WHEN IS AN ALTER EGO A MINIMAL DUALIZING STRUCTURE?

by
Bashet Ullah
M.S., Jahangirnagar University, 2006

THESIS SUBMITTED IN PARTIAL FULFILLMENT OF
THE REQUIREMENTS FOR THE DEGREE OF
MASTER OF SCIENCE
IN
MATHEMATICS

UNIVERSITY OF NORTHERN BRITISH COLUMBIA
August 2018
c Bashet Ullah, 2018

Abstract

We show that a particular five-element algebra is dualized by a particular alter-ego. We also show
that, in some limited sense, the alter ego we chose is minimal. The algebra that we examine is a
{0, 1}-valued unary algebra with 0. It has meet and join as homomorphisms.

i

TABLE OF CONTENTS

Abstract

i

Table of Contents

ii

List of Figures

iv

List of Tables

v

Dedication

vi

Acknowledgement

vii

1

Introduction

1

2

Preliminary and Background Materials
4
2.1 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Quasivarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.4 Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.5 Dualizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.6 The Interpolation Condition and Duality Theorems . . . . . . . . . . . . . . . . . 16

3

An Example of a Finite Algebra and Dualizing Structure
3.1 The Five-Element Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 An Alter Ego . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Morphism Ranges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2 The General Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

Minimality
45
4.1 Non-evaluation morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.2 When does the Map αi Respect Q j for j 6= i? . . . . . . . . . . . . . . . . . . . . 52

ii

18
18
21
22
23
23

5

Conclusion
73
5.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
5.2 Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Bibliography

76

iii

LIST OF FIGURES

2.1
2.2
2.3

The Dual of an algebra is a topological structure. . . . . . . . . . . . . . . . . . . 13
The Dual of a topological structure is an algebra. . . . . . . . . . . . . . . . . . . 14
Dualizability of an algebra occurs when eA : A → E(D(A)) is an isomorphism for
all A. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.1 The Lattice Diagram for the five-element algebra M. . . . . . . . . . . . . . . . .
3.2 Diagram for range(α) = {0, 1, 2, 3, 7}. . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Diagram for range(α) = {0, 1, 2, 3}. . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Diagram for range(α) = {0, 2, 3, 7}. . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Diagram for range(α) = {0, 1, 3, 7}. . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Diagram for range(α) = {0, 2, 3}. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7 Diagram for range(α) = {0, 1, 3}. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8 Diagram for range(α) = {0, 3, 7}. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.9 Diagram for range(α) = {0, 2, 7}. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.10 Diagram for range(α) = {0, 1, 7}. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.11 Diagram for range(α) = {0, 3}. . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.12 Diagram for range(α) = {0, 2}. . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.13 Diagram for range(α) = {0, 1}. . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.14 Diagram for range(α) = {0, 7}. . . . . . . . . . . . . . . . . . . . . . . . . . . .

20
25
28
30
32
33
35
36
37
38
39
40
41
43

The Lattice Diagram for S01 (Q1 ). . . . . . . . . . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q2 ). . . . . . . . . . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q4 ). . . . . . . . . . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q3 ), and S01 (Q4 ). . . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q5 ), S01 (Q6 ), and S01 (Q7 ). . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q8 ), and S01 (Q9 ). . . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q10 ), and S01 (Q11 ). . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q12 ). . . . . . . . . . . . . . . . . . . . . . . . . . .
The Lattice Diagram for S01 (Q13 ). . . . . . . . . . . . . . . . . . . . . . . . . . .

54
57
59
67
68
69
70
71
72

4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9

iv

LIST OF TABLES

2.1
2.2

A is a {0, 1}-valued unary algebra. . . . . . . . . . . . . . . . . . . . . . . . . . . 6
The escalator algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.1
3.2

The five-element algebra M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
Relations Qi in the alter-ego M. Qi omits βi but satisfies the co-ordinate restrictions. 21

4.1 Elements βi , ui , vi for proof of the Lemma 4.1.2 . . . . . . . . . . . . . . . . . . .
4.2 Elements bi , ci , di for proof of the Lemma 4.1.2 . . . . . . . . . . . . . . . . . . .
4.3 Homomorphisms of Q2 on {0, 1}-elements. . . . . . . . . . . . . . . . . . . . . .
4.4 Up-sets table of S01 (Q1 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Homomorphism table of S01 (Q1 ) . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Up-sets table of S01 (Q2 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 Homomorphism table of S01 (Q2 ) . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8 Up-sets table of S01 (Q4 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.9 Homomorphism table of S01 (Q4 ) . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.10 The map αi respects Q j for j 6= i (Part I). . . . . . . . . . . . . . . . . . . . . . . .
4.11 Where αi respects Q j (Part II). . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.12 Where αi respects Q j (Part III). . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v

