 Remark. (1) The convese of part (1) in Theorem 8-8'is

not true. The space X in Remark 8-5 is irreducible between
1 1
points (- ,1) and (- ,1) but is not a D-space.

(2) In application 8-11, there are two D-spaces which

are not locally connected.

8-10 Theorem. Let S be a D-space with a D-point p whose max-

imal level set M contains more than one point. If S satisfies








S = S / K. then the D-point p is either a left or a right

identity for S. Furthermore, if v is a midunit for S, then

v = p.

Proof. It follows from Theorem 1-6 that there exists an

idempotent e which is a p-maximal element of S since S is
2
compact and S = S. It follows from Theorem 1-7 that e is not

a weak cut point of S since, by assumption S / K. Thus e

must be the only point of S which is not a weak cut point of.

namely, e = p. Since by part (2) of Theorem 8-8 we know that

every point, except the D-point p, is a weak cut point of S.

Also it follows from Theorem 1-7 that e is the only p-maximal

element S. This implies that SeS = S, since e is an idempotent

and hence J(e) = SeS.

On the other hand we know that each eS and Se is a sub-

continuum and containsthe D-point p, so that either eS C Se

or Se C eS. Without loss of generality we may assume

eS C Se. Then we have

S = SeS C S(Se) = S e = Se.

In this case e is a right identity. Similarly if Se C eS

then e is a left identity.

Furthermore, if v is a midunit of S, then v = p, since

v is also a p-maximal idempotent in S.

8-11 Application. (1) Let the underlying space of S be

defined as S = ((x,y) y = sin(-), 0 < x )U

((x,y)I x = 0, -1 y s 1), with the usual topology as seen

in the following figure.






Y











Figure 8-2

2 1
If S satisfies S = S 4 K, then v = (-,0) is either a

left or a right identity for S. Since S is a D-space with

D-point v = (-,0) its corresponding maximal level set is

M = ((x,y) x = 0, -1 s. y 1].

(2) Let S = ((e2 ite-t)It E [0,O])]U[C x (0)] where

C is a unit circle, with the usual topology as seen in the

following figure.











Figure 8-3

2
If S satisfies S2 = S K then p is either a left or a

right identity for S. Since S is a D-space with a D-point

p, its corresponding maximal level set is M = C x (0].

We end this chapter with an example. This is an example

of a continuum semigroup on a triod, probably the simplest

continuum one can find which is not a D-space, which satisfies

S2 = S K and has a proper midunit.







8-12 Example. This example is constructed as follows. Let

T = [v.a.b.0] with multiplication defined in the Table 4-1.

Then T is a semigroup, with discrete topology on it, with a
2
proper midunit v and satisfies T = T / K = (0}. Let

I = [0.1] denote the closed real unit interval with the usual

topology. Let SO = T x I with product topology and coordinate-

wise multiplication. Then SO is a semigroup with proper

midunit (vl) and satisfies S = S. If we let

S1 = ((v,0),(a,0),(b.0)) U [(0) x I], then S1 is a closed

ideal in SO. Then the Rees quotient S = S i/S1 is a semigroup
2
with proper midunit (v,l) and zero satisfying S = S. and is

homeomorphic to a triod as in the following figure [21].



(v,l) (a,l) (b,l)




(0,0)


Figure 8-4

It is not difficult to see that this space is not a

D-space, because none of the points in S can be a D-point.














CHAPTER IX


EXAMPLES OF SEMIGROUPS ADMITTING NO PROPER MIDUNIT


In this chapter we consider S to be a circle with a

closed interval issuing from a point of the circle as seen in

Figure 9-2. We prove that if S is a semigroup with midunit v

and satisfies S = SEUES 4 K, then v is either a left or a

right identity for S and the circle is contained in K. The

way of approaching this result is different from the one

described in the last chapter. In working on this problem we

found the interesting cohomological result given in Theorem

9-1. We also give two examples of semigroups which, due to the

topological structure of their underlying spaces, do not

admit the structure of a semigroup S satisfying S = S

unless the multiplication on S is either left or right zero

multiplication. We proceed with the main theorem of this

chapter.

9-1 Theorem. Let S be a continuum semigroup with zero satis-

fying S = ESUSE. If eS n fS and Se 0 Sf are connected for

each pair of idempotents e and f of S, then H (S) = 0.

Proof. Suppose the conclusion is not true; that is,

there exists such a semigroup S such that H (S) f 0. Let h be

a nonzero element of H (S). Let F be a closed subset of E







which is minimal relative to the condition hIFSUSF i 0. The

existence of F is guaranteed by the assumption S = ESUSE and

the Reduction Theorem 1-2. It is claimed that F cannot be a

singleton set. So we assume that F = (e] where e is a idem-

potent such that h eSUSe / 0. By the Mayer-Vietoris Sequence

Theorem 1-1, we have the exact sequence,

H(eSe) a H (eSUSe) t x t H (eS) x H (Se) i i H (eSe).

Since eSe is a subcontinuum semigroup with identity e and zero,

one has H(eSe) = 0 and Hl(eSe) = 0 [19]. Therefore t*x t

is an isomorphism from H (eSUSe) onto H (eS) x H (Se). But

eS and Se are subcontinuum semigroups with zero and left

and right identity, respectively, and hence H (eS) = 0 and

H (Se) = 0 [19]. Thus H1(eSUSe) = 0, which contradicts the

fact that hLeSUSe / 0. So one may write F as union of two

closed proper subsets A and B, that is F = AUB. Then we have

FS U SF = (A U B)S US (A U B)

= (AS U BS) U (SA U SB)

= (AS U SA) U (BS U SB)

and (AS U SA) n (BS U SB) = (AS n BS) U (SA N BS) U (AS n SB) U

(SA n SB).

