CONTINUUM SEMIGROUPS WITH MIDUNIT
by
HUNGTZAW 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 pCLASS OF MIDUNITS IN A COMPACT SEMIGROUP..15
IV FINITE SEMIGROUPS WITH MIDUNIT..................20
V THE CODIMENSION OF THE yCLASS OF MIDUNITS IN
A CONTINUUM SEMIGROUP...............................28
VI THE COMPLEMENT OF THE pCLASS OF MIDUNITS AND
MINIMAL CONTINUUM SEMIGROUPS WITH MIDUNIT........34
VII A METHOD OF CONSTRUCTING SEMIGROUPS WITH
MIDUNIT................ ........ .............39
VIII SEMIGROUPS ON DSPACES..........................44
IX EXAMPLES OF SEMGROUPS ADMITTING NO PROPER
MIDUNIT............. ... ........................ 54
X SEMIGROUPS WITH MIDUNIT ON THE TWOCELL......... 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
Hungtzaw 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 pclass of the
midunits, then J is the unique maximal pclass 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
pclass, but S does not have a midunit. The cohomology used
is that of AlexanderWallaceSpanier 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 pclass 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 pmaximal idempotent,
J is the 5class 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 twocell 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 twocell 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 pclass 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 pclass 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 gclass 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 pclass
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 pclass 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 pmaximal idempotent, J is the pclass 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 Dspaces, 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
twocell with midunit and zero satisfying S = S. The
cohomology theory used is that of AlexanderWallaceSpanier
with arbitrary coefficient group unless otherwise specified.
Let S be a semigroup on the twocell 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
onetoone, 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 r1 = {(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 quasiorder is a reflexive
transitive relation. A partial order is an antisymmetric
quasiorder. An equivalence relation is a symmetric
quasiorder. 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 AlexanderWallace
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].
11 MayerVietoris 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).
12 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.
13 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].
14 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. twosided) 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 twosided
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.
15 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 pmaximal, then a is Imaximal as well as Rmaximal
[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].
17 Theorem. Let S be a continuum semigroup satisfying
S2 = S. If e is a pmaximal 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
18 Theorem. Let S be a continuum semigroup with S = S,
a an fmaximal 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
pmaximal then a is maximal. It follows from Theorem 16.
2
that if S is a continuum semigroup satisfying S = S ; K,
and b is a pmaximal element of S, then b is not a local
separating point of S.
Also under some conditions on S, a pmaximal idempotent
cannot have a twodimensional Euclidean neighborhood as
follows.
19 Theorem. Let S be a continuum semigroup satisfying
S = ESE, and suppose e is a pmaximal 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 onesided 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.
21 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 oneside1 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].
22 Lemma. Let S be a semigroup with midunit. If a and b
are midunits of S, then ab is an idempotent midunit.
23 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 22 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 onetoone 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.
24 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 22 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 onetoone
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, twosided) ideal containing b is
denoted by L(b), (R(b), J(b)), and the Mclass which contains
b is denoted by Hb. If b is an idempotent then Hb is the
maximal subgroup of S which contains b.
25 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 11class. 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.
26 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 25, 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 24.
CHAPTER III
THE 9CLASS 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 gclass 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 gclass
as described in Proposition 31. From Theorem 16 we have
that if S is a compact semigroup satisfying S = S and z is
a 9maximal element of S then there exists a pair of idempo
tents e and f such that z = ezf.
31 Proposition. Let S be a compact semigroup with midunit
satisfying S = S. Then
(1) There is cnly one maximal 2class in S and this
9class 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 9class,
which is the only maximal 9class in S. Let e and f be two
idempotents such that z = ezf. Then z = ez so that by Lemma
23 we know that z is an idempotent.
16
32 Remark. Without the algebraic condition S2 = S, none of
the results in Proposition 31 remain true. Recall in Example
2 1 1
21 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 yclass.
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].
33 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 15. 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 yclass of the midunits, M be the set of all
midunits, and H be the Hclass of midunit v.
34 Proposition. x E J iff SxS = S.
Proof. (=) If x is an element of J, then by Theorem 16
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
5class 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 31.
35 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 34 we have that SxS = S.
Then SxS = SxvS since v is a midunit; thus SxvS = S, which
implies that xv e J by Proposition 34. Hence we have
L(xv) C L(v) C J(xv), v'* ",h implies that L(xv) = L(v) by
Lemma 33. Similarly we can prove R(vx) = R(v). Hence it
follows that vxv H .
v
36 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
35. From Proposition 31 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 34.
(=) It follows easily from Proposition 34.