48
49
51
55
55
56
58
60
60
64
65
66

Dedication
To my late Grandfather.

vi

Acknowledgement
First of all, all praise be to the Almighty God. I could never done my thesis without His grace and
mercy. Then I would like to thank my original Supervisor Dr. Jennifer Hyndman for taking care
of me past few years. I sincerely appreciate all the help she gave. Moreover, I want to special thank
my current Supervisor Dr. David Casperson for willingly helping me with my thesis. There are
no words to express how much help he has been to me. I am truly grateful for all his help. Further, I
thank my committee member Dr. Jernej Polajnar and former committee member Dr. Iliya Bluskov
for their useful help. I thank the University of Northern British Columbia for all help. I am also
thankful to Joya Danyluk, Brett Kelly, Jean Bowen, Brian Schaan, Akintunde Akinola, Andrew
Agbonigha, Erin Beveridge and Marva Byfield for their practical suggestions. I appreciate my
respected parents, parents-in-law, my beloved wife Dilruba Akter and all my family members for
their unconditional love and care throughout. Finally, I wish to thank all my friends for their
encouragement and support.

vii

Chapter 1

Introduction

A new branch of mathematics, called universal algebra, was first developed in Garrett Birkhoff’s
1935 paper, “On the structure of Abstract Algebra” [15]. He stated a famous theorem, now call
the Birkhoff’s theorem. It says that the equational classes of algebras are closed under H, S, and
P; where H, S, and P stand for the closure operations of homomorphism, subalgebra, and product
respectively. Duality starts with Birkhoff for finite distributive lattices. Later Duality theories
were developed. These included Pontryagin’s duality for Abelian groups and Stone’s duality for
Boolean algebras [10]. In 1975 H.A. Priestley provided a duality theory for distributive lattices [8].
An overview of duality theory from 1980 to 1992 was submitted by Brian A. Davey in his paper,
“Duality Theory on Ten Dollars a Day” [9].
Now we discuss work related to dualizability, full dualizability, strong dualizability, unary
algebras, unary algebras with zero, rank, and finitely based quasi-equational theories. In 2000
Jennifer Hyndman and Ross Willard displayed the first known example 3-element algebra which
is dualizable but not fully dualizable by any alter-ego of bounded cardinality [2]. In 2002 Jennifer
Hyndman showed that mono-unary algebras have rank at most two and are thus strongly dualiz1

able [7]. In 2003 Clark, Davey, and Pitkethly classified of three-element unary algebras that are
dualizable [13]. In 2004, J. Hyndman showed that a finite unary algebra with a certain positive
primitive formula does not have enough algebraic operations, and consequently does not have a
finite basis for its quasi-equation [5]. In 2005 J. Hyndman and J.G. Pitkethly showed that, within
the class of three-element unary algebras, there is a tight connection between a finitely based quasiequational theory, finite rank, enough algebraic operations and a speacial injectivity condition [6].
Under the supervision of Jennifer Hyndman, in 2006 E. Beveridge generalized the work of J. Hyndman and R. Willard [2] and defined escalator algebras that have infinite rank, and are dualizable
but not strongly dualizable in her master’s thesis [16]. In 2006 E. Beveridge, D. Caspersion, J.
Hyndman, and T. Niven provided sufficient conditions on a finite algebra to prove that certain
unary algebras are not strongly dualizable [18]. In 2015 D. Casperson, J. Hyndman, J. Mason, J.B.
Nation, and B. Schaan looked at {0, 1}-valued unary algebras with zero with respect to having a
finite for the quasi-equations [3]. Under the supervision of Jennifer Hyndman, in 2014 B. Schaan
provided results regarding the natural dualisability of certain {0, 1}-valued unary algebras with
zero in his master’s thesis [19]. In 2014 D. Casperson, J. Hyndman, and B. Schaan defined tangled
functions which need to be included for an extension of the algebra to be dualizable [4]. Under
the supervision of Jennifer Hyndman, in 2016 J. Danyluk looked at {0, 1}-valued finite unary algebras with zero where meet is defined on the algebra and join is partially defined in her master’s
thesis [20].
Our result fits in to the works by Beveridge [16] and Schaan [19]. In this thesis we look for
when an alter ego is a minimal dualizing structure. In Chapter 2, we present preliminary materials
such as notations, definitions, examples, theorems for algebras, lattices, quasivarieties, topology,
and dualizability. In Chapter 3, we define a five-element algebra, an alter ego and conclude that our
defined alter ego satisfies the interpolation condition relative to the algebra and thus dualizes the
algebra. In Chapter 4, we show that the set of relations is a necessary part of a dualizing structure
2

for a five-element algebra by finding a specific morphism that is a non-evaluation morphism. In
Chapter 5, we summarize the main points of this document and give our future research idea.