Since AS n BS = U(eS n fSI e e A and f EB), AS n BS is a union

of connected sets with a common point zero, hence AS n BS is

a connected set. Similarly we can prove SA A SB is a connected

set. It is easy to see that SA n BS = BSA and SA n SB = ASB

are connected sets. Therefore (AS U SA) n (BS U SB) is a

connected set in S.

Consider the following portion of the Mayer-Vietoris

exact sequence.








no ((AS U SA) 11 (BS U SB)) A H (FS SF) t t

H (AS U SA) x H (BS U SB).

By the minimal condition on F we have hlAS U SA = 0 and
'*
hIBS U SB = 0, i.e., t x t (h FS U SF) = (h]AS U SA,

hIBS U SB) = (0,0). On the other hand H ((AS U SA) n

(BS U SB)) = 0, since (AS U SA) n (BS U SB) is a connected
*
set. This implies that t x t is injective so that

t x t (hIFS U SF) / (0,0), since hlFS U SF / 0. This is

a contradiction and hence H (S) = 0. The proof is complete.

9-2 Remark. Without the condition that eS n fS and Se 0 Sf

are connected for each pair of idempotents e and f of S, the

conclusion of Theorem 9-1 is false by an example of Hudson [7].

This example is of a semigroup S = SE which is a continuum,

has a zero and H (S) G for any coefficient group G. The

underlying space of S is a circle with two closed intervals

issuing from a common point of the circle as seen in the

following figure.

q
e


p 0


r

Figure 9-1


The points e and f are idempotents and 0 is the zero for

S. Under the multiplication defined there, one has Se to be

the arc from p to e containing q and 0, and Sf to be the







arc from p to f containing r and 0. Also we have Se n Sf =
[p.01 which is a disconnected set.
If S is a continuum semigroup with zero satisfying
S = ESE, then it is not difficult to prove that all left and
right ideals of S are connected. In particular, eS n fS and
Se n Sf are connected for each pair of idempotents e and f,
since eS n fS and Se n Sf are right and left ideal of S,
respectively. The following theorem due to Cohen and Koch
[4] follows easily from Theorem 9-1.
9-3 Theorem. If S is a continuum semigroup with zero satis-
fying S = ESE, and if I is a closed ideal of S, then H (I) = 0.
In particular, H (S) = 0.
The following example is an application of Theorem 9-1.
9-4 Example. Let S be a circle C with a closed interval
I = [0,1] issuing from a point of the circle, as illustrated
in the following figure.



p


0-

Figure 9-2


Suppose S = ESUSE. Then the following statements hold.
(1) If S = K, then S has either left zero or right
zero multiplication.
(2) K contains the circle C.





58

(3) If S / K and v is a midunit for S, then v is either

a left or a right identity for S and is the endpoint

of the protruding segment.

We prove this result as follows.

(1) If S = K then since S admits neither a non-trivial

topological factorization nor a topological group, there are

only two possible multiplications on S; namely the left zero

or right zero multiplication.

(2) In order to prove that K contains the circle C, we

show first S does not admit the structure of a semigroup

satisfying S = ESUSE with zero. It is claimed that both

eSnfS and SeOSf are connected for each pair of idempotents e

and f of S. Assume e and f are a pair of idempotents such

that eSnfS is a disconnected set and consider the following

portion of the Mayer-Vietoris exact sequence.

H(eS) X H(fS) i -i' Ro(esnfS) A H (eSUfS) t*

H (eS) x H (fS) -

where fo(eS) = H (eS) = 0 and H (fS) = 0, since

both eS and fS are subcontinuum semigroups with zero and left

identity [19). Also we know that H(eSnfS) # 0, since by

assumption eS n fS is a disconnected set. It follows that

H (eSUfS) 0 since A is an isomorphism. It is not difficult

to check that a subcontinuum B of S satisfies H (B) / 0 only

if C C B. Hence eSUfS is either the circle C or it is the

circle C with part of the closed interval [O,p] of I for

some point p, as in Figure 9-2.






59

Suppose that neither e E fS nor f e eS. If we consider

T = eSLfS, then T is a subcontinnum semigroup satisfying

T = T2 and has e and f as two R-maximal idempotents. It

follows from Theorem 1-8 that neither e nor f can be a locally

separating point of T. Since there is only one possible non-

locally separating point in T (which is denoted by p in

Figure 9-2). it follows that e = f = p. This contradicts

the assumption that neither e c fS nor f E eS. Therefore

either e e fS or f E eS. Without loss of generality we may

assume e e fS. Then we have that eSnfS = fS, which is again

a contradiction since by assumption eSnfS is disconnected.

It follows that eSnfS and SenSf are connected for each pair

of idempotents e and f of S. By Theorem 9-1 we have H (S) = 0.

But we know that H (S) / 0 and hence S does not admit struc-

ture of a semigroup with zero satisfying S = ESUSE.

In order to prove that K contains the circle, we assume

C is not contained in K. Then the underlying space of the

Rees quotient S/K is homeomorphic to S again. Let E denote

the set of idempotents in S/K, and let 4: S S/K be the

natural homomorphism. Then 4(E) = E and it follows that

I(S) = *(E)*(S) U 4(S)y(E) and has zero. By preceding argu-

ment we know this is impossible, thus C is contained in K.