37 Proposition. M = J n E.
Proof. Let v be a midunit and u an element of J n E.
It follows from Proposition 35 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 23 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 31.
38 Proposition. J is a completely simple subsemigroup.
Proof. Since J is a 9class in a compact semigroup
hence J is a closed set in S. From Proposition 36 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 37. 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.
39 Theorem. Let S be a compact semigroup with midunit
satisfying S = S. Let J be the gclass of the midunits and
M be the set of all midunits of S. Then
(1) J is the unique maximal 9class 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 24,
Proposition 31, and Proposition 37. Analogous to the
proof of Proposition 24, 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 26.
CHAPTER IV
FINITE SEMIGROUPS WITH MIDUNIT
In this chapter we study finite semigroups which satisfy
2
S = S with a unique maximal pclass which is a completely
simple subsemigroup. Recall that we proved in Theorem 39
that in a compact semigroup S with midunit satisfying S = S,
the pclass 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 pclass, 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.
41 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
42 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.
43 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).
44 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 = m1. If
v is neither a left identity nor a right identity we know
that vS\vSv i and Sv\vSv / n by Proposition 43. 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 + (ml) = m + 1.
That is a contradiction since vS U Sv is a subset of S.
Therefore
n(vSv) < n(S) 1.
45 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 44 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 43 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 44 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
42, 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 pclass of the pmaximal element v in S. Thus
both a and b are pmaximal elements of S. By Theorem 16,
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.
46 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 41
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 pclass which is a
completely simple subsemigroup, but S does not have a mid
unit.
47 Example. Let S = (v,a,b,c,0] with multiplication as in
the following table.
Table 42
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 9class 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 PCLASS OF
MIDUNITS IN A CONTINUUM SEMIGROUP
This chapter is devoted to proving that the pclass 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 hF # 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].
51 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 11classes in
compact semigroups.
52 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 15. 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 Mclass 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 hclass of a in S with cd(Ha) = n, then
Hn(Ha;Z) / 0.
53 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 52 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 51. 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 Hclasses.
a
We know that the hclass 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
hclasses. 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
51 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.
54 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 13, 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*.
Hnl(R) Hnl([R])
k* *
Hn1 ( ) (RA)* Hn1 ([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) > nl. But we know that
cd(X)cd(X x Y) = n, therefore cd(X) = cd(X x Y). Hence the
proof is complete.
55 Corollary. Let S be a compact semigroup with midunit of
2
finite codimension satisfying S = S T K. Let J be the
yclass of the midunits, M be the set of all midunits and H
be the Mclass of midunit v. If H is totally disconnected
for some midunit, then cdJ = cdM.
Proof. It is not difficult to prove from Theorem 39
that J is homeomorphic to the space M x H The conclusion
follows from Proposition 54.
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.
56 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 pclass J of
midunits is totally disconnected.
Proof. As before we let M be the set of all midunits
and H be the Ilclass of the midunit v. From Thcore.m 53
we know that Hv is totally disconnected for each midunit v.
In order to prove that J is totally disconnected it suffices
by Corollary 55 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
gclass 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 MayerVietoris Sequence Theorem 11 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 yCLASS 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 pclass 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 65) 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.
61 Proposition. Let S be a continuum semigroup satisfying
2
S = S / K with midunit and J be the pclass of the midunits.
Then S\J is a dense connected maximal ideal of S.
Proof. From Proposition 36 and Theorem 39 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 36 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.
62 Proposition. Let S be a continuum semigroup with midunit
satisfying S2 = S / K and J be the yclass 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 61 we know that S\J is connected and
[S\J]* = S. Therefore S\N is connected.
63 Proposition. Let S be a continuum semigroup with mid
2
unit satisfying S = S K and J be the pclass 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 36 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.
64 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.
65 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].
66 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 pclass 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 39
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 Jmaximal idempotent, J is the 1class 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.
71 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,
twoside) ideal of S, then N is a right (left, twosided)
ideal of (S;*).
72 Proposition. The following two statements are equivalent:
(1) N is a minimal right (left, twosided) ideal in S.
(2) N is a minimal right (left, twosided) 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, twosided) ideal of S which
contains z, and L(z),(R(z), J(z)) be the smallest left (right,
twosided) 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.