3

Chapter 2

Preliminary and Background Materials

2.1

Algebras

In this section we discuss the concept and notions of algebras from universal algebra. All materials
came from [1] which are used throughout this paper.
First we start with definitions and examples of algebras. If A is a non-empty set, called the
universe or underlying set of A and F is a language or type of algebras then an ordered pair
hA, Fi
is written A and is called an algebra of type F , where F is a family of finitary operations on A
indexed by F such that there is an n-ary operation f A on A if for every n-ary function symbol
f ∈ F . This non-negative integer n is called the arity of f and f A ’s are called fundamental
operations of A. An operation whose arity is one is called unary. Similarly, an operation f on A
is called a binary or nullary if its arity is two or zero respectively. If all of an algebra’s operations
are unary, then an algebra A is also called unary.
4

For example an algebra hG, ·G , −1G , 1i or hG, ·,−1 , 1i with a binary, a unary, and a nullary
operation respectively can define a group if the following three identities:

• x1 · (x2 · x3 ) = (x1 · x2 ) · x3 ,
• x1 · 1 = x1 = 1 · x1 , and
• x1 · x1−1 = 1 = x1−1 · x1
are true for all x1 , x2 , x3 ∈ G where the language of type F is {·,−1 , 1}. If x1 · x2 = x2 · x1 then the
group G is called Abelian or commutative.
Similarly, a semigroup is an algebra hG, ·i satisfying
x1 · (x2 · x3 ) = (x1 · x2 ) · x3
for x1 , x2 , x3 ∈ G.
For another example, we define a ring of type F = {+, ·, −, 0} to be an algebra hR, +, ·, −, 0i
where + and · are binary, − is unary, and 0 is nullary which satisfies the following four properties:

• The algebra hR, +, −, 0i is a commutative group,
• x1 · (x2 · x3 ) = (x1 · x2 ) · x3 [ the algebra hR, ·i is a semi-group ],
• x1 · (x2 + x3 ) = (x1 · x2 ) + (x1 · x3 ), and
• (x1 + x2 ) · x3 = (x1 · x3 ) + (x2 · x3 )

for all x1 , x2 , x3 ∈ R.
We often write A = hA, f1 , . . . , fk i for hA, Fi and say the algebra is of finite type if F =
h f1 , . . . , fk i is finite. If for all f in F the operation f is unary, then A = hA, Fi is called a unary
5

algebra. A unary algebra A = hA, Fi is called a {0, 1}-valued unary algebra if the range of f is
contained in {0, 1}, for each f ∈ F which is not identity map.
Example 2.1.1. An algebra A = h{0, 1, 2, 3}, f1 , f2 i shown in the Table 2.1 contains two unary
operations f1 and f2 which take each element in {0, 1, 2, 3} to either 0 or 1.

f1

f2

0

0

0

1

1

1

2

0

1

3

1

0

Table 2.1: A is a {0, 1}-valued unary algebra.

We define the rows of A of a unary algebra A = hA, Fi of finite type by the following k-ary
relation
Rows(A) = {row(a) | a ∈ A}
where row(a) = h( f1 (a), . . . , fk (a)) | a ∈ Ai and { f1 , . . . , fk } is a fixed enumeration of the nonconstant, non-identity operations in F.
Now we suppose X ⊆ A and f : A → B and α : X → B are two functions such that for all x ∈ X
α(x) = f (x)
then α is called the restriction of f to X, and is denoted f X . If A and B are two algebras of the
same type, B ⊆ A, and f B = f A B for all f ∈ F then B is a subalgebra of A, denoted B ≤ A. Any
subset C of A that is closed under every operation in F is called a subuniverse of A. If B is a
subalgebra of A then B is a subuniverse of A. For every X ⊆ A, the subuniverse generated by
6