(3) It follows from Theorem 1-8 that v has to be the

endpoint of the closed interval I in Figure 9-2. Also

since vSv meets K and C C K, one has [S\K] C vSv, i.e., v is

a two-sided identity for S\K. Therefore, in order to prove

that v is either a left or a right identity for S, it is








sufficient to prove that v is either a left or a right

identity for K. Since K cannot be non-trivially topologically

factored, there are only three possible multiplications on K;

a group multiplication, left zero and right zero multiplica-

tions. In case K is a group with identity w, then (wv.vw) C K
2
since K is an ideal in S. Also we have (wv) = (wvw)v
2 2 2
w v = wv and (vw) = v(wvw) = vw = vw, i.e., wv and vw are

both idempotents in K. But there is only one idempotent in

K, hence vw = wv = w, i.e., v is a two-sided identity for K.

Indeed, v is a two-sided identity for S when K is a group.
2
In case K has left zero multiplication, then kvk = k = k,

and (kv)k = kv for each k in K, which implies that v is a

right identity for K. In case K has right zero multiplication,

we can prove similarly that v is a left identity for S. Hence

the proof is complete.

We conclude this chapter with two examples which, due to

their topological structure, do not admit the structure of a

topological semigroup S satisfying S = S unless S has left

or right zero multiplication. The notion of midunit is used

in the proof. The first example is due to McCharen [11]. We

give his result here without proof.

9-5 Example. Let S be the space of tangent circles converging

to a point e, as illustrated in the following figure.

















Figure 9-3


Then S does not admit the structure of a topological
2
semigroup satisfying S = S unless S'has of left or right

zero multiplication.

Recall that a continuum X is I-semi-locally connected

at point p, if for any open neighborhood U'of p there exists

an open set V such that p e V C U and X\V is connected. It

is not difficult to prove that if X is a continuum and p is a

marginal point of X, then X is I-semi-locally connected at p.

9-6 Example. Let S be defined by

S = ((0,y) -2 y 1) U ( (x,-2) 0 x -1) U (-,y) -2y0]O
1 1
UT where T = ((x,sin) 0 < x s -], as illustrated in
x TT
the following figure.
Y


_3 3 2 c x







Figure 9-4

2
Then S does not admit a semigroup satisfying S = S unless

S has either left or right zero multiplication.








We may assume S / K, since otherwise we are finished,

since this space possesses no non-trivial topological factori-

zation and, being nonhomogeneous, cannot be a group. By a

sequence of observations we shall arrive at a contradiction.

It is first claimed that S does not admit the structure of a

topological semigroup S with midunit and zero also satisfying

S = S. If v is a midunit of S then from Proposition 3-1 we

have that v is a Y-maximal idempotent in S. By Theorem 1-8

we know that a 1-maximal element cannot be a local separating

point. It follows that v must be located at the point

p = (0,1), since p is the only non-local separating point in

S. The sets vS and Sv are both continuum semigroups in S

with zero, also with left identity and right identity,

respectively. Hence H (vS) = 0, and H (Sv) = 0 [19]. Also

it is not difficult to see that HI () / 0, so that both vS

and Sv are proper subcontinua of S and contain a common

point p = (0,1).

We try to prove that either vS C Sv or Sv C vS. Let us

consider N = (c c = (-!-,0), n = 1,2,3,...) as illustrated

in Figure 9-4. It is claimed that both vS and Sv could

contain only finite many points of N. If vS contains infinitely

many points of N, then it follows that vS contains the subset

((x,sin-) 10 < x < c) of T for some c, since vS is a continuum

containing p. Therefore vS is not I-semi-locally connected

at p, otherwise it contradicts the fact that v is marginal in

vS [11]. Hence both vS and Sv contain only finitely many

points of N. Then it is easy to see that vS C Sv or Sv C vS.








Without loss of generality we may assume Sv C vS, then
2
S = S = (Sv)S C (vS)S = vS C S; thus S = vS. Then S is a

continuum semigroup with left identity and zero. Therefore

one has H (S) = H (vS) = 0 [19]. This contradicts the fact

that H (S) / 0.

Now we may assume K is not a singleton set. We consider

the Rees quotient S/K, and let *: S S/K be the natural

homomorphism. Then it is evident that ((S) = S/K is a con-

tinuum semigroup satisfying (S/K)(S/K) = S/K and contains

a midunit t(v) and a zero. From the topological point of

view we know that S/K is either homeomorphic to circle or to

the original space as seen in Figure 9-4. But the case that

S/K is homeomorphic to a circle cannot happen because every

point on a circle is a local separating point. Hence S/K is

homeomorphic to the original space S. From the preceding

argument it follows that S/K does not admit a semigroup S/K
2
with midunit and zero and satisfying S = S.

Now we are able to prove that S does not admit a semi-
2
group satisfying S = S unless S has either left or right

zero multiplication. It follows from Theorem 1-8 that no

P-maximal element is a local separating point of S and hence

p must be the unique p-maximal element of S. By Theorem 1-6

we know p must be an idempotent. Therefore one has S = SeS.

The conclusion follows from Proposition 7-1, Proposition 7-2

and preceding argument.