73 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
74 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 9class 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.
75 Theorem. Let S be a compact semigroup with minimal ideal
K. Let v be a 9maximal idempotent of S such that its
pclass 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 pmaximal
idempotent in T. Let J be the yclass 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 pclass of v in (T;*), then by Theorem 39,
(J;*) is isomorphic to a cartesian product of a rectangular
band and H with coordinatewise multiplication. By Proposi
tion 74 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 72 and Theorem 56.
CHAPTER VIII
SEMIGROUPS ON DSPACES
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 Dspace (to be defined in Definition
86) 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 21 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 21 satisfies, is S2 = S K. It is not difficult
to prove the following proposition.
81 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
82 Definition. Let X be a continuum. A point p of X is a
Dpoint 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 Dpoint 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 nonempty 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 quastorder 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 quasiorder 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.
83 Lemma. Let X be a continuum with a Dpoint 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 quasiorder 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.
84 Proposition. Let X be a continuum with a Dpoint p and s
be the induced quasiorder. 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 quasiorder on X
we have c < b. By Lemma 83 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.
85 Remark. There exists a continuum with a Dpoint whose
induced quasiorder 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 81
Then X is a continuum with a Dpoint p = ( 1). But
the induced quasiorder 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 81 we know that is
not a closed relation.
86 Definition. A continuum X with a Dpoint p is called a
Dspace if the induced quasiorder s is a closed relation on
X.
We have seen in Remark 85 that X is not locally connect
ed at p, while the induced quasiorder s is not closed. But
if X is a Dspace, then X is locally connected at its Dpoint
p.
87 Proposition. If X is a Dspace with Dpoint p, then X is
locally connected at p.
Proof. Let V be an open proper subset containing the
Dpoint 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 Dspace with a Dpoint 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.
88 Theorem. Let X be a Dspace with Dpoint p. Then the
following statements hold.
(1) X is irreducible between its Dpoint 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 Dpoint 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 Dpoint, 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 Dpoint
p, but neither of them contains the other as a subset, which
contradicts the fact that p is a Dpoint. 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.
89 Remark. (1) The convese of part (1) in Theorem 88'is
not true. The space X in Remark 85 is irreducible between
1 1
points ( ,1) and ( ,1) but is not a Dspace.
(2) In application 811, there are two Dspaces which
are not locally connected.
810 Theorem. Let S be a Dspace with a Dpoint p whose max
imal level set M contains more than one point. If S satisfies
S = S / K. then the Dpoint 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 16 that there exists an
idempotent e which is a pmaximal element of S since S is
2
compact and S = S. It follows from Theorem 17 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 88 we know that
every point, except the Dpoint p, is a weak cut point of S.
Also it follows from Theorem 17 that e is the only pmaximal
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 Dpoint 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 pmaximal idempotent in S.
811 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 82
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 Dspace with
Dpoint v = (,0) its corresponding maximal level set is
M = ((x,y) x = 0, 1 s. y 1].
(2) Let S = ((e2 itet)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 83
2
If S satisfies S2 = S K then p is either a left or a
right identity for S. Since S is a Dspace with a Dpoint
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 Dspace, which satisfies
S2 = S K and has a proper midunit.
812 Example. This example is constructed as follows. Let
T = [v.a.b.0] with multiplication defined in the Table 41.
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 84
It is not difficult to see that this space is not a
Dspace, because none of the points in S can be a Dpoint.
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 92. 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
91. 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.
91 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 12. 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 MayerVietoris Sequence
Theorem 11, 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 MayerVietoris
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.
92 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 91 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 91
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 91.
93 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 91.
94 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 92
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 nontrivial
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 MayerVietoris 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 92.
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 Rmaximal idempotents. It
follows from Theorem 18 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 92). 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 91 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 18 that v has to be the
endpoint of the closed interval I in Figure 92. Also
since vSv meets K and C C K, one has [S\K] C vSv, i.e., v is
a twosided 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 nontrivially 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 twosided identity for K.
Indeed, v is a twosided 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.
95 Example. Let S be the space of tangent circles converging
to a point e, as illustrated in the following figure.
Figure 93
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 Isemilocally 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 Isemilocally connected at p.