X is
SgA (X) = ∩{B : X ⊆ B and B is a subuniverse of an algebra A}.
Note that SgA (X) is a subuniverse.
Let A and B be two algebras of the same type F . A mapping α : A → B is called a homomorphism from A to B if
α( f A (a1 , . . . , ak )) = f B (α(a1 ), . . . , α(ak ))
for each k-ary f in F and each sequence a1 , . . . , ak from A. An isomorphism is a homomorphism
which is one-to-one and onto.
If A is a non-empty set and n is a positive integer, then an n-ary relation on A is a subset of An ,
the set of all n-tuples of A. If α : A → B is a homomorphism, then the kernel of α, written ker(α),
is defined by
ker(α) = {ha, bi ∈ A2 : α(a) = α(b)}.
The kernel of α is a binary relation on A.
Let A be an algebra of type F . For every 1 ≤ i ≤ k, we have a k-ary projection operation πi
on a non-empty set A defined by
πi (a1 , . . . , ak ) = ai
for all ai ∈ A. For a k-ary operation β , and k j-ary operations {γi }ki=1 , there is a j-ary operation α
defined by composition of operations, that is,
α(a1 , . . . , a j ) = β (γ1 (a1 , . . . , a j ), . . . , γk (a1 , . . . , a j )).

Moreover, all the operations constructed by the composition of operations using the basic operations of A and the projections on A are called the term functions or term operations of A.

7

2.2

Lattices

In this section we introduce the idea of a lattice that came from [1]. We define lattice in two
standard ways which are based on the same (algebraic) footing as groups or rings, and the notion
of order.
The first way is to define a lattice as an algebra
L = hL, ∧, ∨i
where L is a nonempty set with two binary operations ∧ (meet) and ∨ (join) on L in which the
following identities hold:

• Commutative: a ∧ b = b ∧ a and a ∨ b = b ∨ a
• Associative: a ∧ (b ∧ c) = (a ∧ b) ∧ c and a ∨ (b ∨ c) = (a ∨ b) ∨ c
• Idempotent: a ∧ a = a and a ∨ a = a
• Absorption: a = a ∧ (a ∨ b) and a = a ∨ (a ∧ b)

for all a, b, c ∈ L.
Moreover if the above commutative, associative and idempotent identities are true for only one
binary operation ∧ on a set S, then the algebra
S = hS, ∧i
is called a ∧-semilattice.
The second way to define a lattice comes from the notion of a partial order. A binary relation
≤ on a set X is called a partial order on X if for each x, y, z ∈ X, the following three identities hold
in X:
8

• Reflexivity: x ≤ x,
• Antisymmetry: x ≤ y and y ≤ x imply x = y, and
• Transitivity: x ≤ y and y ≤ z imply x ≤ z

We also say that ≤ on a set X is a total order on A if x ≤ y or y ≤ x for each x, y ∈ X. A partially
ordered set is defined by a set X with a partial order on X. We write x < y to mean x ≤ y but x 6= y
in X. Note that every lattice has a partial order given by defining x ≤ y to mean x ∧ y = x. When
the relation is a total order we call the partially ordered set totally ordered.
Let S be a subset of a partially ordered set X. An element x ∈ X is an upper bound for S if
s ≤ x for every s ∈ S. An element x ∈ X is the supremum of S, written sup S, if x is an upper
bound of S, and s ≤ y for every s ∈ S implies x ≤ y. A supremum is unique if it exists. Similarly,
an element x ∈ X is a lower bound for S if x ≤ s for every s ∈ S. An element x ∈ X is called the
infimum of S, written inf S, if x is an lower bound of S and y ≤ x for all lower bounds y of S. An
infimum is unique if it exists. A partially ordered set X is a lattice if and only if for every x, y ∈ X
both sup {x, y} and inf {x, y} exist in X.

2.3

Quasivarieties

First we define isomorphic, direct product and homomorphic images, and then finally look at
quasivariety.
If a mapping α is an isomorphism from an algebra A1 to another algebra A2 , then A1 is isomorphic to A2 , written A1 ∼
= A2 .
The direct product of two algebras A and B of the same type F is the one algebra whose
9

universe is the set of cartesian product of A and B and the operations are defined by
f A×B (ha1 , b1 i, . . . , hak , bk i) = h f A (a1 , . . . , ak ), f B (b1 , . . . , bk )i
for ai ∈ A, bi ∈ B, 1 ≤ i ≤ k and for f ∈ Fk (the subset of k-ary function symbols in F is denoted
by Fk ).
For i in some set I, if (Ai )i∈I is a collection of algebras of type F , then the product algebra,
written A = ∏i∈I Ai , is the algebra with universe ∏i∈I Ai and whose operations are defined by
f A (a1 , . . . , ak )(i) = f Ai (a1 (i), . . . , ak (i))
for f ∈ Fk and ha1 , . . . , ak i ∈ ∏i∈I Ai .
Now we can define a quasivariety ISP(M) generated by an algebra M to be the class of
algebras that are obtained by taking all isomorphic copies of subalgebras of products of M, that is
ISP(M), where I is the isomorphic copies, S is a subalgebras, and P is the products of an algebra M.
Similar to Birkhoff’s theorem which says HSP(M) is equationally defined, we note that ISP(M)
is quasi-equationally defined. This explains the interest in determining when the quasi-equational
theory of quasivariety is finitely based.