CHAPTER X


SEMIGROUPS WITH MIDUNIT ON THE TWO-CELL


In this chapter we investigate the structure of those
2
semigroups with midunit and zero satisfying S = S on a top-

ological two-cell. We let S be a two-cell with boundary B

in a complex plane T, M be the set of all midunits, J be the

;-class of midunits and Hv the maximal subgroup containing

the midunit v. We prove that if H1(H ) / 0 for some midunit

v, then M = (v], J = H = B, and v is a two-sided identity

for S. If H (M) / 0, then J = M = B, and all midunits are

either left identities for S or all are right identities for

S. It follows that cd(J) 1. We conclude this chapter with

an example of a semigroup S with zero on a two-cell which

satisfies S = S and contains a proper midunit.

Let C be a complex plane with the usual topology, and X

be the unit two-cell in T; that is, X = {zlz e C and Iz < 1).

Let B be the boundary of X in C; that is, B = [zl Izl = 1).

We proceed with a topological lemma.

10-1 Lemma. Let X be the unit two-cell with boundary B in

a complex plane E. If A is a closed subset of X such that

B C A and H1(A) = 0, then A = X.

Proof. If A / X, then there exists an element z0 e X\A.
64





65

Without loss of generality we may assume z = (0.0). If the

function p: A B is defined by r(z) = then p is con-

tinuous. It follows that B is a retract of A, since
b b
n(b) = b = b for each b in B. If we let i be the

inclusion map from B into A, then its induced homomorphism
* 1 1
i : H (A) H (B) is an epimorphism [13]. Therefore

H1(A) 4 0, since H (B) / 0. But this is a contradiction,

since by assumption H (A) = 0. The proof is complete.

It is well known that if A is a closed subset of the

complex plane 1, then A cuts T if and only if H1(A) / 0 [13].

10-2 Lemma. Let X be the unit two-cell with boundary B in

a complex plane E. If A is a closed subset of X such that

X\A is a dense connected subset of X and H (A) / 0, then

B C A.

Proof. Since by assumption H (A) / 0, we know that A

cuts the complex plane {. Then C\A can be written as a

union of a pair of nonempty separated sets P and Q; that is,

C\A = P U Q. Without loss of generality we may assume that

X\A C P, since by assumption X\A is a connected set. Then it

follows that (X\A)* C P*, which is equivalent to X C P*, since

by assumption we know that X\A is a dense set in X. It is

claimed that P* = P U A. That P U A is a subset of P* is easy

to see since A C X C P*. On the other hand, since P* 0 Q = O

and P U A U Q = T, we have that P* is a subset of P U A.

Therefore P* = P U A, and Q = {\P*. If we let F[P*] be the

boundary of P* in C, then F[P*] = P* n (r\P*) = P* n Q* C A.

and F[P*] = P*\P*O C P*\XO C B U (P*\X). This implies that







F[P*] C A n [B U (P*\X)] = (A 0 B) II [A N (P*\A)] = A N B,

since A n (P*\X) = D. It is true that U\F[P*] = P*o U Q,

because P* U Q = D and F[P*] = P*\P*o On the other hand

P*O is a non-empty set, since Xe is a nonempty subset of

P*o. This implies that F[P*] cuts the complex plane C, since

X\F[P*] can be written as a union of a pair of nonempty sep-
*o
arated sets P and Q. But no subset of B cuts T except B

itself. This implies that B C F[P*] and therefore B C A

since we proved that F[P*] C A n B.

10-3 Theorem. Let S be a two-cell with boundary B in the

complex plane T with the usual topology. Suppose S is a semi-
2
group with midunit and zero satisfying S = S. Let M be the

set of all midunits J be the p-class of the midunits, and

H be the maximal subgroup of the midunit v. Then the fol-

lowing statements hold.

(1) If H1(H ) / 0 for some midunit v, then J = H = B,

M = (v], and v is a two-sided identity for S.

(2) If H (M) / 0, then J = M = B, and all midunits are

either left identities for S or all are right

identities for S.

(3) cd(J) 1.

Proof. (1) By Proposition 6-1 and Proposition 6-2 we

know that S\H is a dense connected subset of X. It follows

from Lemma 10-2 that B C Hv, since by assumption we know that

H (Hv) / 0. Therefore B C vSv, since H C vSv. On the other

hand we know that H (vSv) = 0, since vSv is a continuum sub-

semigroup with two-sided identity v and zero [19]. It








follows from Lemma 10-1 that S = vSv. In order to prove
2
M = [v), let u be a midunit of S. Then we have vuv = v = v

and also vuv = u, since v is a two-sided identity for S,

therefore u = v and M = (v]. Since v is a two-sided identity

for S, we know that each point of H is a marginal point of

S [6; p. 168]. This implies that H C B. But since we have

proved that B C Hv, one has H = B. It follows easily from

part (3) of Theorem 3-9 that J = H v. The proof of (1) is

complete.

(2) By Proposition 6-1 and Proposition 6-2 we know that

S\M is a dense connected subset of X. It follows from Lemma

10-2 that B C M, since by assumption we know that H (M) / 0.

Therefore B C MSM, since M C MSM. On the other hand we know

that H (MSM) = 0, since MSM is a continuum subsemigroup with

zero and M is a set of idempotents in S [4]. By Lemma 10-1

we know that S = MSM. It follows from Theorem 1-9 that

M C B, since each midunit is a y-maximal idempotent element of

S. But since we have proved that B C M, one has M = B. By

Proposition 2-4 we know that M is a rectangular band; i.e.,

M M Mv x vM with coordinatewise multiplication while Mv is a

subsemigroup with left zero multiplication and vM is a sub-

semigroup with right zero multiplication, where v is a midunit.