96 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 94
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 nontrivial 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 31 we
have that v is a Ymaximal idempotent in S. By Theorem 18
we know that a 1maximal 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 nonlocal 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 94. 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 Isemilocally 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 94. 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 18 that no
Pmaximal element is a local separating point of S and hence
p must be the unique pmaximal element of S. By Theorem 16
we know p must be an idempotent. Therefore one has S = SeS.
The conclusion follows from Proposition 71, Proposition 72
and preceding argument.
CHAPTER X
SEMIGROUPS WITH MIDUNIT ON THE TWOCELL
In this chapter we investigate the structure of those
2
semigroups with midunit and zero satisfying S = S on a top
ological twocell. We let S be a twocell 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 twosided 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 twocell which
satisfies S = S and contains a proper midunit.
Let C be a complex plane with the usual topology, and X
be the unit twocell 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.
101 Lemma. Let X be the unit twocell 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].
102 Lemma. Let X be the unit twocell 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 nonempty 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.
103 Theorem. Let S be a twocell 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 pclass 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 twosided 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 61 and Proposition 62 we
know that S\H is a dense connected subset of X. It follows
from Lemma 102 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 twosided identity v and zero [19]. It
follows from Lemma 101 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 twosided identity for S,
therefore u = v and M = (v]. Since v is a twosided 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 39 that J = H v. The proof of (1) is
complete.
(2) By Proposition 61 and Proposition 62 we know that
S\M is a dense connected subset of X. It follows from Lemma
102 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 101
we know that S = MSM. It follows from Theorem 19 that
M C B, since each midunit is a ymaximal idempotent element of
S. But since we have proved that B C M, one has M = B. By
Proposition 24 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 nontrivial
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 39 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 55, 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 61 and Proposition 62
we know that S\N is a dense connected subset of X. It follows
from Lemma 102 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 101 we know that
S = MSM. It follows from Theorem 19 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.
104 Remark. (1) In [12]. Mostert and Shields give a
description of a semigroup on the twocell where the boundary
of the twocell relative to the plane is a group,corresponding
to the situation in part (1) of Theorem 103.
(2) In [10], Lester gives a description of a semigroup
on the twocell where the multiplication satisfies the fol
lowing two conditions: (1) for x and y in the boundary B of
the twocell to the plane xy = x, and (2) there is a zero
but no other idempotent in the interior of.the twocell.
Then there exists an Isemigroup 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 103.
We end this chapter with an example of a semigroup
S with zero and a proper midunit on a twocell which satisfies
2
S = S.
105 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 101. 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 twocell.
(a,l)
(b,')
1
Figure 101
CHAPTER XI
MARGINALITY OF THE MIDUNITS
IN A CONTINUUM SEMIGROUP
It is a wellknown 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 21.
111 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.
112 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 yclass of the midunit v,
then it follows from Proposition 36 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
35 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 112.
113 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 112 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.
121 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 92. 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
122 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 12 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 MayerVietoris Sequence Theorem 11 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.
123 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 122.
(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 122, 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 71 and Proposition
72 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 123, 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
91. The endpoints of those intervals are each midunits for
S (use of Proposition 42 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 twosphere with four intervals issuing
from a common point from the sphere. A subsemigroup consisting
of the twosphere 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 124). 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.
124 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.
125 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.
126 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 14, 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 126.
127 Corollary. If S is a continuously factorable continuum
semigroup with midunit and zero, then S is acyclic.
128 Remark. Corollary 127 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), 346351.
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),
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4. Cohen, H. and Koch, R. J., "Acyclic semigroups and multi
plications on twomanifolds," Trans. Amer. Math. Soc.
118(1965), 420427.
5. Faucett, W. M., Koch, R. J., and Numakura, K., "Comple
ments of maximal ideals in compact semigroups," Duke Math.
22(1955), 655662.
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), 648653.
8. Koch, R. J. and Wallace, A. D., "Maximal ideals in compact
semigroups," Duke Math. J. 21(1954), 681686.
9. Koch, R. J. and Wallace, A. D., "Admissibility of semi
group structures on continue," Trans. Amer. Math. Soc.
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10. Lester, A., "Some semigroups on the twocell," Proc. Amer.
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11. McCharen, J. D., Maximal Elements in Compact Semigroups,
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14. Sigmon, K., "A strong homotopy axiom of Alexander
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20. Wallace, A. D., Project Mob, Univ. of Florida, 1965.
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BIOGRAPHICAL SKETCH
Hungtzaw 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