2.4

Topologies

In this section we state the concept of topological space [11] and some algebraic operation, and
then define topological quasi-variety generated by an alter ego, and some definitions of maps.
The collection of subsets of a set X is called a topology T of a space X if the following three
properties hold:

10

• The empty set and X are in T ,
• If {Ti }i∈I ⊂ T for any set I, then

i∈I Ti ∈ T ,

S

• If {Ti }ni=1 ⊂ T is a finite set of sets, then

i=1 Ti ∈ T .

Tn

The ordered pair (X, T ) is called topological space where X is a set and T is a topology of X
whose elements are called open sets of X. The closed sets are the complements to the open sets
S ∈ T . The discrete topology is the topology where all sets are open (and hence closed).
For some finite and non-zero n, an n-ary relation on an algebra A is called an algebraic relation on A if it is the universe of a subalgebra of An . A homomorphism from An to A is called
algebraic (total) operation on A for finite and non-negative n. An algebraic partial operation is
also defined by the homomorphism from a subalgebra of An for some finite non-negative n to an
algebra A.
Let M be a finite algebra. If G is a set of algebraic total operations on M, H is a set of algebraic
partial operations on M, R is a set of algebraic relations on M, and T is the discrete topology on
M, then the structured topological space
M = hM; G , H , R, T i
is called an alter ego of M. Both the algebra M and the alter ego M have the same underlying
set M.
Now using the knowledge of alter ego, we can define a topological quasi-variety, ISc P+ (M),
which consists of all isomorphic copies of topologically closed substructures of non-zero powers
of the alter ego M, where I is all isomorphic copies, Sc is all closed sub-structured topological
spaces and P+ is all non-zero powers of the alter ego M. See [10] for full definitions.

11

Moreover, for a function α from a topological space U to a topological space V if the set
α −1 (S) is an open subset of X for all open subsets S ∈ V then α from U to V is called a continuous function. For U, V ∈ ISc P+ (M) a morphism is a continuous map α : U → V which
preserves the total operations, partial operations, and relations on M. This map is called an embedding if α : U → α(U) is isomorphism and α(U) is a substructure of V. The map α is called a
homeomorphism when α is a bijection and both α and α −1 are continuous.

2.5

Dualizability

In this section we introduce the basic idea of natural duality theory, and we also look into what it
means for an algebra to be dualisable.
Let M = hM, F i be a finite algebra. For an alter-ego M = hM; G , H , R, T i, and for any
algebra A ∈ ISP(M), define the dual of A (with respect to M) to be
D(A) = hHom(A, M); G , H , R, T i
seen as a topologically-closed substructure of MA , where Hom(A, M) is the set of all homomorphisms from A to M. Note Hom(A, M) is a subset of MA . This follows from G , H , and R being
algebraic. It turns out that D(A) is in the topological quasivariety ISc P+ (M). The proof of this
non-trivial fact is given in Chapter 1 of [10]. See Figure 2.1 on the next page.
Likewise, for any topological structure X ∈ ISc P+ (M), define the dual of X to be
E(X) = hHom(X, M); F i
seen as a subalgebra of MX , where Hom(X, M) is the set of all M-morphisms from X to M. We
have Hom(X, M) is a subset of MX . The proof that E(X) ∈ ISP(M) is given in Chapter 1 of [10].
See Figure 2.2 on page 14.
12

ISc P+ (M)

ISP(M)

A

D

Algebra

D(A)

Topology

Figure 2.1: The Dual of an algebra is a topological structure.
If an algebra A is in ISP(M) then the evaluation map is a natural embedding eA : A →
E(D(A)), defined by
eA (a)(x) = x(a)
for all elements a ∈ A and homomorphisms x ∈ D(A). The map eA (a) : Hom(A, M) → A is called
an evaluation.
Similarly, for a structure X ∈ ISc P+ (M) there is a natural embedding εX , called an evaluation
map from an algebra X to D(E(X)), defined by
εX (x)(a) = a(x)
for all homomorphisms x ∈ X and morphisms a ∈ E(X).
Furthermore, a map β : D(A) → M is given by evaluation at a on B where B ⊆ D(A) if
β B = eA (a) B for a ∈ A.
If for each algebra A in ISP(M) we have A is isomorphic to a double dual of A, (ED(A))
through the evaluation map eA , then an alter ego M is said to yield a duality on the algebra M. If