Since B is a unit circle which does not admit a non-trivial

topological factorization, one has either M e Mv or M E vM [9].

In case M has left zero multiplication, it is claimed that

all midunits are right identities for S. In order to prove

this we let x be an element of S. Then there exists a pair







of midunits v and u such that x = vxu since S = MSM. If w is

a midunit, then xw = vxuw = vx(uw) = vxu = x, which implies

that w is a right identity of S. Hence in this case all

midunits are right identities for S. In case M has right

zero multiplication we could prove analogously that each mid-

unit is a left identity for S. It is true that if S is a

continuum semigroup with zero satisfying S = vS(Sv) for some

idempotent v, then each point of H is marginal [11, p.6].

This implies that H = (v) for each midunit v, since B = M

and each midunit is either a left or a right identity for S.

It follows from Theorem 3-9 that J = M. The proof of part

(2) is complete.

(3) If H (H ) v 0 for some midunit v, then by (1) we

have that J = B so cd(J) = 1. We assume, therefore, that

H (H ) = 0 for each midunit v; that is, Hv is totally discon-

nected for each midunit v. From Corollary 5-5, it is suf-

ficient to prove that cd(M) & 1. In order to prove cd(M) 1,

we let N be a closed subset of M such that i*: H (M) H (N)

is not epimorphic, where i is the inclusion map from N into M.

Therefore H (N) / 0. By Proposition 6-1 and Proposition 6-2

we know that S\N is a dense connected subset of X. It follows

from Lemma 10-2 that B C N. Therefore B C M, since N is a

subset of M. On the other hand we know that H (MSM) = 0,

since MSM is a continuum subsemigroup with zero and M is a

set of idempotents in S [4]. By Lemma 10-1 we know that

S = MSM. It follows from Theorem 1-9 that M C B, since each

midunit is a ;-maximal idempotent of S. Then we have








M C B C N C M, which implies that M = N. This contradicts

the fact that i* is not epimorphic. This completes the proof.

10-4 Remark. (1) In [12]. Mostert and Shields give a

description of a semigroup on the two-cell where the boundary

of the two-cell relative to the plane is a group,corresponding

to the situation in part (1) of Theorem 10-3.

(2) In [10], Lester gives a description of a semigroup

on the two-cell where the multiplication satisfies the fol-

lowing two conditions: (1) for x and y in the boundary B of

the two-cell to the plane xy = x, and (2) there is a zero

but no other idempotent in the interior of.the two-cell.

Then there exists an I-semigroup T in S such that S = BT.

Also for e and f in B and s in T (es)(ft) = e(st); and if

es = ft then s = t. This corresponds to a special case of

part (2) of Theorem 10-3.

We end this chapter with an example of a semigroup

S with zero and a proper midunit on a two-cell which satisfies
2
S = S.

10-5 Example. Let (a,b) be a discrete space with left zero

multiplication and I = [0,1] denotes the real unit closed

interval with usual multiplication. Then (a,b) x I, with

coordinatewise multiplication, is a compact semigroup. If
1
we set TO = ([a) x I) U ((b) x [0,-]), then TO is a closed

subsemigroup of (a,b] x I. It is not difficult to see that

10 = ((a,0),(b,0)) is a closed ideal in T0. If we let SO be

the Rees quotient /{ (a,0),(b,0)), then SO is a semigroup





70

on a closed unit interval with zero and with a right identity

(a.l) at one endpoint which is not simultaneously a left

identity for S0, as seen in Figure 10-1. Let S = SO x SO

with multiplication be defined by (x,y)(z.w) = (xz,wy). Then
2
it can be verified that S = S with v = ((a.l), (a,l)) is a

proper midunit. Since S = SO x SO, the underlying space of

S is homomorphic to a two-cell.





(a,l)


(b,')
1


Figure 10-1














CHAPTER XI


MARGINALITY OF THE MIDUNITS
IN A CONTINUUM SEMIGROUP

It is a well-known result that if S is a continuum

semigroup with identity which is not a group, then each point

of the maximal group of the identity is marginal in S [6;p.168].

In this chapter we investigate the marginality of the midunits
2
in a continuum semigroup S satisfying S2 = S K. Recall

that a point p in a space X is marginal if and only if for

each open set U containing p, there exists an open set V such

that p E V C U and i*: H*(X) H*(X\V) is an isomorphism,

where i is the inclusion map from X\V into X.

Without the condition S = S / K, a midunit may not be a

marginal point, as it is seen in the following example which

was mentioned in Example 2-1.

11-1 Example. Let S be the closed unit interval [0,1] with

the usual topology. If ab = minimum fa,b,-], then S is a
1
semigroup with 1 as a midunit and 0 as zero. It is easy to
2 1
see that S / S and is not a marginal point.

We proceed with a lemma.

11-2 Lemma. Let S be a compact semigroup with midunit v
2
satisfying S = S / K. Then the following two statements are

equivalent.








(1) H C (vSv)

(2) H n (Sv\vSv)* = C and Hv n (vS\vSv)* = F.

Proof. (1) (2) We know that (vSv)O n (Sv\vSv) =

since (vSv) C vSv. which implies that (vSv) n (Sv\vSv)* =

because (vSv)0 is open. Therefore Hv n (Sv\vSv)* = i, since

by assumption we have H C (vSv) Similarly we can prove

that H n (vS\vSv)* = o.