13

ISc P+ (M)

ISP(M)

E(X)
X
E

Algebra

Topology

Figure 2.2: The Dual of a topological structure is an algebra.
there is an alter ego M such that for every A ∈ ISP(M) we have that
A∼
= ED(A)
via eA , then an algebra M is said to be dualizable. See Figure 2.3.
D

ISP(M)

ISc P+ (M)

A

D(A)
eA

ED(A)
E
Algebra

Topology

Figure 2.3: Dualizability of an algebra occurs when eA : A → E(D(A)) is an isomorphism for
all A.

Moreover, we say that an algebra M is called non-dualizable when no alter ego dualizes M.
14

Example 2.5.1. (Although bounded distributive lattices are not used to this document, we give this
as a example in the area of natural duality theory) H.A Priestley first found that the two elements
bounded distributive lattice,
M = h{0, 1}; ∨, ∧, 0, 1i
is fully dualized by the alter ego
M = h{0, 1}; ≤, T i
where T is a discrete topology, and the partial operation ≤ is defined on {0, 1} with 0 ≤ 1.
See [8, 9].
Example 2.5.2. J. Hyndman and R. Willard showed that the algebra M = h{0, 1, 2}; f , gi (where
f , g are unary operations shown in the following Table 2.2)

f g
0 0 1
1 0 2
2 1 2
Table 2.2: The escalator algebra

is dualized by the alter ego
M = hM, ∧, ∨, E, R, T i
where ∧ and ∨ are the lattice meet and join operations on an order chain hM, ≤i with 0 < 1 < 2,
and E ⊆ M 2 and R ⊆ M 4 , defined by
E = {(x, y) : x ≤ y and (x, y) 6= (0, 2)},
R = {(x, y, z, w) : x ≤ y ≤ z ≤ w and x = y or z = w},
and T is a discrete topology. However, M is not fully dualizable. See [2].

15

2.6

The Interpolation Condition and Duality Theorems

In this section we define the interpolation condition and state some duality theorems which are
helpful for proving that an algebra is dualizable.
First, we provide the following lemma which is explicitly proved in [14] and was used to prove
Theorem 2.6.3.
Lemma 2.6.1. [14] Let M be a finite algebra and A ∈ ISP(M), and let α : D(A) → M and n ∈ N.
Then α preserves the set of n-ary algebric relations on M, denoted Rn , if and only if α agrees with
an evaluation on each subset of D(A) with at most n elements.

We say that the alter ego M satisfies the Interpolation Condition relative to M if for each
n ∈ N and each closed substructure X of Mn , every morphism α : X → M extends to a term
function t : M n → M of the algebra M.
Now we need the definition of a total structure before we introduce the next theorem. An alter
ego M is said to be a total structure if the only algebric operations in M are the total ones. If M
is a total structure with finitely many algebric relations and M satisfies the interpolation condition
relative to M, then the Second Duality Theorem says that M is dualizable.
Theorem 2.6.1 (Second Duality Theorem). ( [10]) Assume that an alter ego M is a total structure
in which R is finitely many algebraic relations. If the interpolation condition holds, then M yields
a duality on an algebra M.

The next theorem allows the use of partial operations in the type of M and was used to prove
Theorem 2.6.3 [14].
Theorem 2.6.2 (Duality Compactness Theorem). ( [10, 12]) Let M be a finite algebra and let M
16

be an alter ego of M with finite type. If M yields a duality on each finite algebra in ISP(M), then
M dualizes M.

The following theorem is also helpful to prove the dualisability of a finite algebra.
Theorem 2.6.3. [14] Let M be a finite algebra which has binary homomorphisms ∧ and ∨ such
that hM; ∧, ∨i is a lattice. Then the alter ego M := hM, ∧, ∨, R2|M| , T i yields a duality on ISP(M).

We now have the necessary notation and background to formulate the results of this document.

17

Chapter 3

An Example of a Finite Algebra and
Dualizing Structure

In this chapter we define a five-element algebra which is our central example in our thesis. Also
we define an alter ego for our algebra. Then we show that the alter ego satisfies the interpolation
condition relative to the algebra and thus dualizes the algebra. We look at this particular alter ego
because it may turn out to be minimal.

3.1

The Five-Element Algebra

In this section we give an example of a finite algebra, and lattice diagram for the algebra. This
five-element algebra is our central example in this paper.
Let M be the algebra hM, p, q, r, 0̄i where p, q, r and 0̄ are the unary operations defined on the
underlying set M = {0, 1, 2, 3, 7} as shown in Table 3.1 on page 20.

18

From the concept of the row of an algebra which is introduced in [3], we have the row of an
algebra M as follows:
row(0) = (p(0), q(0), r(0))
= (0, 0, 0)
or, row(0) = 000,
row(1) = (p(1), q(1), r(1)) = 001,
row(2) = (p(2), q(2), r(2)) = 010,
row(3) = (p(3), q(3), r(3)) = 011,
and row(7) = (p(7), q(7), r(7)) = 111.
Moreover, we have Rows(M) as follows:
Rows(M) = {row(0), row(1), row(2), row(3), row(7)}
= {000, 001, 010, 011, 111}.
Note that neither 0̄ nor id are used in computing rows.
Now we use Rows(M) to draw the lattice diagram induced by M. (Note that we use Rows(M)
to define the lattice order on M, and that meet ∧ and join ∨ are homomorphisms M2 → M.) This is
done by drawing a Hasse diagram using the point-wise order induced by the elements in Rows(M)
by the order induced by 0 < 1. This lattice diagram for this algebra shown in Figure 3.1 on the
following page is drawn by looking at each row of M and comparing them coordinate-wise.

19

p q r 0̄
0 0 0 0 0
1 0 0 1 0
2 0 1 0 0
3 0 1 1 0
7 1 1 1 0
Table 3.1: The five-element algebra M

7 (111)

3 (011)

2 (010)

1 (001)

0 (000)

Figure 3.1: The Lattice Diagram for the five-element algebra M.

20

3.2

An Alter Ego

In this section we define an alter ego which is our central dualizing structure. First we introduce
the relations Qi for 1 ≤ i ≤ 13. These are gathered together in Table 3.2. Here k is the arity of the
relation being defined. The relation Q0i is all k-tuples that satisfy the co-ordinate restrictions. The
relation Qi is the largest subuniverse of Q0i that excludes βi , given in the fifth column of the tables.

Range(α)

Lemma k Name

βi

Co-ordinate restrictions

{0, 7}

3.3.15 2

Q1

h0, 7i

Y1 ≤ Y2

{0, 1}

3.3.14 4

Q2

h0, 1, 1, 0i

Y1 ≤ Y2 ≤ Y3 ;

Y1 ≤ Y4

{0, 2}

3.3.13 4

Q3

h0, 2, 2, 0i

Y1 ≤ Y2 ≤ Y3 ;

Y1 ≤ Y4

{0, 3}

3.3.12 4

Q4

h0, 3, 3, 0i

Y1 ≤ Y2 ≤ Y3 ;

Y1 ≤ Y4

{0, 1, 7}

3.3.11 5

Q5

h0, 1, 1, 7, 0i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ;

Y1 ≤ Y5

{0, 2, 7}

3.3.10 5

Q6

h0, 2, 2, 7, 0i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ;

Y1 ≤ Y5

{0, 3, 7}

3.3.9

5

Q7

h0, 3, 3, 7, 0i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ;

Y1 ≤ Y5

{0, 1, 3}

3.3.8

6

Q8

h0, 1, 1, 3, 3, 0i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤ Y5 ;

Y1 ≤ Y6

{0, 2, 3}

3.3.7

6

Q9

h0, 2, 2, 3, 3, 0i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤ Y5 ;

Y1 ≤ Y6

{0, 1, 3, 7}

3.3.6

7

Q10

h0, 1, 1, 3, 3, 7, 0i Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤ Y5 ≤ Y6 ; Y1 ≤ Y7

{0, 2, 3, 7}

3.3.5

7

Q11

h0, 2, 2, 3, 3, 7, 0i Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤ Y5 ≤ Y6 ; Y1 ≤ Y7

{0, 1, 2, 3}

3.3.4

8

Q12

h0, 1, 1, 3, 3, 0, 2, 2i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤ Y5 ;
Y6 ≤ Y7 ≤ Y8 ≤ Y4

{0, 1, 2, 3, 7}

3.3.3

9

Q13 h0, 1, 1, 3, 3, 7, 0, 2, 2i

Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤ Y5 ≤ Y6 ;
Y7 ≤ Y8 ≤ Y9 ≤ Y4

Table 3.2: Relations Qi in the alter-ego M. Qi omits βi but satisfies the co-ordinate restrictions.