(2) (1) It is first claimed that

H n [(Sv\vSv)*(vS\vSv)*] = [. Otherwise we let x and y

be a pair of elements such that x E (Sv\vSv)*, y c (vS\vSv)*

and xy E H If we let J be the y-class of the midunit v,

then it follows from Proposition 3-6 that [x,y] C J. It is

true that xv = x and vy = y, because x e (Sv\vSv)* C (Sv)* =

Sv and y c (vS\vSv)* C (vS)* = vS. Through use of Proposition

3-5 we know that vyv E Hv. If z is the inverse of vyv in H v,

then we have x = xv = xvyvz = xyvz = xyz c H But this is a

contradiction since Hv (Sv\vSv)* = O. Next we let

V = S\[(Sv\vSv)* U (vS\vSv)* U (Sv\vSv)*(vS\vSv)*], and we
0
want to prove that H C V C (vSv) By assumption we have

that H n (Sv\vSv)* = ] and H n (vS\vSv)* = O. Also we have

proved that H n [(Sv\vSv)*(vS\vSv)*] = O. It follows that

H C V. In order to prove that V C (vSv)o, it is sufficient

to show that V C vSv, since V is open. Otherwise let z be
2
a point such that z e V\vSv. Since S = S and v is a midunit,

z can be written as z = avb for some pair of elements a and b

in S. It is claimed that av vSv. So we assume that av = vav.

Then z = avb = vavb e vS\vSv, which contradicts the fact that








V n (vS\vSv)* = [. This provesthat av j vSv. Similarly we

can prove that vb e vS\vSv. Then

z = avb = avvb c (Sv\vSv)(vS\vSv) C (Sv\vSv)*(vS\vSv)*.

But this contradicts the fact that V n (Sv\vSv)*(vS\vSv)* = F.

The proof is complete.

It has been proved that if A is a closed subspace of X

and p AO, then p is marginal in X iff p is marginal in A

[11; p.3]. Hence the following theorem follows easily from

Lemma 11-2.

11-3 Theorem. Let S be a continuum semigroup with midunit v

satisfying S2 = S / K. If H n [(SvUvS)\vSv]* = O, then each

point of H is marginal in S.

Proof. It is evident that Hv [(SvUvS)\vSv]* = [ is

equivalent to Hv (Sv\vSv)* = O and H n (vS\vSv)* = . By

Lemma 11-2 we have H C (vSv) Therefore each point of Hv

is marginal in S, since each point of H is marginal in vSv.














CHAPTER XII


ACYCLICITY OF CONTINUUM SEMIGROUPS WITH MIDUNIT

In this chapter we are concerned with the acyclicity

of a continuum semigroup with midunit. A continuum is called

acyclic if it has the cohomology of a point space [18]. It has

been known for some time that a continuum semigroup with left

zero and left identity is acyclic [19]. The following example

of a nonacyclic continuum semigroup with a'zero and a midunit

shows that, in the above, one may not replace the existence

of a left identity with the existence of a midunit.

12-1 Example. Let S = (x.y) I (x+l)2 + y2 = 1) U [(x,y) I0sxsl,

y = 0) with the usual topology as seen in the Figure 9-2. Then

S is a nonacyclic continuum, since H (S) / 0. We define

multiplication on S by

(ab,0) if p = (a,0), q = (b,0), a and b are
nonnegative, and ab denotes the usual
pq = multiplication,

(0,0) otherwise.

Under this multiplication, S satisfies S2 S and has a

zero and a midunit (1,0).

To avoid this triviality we restrict ourselves to the

study of the acyclic properties of those continuum semigroups

satisfying S = S and having zero and a midunit. In working

on this problem we found the following two interesting results

which may be useful in attacking this problem from another

point of view. We proceed with a lemma.
74








12-2 Lemma. Let S be a compact semigroup satisfying S = S

Hin+l(S) { 0, and n (I) = 0 for any closed ideal I in S, where n is a

nonnegative integer. If h is a nonzero element of Hn (+S),

then there exists an idempotent e such that hlSeS / 0.

Proof. It has been proved that if S is a compact semi-
2
group satisfying S = S, then S = SES [8]. Through use of

the Reduction Theorem 1-2 it is possible to select a closed

subset F of E, which is minimal relative to hlSFS f 0. Sup-

pose F is not a singleton set, and let A and B be two nonempty

proper closed subsets of F such that F = A U B. It is easy

to see that SFS = SAS U SBS and SAS A SBS is a closed ideal

in S. By the Mayer-Vietoris Sequence Theorem 1-1 we have the

following exact sequence
'*
n(SAS n SBS) A Hn+(sFS) t x t Hn+(AS) x Hn+(SBS)-.

By the minimal condition on F, we have (t x t ) (hISFS) =
'*
(hISAS, hISBS) = (0,0); thus t x t is not monomorphic. But

SAS n SBS is a closed ideal of S, so that "n(SAS n SBS) = 0
'*
from our assumption, which implies that t x t is monomorphic.
'*
This is a contradiction since we proved already that t x t is

not mcnomorophic. Thus F is a singleton set and the proof is

complete.

12-3 Theorem. Let S be a continuum semigroup with zero. Then

the following two statements hold.

(1) If S satisfies S = ESUSE and h is a nonzero element

of H (S), then there exists an idempotent e such

that hjSeS / 0.








(2) If S satisfies S = ESE and h is a nonzero element

of H (S), then there exists an idempotent e such

that hjSeS 1 0.

Proof. (1) It is routine to check that each ideal I of

S is connected since S = ESUSE, so that f(I) = 0. The result

then follows Lemma 12-2.

(2) It has been proved by Cohen and Koch [4] that if S

is a continuum semigroup with zero satisfying S = ESE, and

I is a closed ideal in S, then H (I) = 0. By Lemma 12-2, we

know the proof is complete.

Recall that in Chapter VII we introduced a method of

constructing new semigroups with midunit from given known

semigroups. That is, if S is a semigroup and v is an idem-

potent in S, we define a multiplication "*" on S by a*b = avb.

In particular, we proved in Proposition 7-1 and Proposition

7-2 that if S is a semigroup satisfying SvS = S for an idem-

potent element v of S and S has a zero, then (S;*) is a

semigroup satisfying S*S = S, has v as a midunit, and has the

same zero. For more details the reader is referred to

Chapter VII.

From Theorem 12-3, we see that each example of a con-

tinuum semigroup with zero satisfying either S = ESUSE with

H (S) / 0 or S = ESE with H2(S) / 0 gives rise to a nonacyclic

continuum semigroup with zero and midunit satisfying S = S;

namely (SeS,*).

A. Hudson has given two examples of nonacyclic semigroups

with zero, one satisfying S = SE with H (S) / 0 and the







other satisfying S = ESE with H2(S) V 0 [7]. The underlying

space of the first example is a circle with two intervals

issuing from a common point of the circle, as seen in Figure

9-1. The endpoints of those intervals are each midunits for

S (use of Proposition 4-2 may be helpful in verifying this).

We have then a nonacyclic continuum semigroup with zero and
2
midunit satisfying S = S The underlying space of the

second example is a two-sphere with four intervals issuing

from a common point from the sphere. A subsemigroup consisting

of the two-sphere and two of the issuing intervals will have

the endpoints of the intervals as midunits. This gives a

second example of a nonacyclic continuum semigroup with zero

and midunit satisfying S = S.

It might be interesting to seek some sufficient conditions

for continuum semigroup with zero and midunit under which S

has to be acyclic. Toward this direction we introduce first

a class of semigroups which we will call continuously factor-

able semigroups (see Definition 12-4). After a brief intro-

ductionof such class of semigroups, we prove that if S is a

continuously factorable continuum semigroup, then all of its

cohomology is concentrated in its minimal ideal. It follows

that if S is a continuously factorable continuum semigroup

with zero and midunit, then S is acyclic. We proceed with a

definition.

12-4 Definition. Let (S; m) be a topological semigroup with

multiplication m. Then m and S are called continuously

factorable if there exists a map r: S S x S such that








mon(s) = s for all elements s of S. We call n a continuous

factorization of S.

The following are examples of continuously factorable

semigroups.

12-5 Example. (1) A semigroup with left identity v is con-

tinuously factorable. For we could simply define p(s) = (v,s)

for all s of S. It is not difficult to check that q is a

continuous factorization of S. A dual result holds for a

semigroup with right identity. In particular, a clan is

always continuously factorable.

(2) If S is a semigroup satisfying S.= E, then S is

continuously factorable. Naturally we define T(s) = (s,s)

for all elements s of S to obtain a continuous factorization

of S.

(3) If S is a compact Clifford semigroup, then it is

continuously factorable. Recall that a semigroup is called a

Clifford semigroup if it is the union of groups. It has been

proved that if S is a compact Clifford semigroup, then the

map e: S E, which assigns to each s of S the identity of the

group containing it, is continuous [6; p.44]. If we define

r(s) = (s,e(s)) for all elements of S, then p is a continuous

factorization of S.

We proceed with a theorem showing that all the cohomology

of a continuum semigroup with midunit is concentrated in its

minimal ideal if its multiplication is continuously factorable.

12-6 Theorem. Let (S;m) be a continuously factorable continuum





79

semigroup with midunit. If K is its minimal ideal, then the

map i: K S induces an isomorphism i*: H*(S) H*(K).

Proof. If T = vSv, where v is a midunit of S, then T

is a clan with identity v. Since T n K / 0, there is an

idempotent f in T n K. Let the two maps X and Xf from

S x S into S x S be defined by v ((x,y)) = (xv,y) and

Xf((x,y)) = (xf,y) for all elements (x,y) of S x S. On the

other hand we consider a map F: (S x. S) x T S x S defined

by F((x,y),t) = (xt,y) for all elements (x,y) in S x S and

elements t in T. Then for all elements (x,y) of S x S we

have F((x,y),v) = (xv,y) = X((x,y)) and F-((x,y),f) = (xf,y) =

Xf((x,y)). By applying Homotopy Theorem 1-4, we obtain
*
f = : H (S X S) H (S X S).

Now let us consider the following diagram
X
S 7 S x S S x S m S, where m is the
xf

multiplication of S and r is a continuous factorization for m.

For convenience we write 1 componentwise as r(s) = (l(s)) 72(s))

Then moX op(s) = mo v(( l), (s) )

= m('1 (s)v,12 (s))

= p1(s)v 12(s)

= p1(s)p2 (s)

= mop (s)

= s

that is, moXv o is the identity map IS on S. On the other

hand we have

(moX fop)* = *oXf*om*

= r*oX *om*







= (moxvor)*

= (I S)* = IH*(S) '

which implies that (moxfo9)*: H*(S) H*(S) is an isomorphism.

Since moXfop(s) = Cl(s)fr2(s) e SfS for all s of S, we

may define t: S SfS by t(s) = moXfofl(s) = nl(s)fr2(s) for

all s of S. If i is the inclusion map of SfS into S, then it

can be proved that iot(s) = moXfor(s) for all s of S. We

then have IH.(S) = (mofo7)* = (iot)* = ,*oi*, which implies
H*(S) fr
that i*: H*(S) H*(SfS) is a monomorphism, since IH*(S) is

an isomorphism. It is easy to see that K = SfS, since

f e K and K is a minimal ideal of S. Since K is a retract of

S [17], we also have that i* is also a epimorphism. The

proof is complete.

The following corollary follows easily from Theorem 12-6.

12-7 Corollary. If S is a continuously factorable continuum

semigroup with midunit and zero, then S is acyclic.

12-8 Remark. Corollary 12-7 shows that the examples of

nonacyclic continuum semigroups with zero and midunit given

earlier in this chapter are not continuously factorable.













REFERENCES


1. Ault, J. E., "Semigroups with midunits," Semigroup Forum,
6(1973), 346-351.

2. Clifford, A. H. and Preston, G. B., The Algebraic Theory
of Semigroups, Vol. 1, Mathematical Surveys, 7, Amer.
Math. Soc., 1961.

3. Cohen, H., "A cohomological definition of dimension for
locally compact Hausdorff spaces," Duke Math. J. 21(1954),
209-224.

4. Cohen, H. and Koch, R. J., "Acyclic semigroups and multi-
plications on two-manifolds," Trans. Amer. Math. Soc.
118(1965), 420-427.

5. Faucett, W. M., Koch, R. J., and Numakura, K., "Comple-
ments of maximal ideals in compact semigroups," Duke Math.
22(1955), 655-662.

6. Hofmann, K. H. and Mostert, P. S., Elements of Compact
Semigroups, Charles, E. Merrill, Inc. Columbus, Ohio, 1966.

7. Hudson, A. L., "Example of a nonacyclic continuum semi-
group with zero and S = ESE," Proc. Amer. Math. Soc.
14(1963), 648-653.

8. Koch, R. J. and Wallace, A. D., "Maximal ideals in compact
semigroups," Duke Math. J. 21(1954), 681-686.

9. Koch, R. J. and Wallace, A. D., "Admissibility of semi-
group structures on continue," Trans. Amer. Math. Soc.
88(1958), 277-287.

10. Lester, A., "Some semigroups on the two-cell," Proc. Amer.
Math. Soc. 10(1959), 648-655.

11. McCharen, J. D., Maximal Elements in Compact Semigroups,
Dissertation, Louisiana State University, 1969.

12. Mostert, P. S., and Shields, A. L., "On the structure of
semigroups on a compact manifold with boundary," Ann. of
Math. 65(1957), 117-143.

13. Sigmon, K., Algebraic Topology Notes, Univ. of Florida,
1971.






82

14. Sigmon, K., "A strong homotopy axiom of Alexander
cohomology," Proc. Amer. Math. Soc. 31(1972), 271-275.

15. Stadtlander, D., Topological Algebra Notes, Univ. of
Florida, 1972.

16. Wallace, A. D., "Inversesin Euclidean Mobs," Math J.
Okayama Univ., 3(1953), 23-28.

17. Wallace, A. D., "Retractions in semigroups," Pacific J.
Math. 7(1957), 1513-1517.

18. Wallace, A. D., "A theorem on acyclicity," Bull. Amer.
Math. Soc. 67(1961), 123-124.

19. Wallace, A. D., "Acyclicity of compact connected semi-
groups," Fund. Math. 50(1961), 99-105.

20. Wallace, A. D., Project Mob, Univ. of Florida, 1965.

21. Whyburn, G. T., Analytic Topology, Amer. Math. Soc.,
New York, 1952.

22. Yamada, Miyuki, "A note on middle unitary semigroups,"
Kodai. Math. Sem. Rep. 7(1955), 49-52.













BIOGRAPHICAL SKETCH


Hung-tzaw Hu was born November 5, 1941, at Chungking,

China. In June, 1967 he received the degree of Bachelor of

Science with a major in Mathematics from the National Taiwan

University. After his graduation he served as Mathematics

Instructor during his military service in the Army of the

Republic of China. Following his discharge from the army,

September, 1968, he came to the United States and enrolled

in the Graduate School of the University of Florida. In

June, 1970, he received the degree of Master of Science

with a major in Mathematics from the University of Florida.

From June, 1970 until the present time he has pursued his

work toward the degree of Doctor of Philosophy.









I certify that I have.read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




N. Sigmoh, Chairman
Assistant Professor of
Mathematics



I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




T. T. Bowman
Assistant Professor of
Mathematics



I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




J. A. Draper /
Assistant Professor of
Mathematics



I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor f Philosophy.



E. E. Shult
Professor of Mathematics









I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




D. S. Stadtlander
Assistant Professor of
Mathematics



I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.




Mark C. Yang .
Assistant Professor of
Statistics ( /



This dissertation was submitted to the Department of Mathematics
in the College of Arts and Sciences and to the Graduate
Council, and was accepted as partial fulfillment of the re-
quirements for the degree of Doctor of Philosophy.

August, 1974


Dean, Graduate School




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