21

More precisely, we define sets Q0i and Qi by

and

Q0i = {

hY1 , . . . ,Yk i ∈ M k : The co-ordinate restrictions hold on hY1 , . . . ,Yk i

Qi = {

x ∈ Q0i : βi ∈
/ SgMk ({x}) }

}

where the co-ordinate restrictions and βi values are given in Table 3.2 on the preceding page.
Note that Qi and Q0i are subuniverses of Mk and Qi is a subalgebra of Q0i .
Lemma 3.2.1. For i 6= 2, we have Q0i = Qi ∪ {βi }.

Proof. If x ∈ Q0i \ Qi then by definition, βi = f (x) for f ∈ {p, q, r, 0̄, id} but βi ∈
/ {0, 1}k , so f = id,
so x = βi .

Let Q be the set of Qi . Let M := hM, ∧, ∨, 0, Q, T i be an alter ego for the algebra M. In the
alter ego M, {∧, ∨, 0} is a set of algebraic total operations on M, Q = {Q1 , . . . , Q13 } is a set of
algebraic relations on M, and T is the discrete topology on M. (Here the empty set (0)
/ is the set
of algebraic partial operations on M.) Note that ∧ and ∨ are homomorphisms, and 0 is the zero
homomorphism. Also note that all Qi ’s are subalgebras and are closed under p, q, r and 0̄.

3.3

Morphism Ranges

In this section we enumerate all possible cases for range(α), where α : X → M is a morphism with
respect to the alter ego M := hM, ∧, ∨, 0, Q, T i and X is finite. All possible cases for range(α)
are shown in Table 3.2 on the previous page. This list is exhaustive because range(α) is closed
under meet ∧, join ∨ and contains 0. Note that the alter ego does not contain the unary relation
{0} and for any α : D(A) → M we have 0 (the zero homomorphism) in D(A). We know that 0 is

22

the unary operation M → M defined by the homomorphism 0(x) = 0, and α(0(x)) = 0(α(x)) = 0.
So 0 ∈ range(α).
We wish to show that M satisfies the Interpolation Condition relative to M. That is, we have
that a morphism α : X → M is the restriction to the substructure X of an n-ary term operation of
M for each finite n, X ≤ Mn , and α ∈ Hom(X, M).

3.3.1

Notation

In the following lemmas we consistently use the following notation. We let X be a topologicallyclosed substructure of Mn where n is some fixed finite positive integer; α : X → M is a morphism
in the topological quasivariety.
For j ∈ range(α) ⊆ M, we set X j = {x ∈ X : α(x) = j}, and set a j = (X j ) and b j = (X j ).
V

W

Note that a j and b j are well defined, as these are finite meets and joins respectively.

3.3.2

The General Argument

To show that the term interpolation condition holds, in each case we must show that α(x) = t(x) for
x ∈ X where t is some term on M n . Because the algebra is unary, so are all terms, and we must have
t(x) = f (πi (x)) (or t = f ◦ πi ) for some co-ordinate i, where 1 ≤ i ≤ n, and some f ∈ {p, q, r, 0̄, id}.
In each of the following lemmas, we shall find a co-ordinate i (where 1 ≤ i ≤ n ), and argue
that there must be a function f ∈ {p, q, r, 0̄, id} such that for all j ∈ range(α) we have
j = f (πi (a j )) = f (πi (b j )).

23

Lemma 3.3.1. For all j ∈ range(α), if f (a j (i)) = f (b j (i)) = j for some co-ordinate i, where 1 ≤
i ≤ n, and some f ∈ {p, q, r, 0̄, id} then we have α(x) = f (x(i)) for every x ∈ X, and so α(x) = t(x)
for x ∈ X where t is some term on M n .

Proof. We have for x ∈ X j that
a j ≤ x ≤ b j,
which means that each co-ordinate respects the order, so πi (a j ) ≤ πi (x) ≤ πi (b j ). Because meet
and join are homomorphisms, f is order preserving, and thus
f (πi (a j )) ≤ f (πi (x)) ≤ f (πi (b j ));
whence
f (πi (x)) = j = α(x).

Because f = id is a possibility, it certainly suffices to show that
j = πi (a j ) = πi (b j )

for j ∈ range(α).

Now we analyze all possible range (α) and find the term function of M for each range one by one.
Lemma 3.3.2. If range(α) = {0}, then α = 0̄ ◦ π1 X .

Proof. Suppose range(α) = {0}. Since 0̄ is the constant 0 function on the algebra M, we have
α = 0̄ ◦ π1 X .
Lemma 3.3.3. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 1, 2, 3, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q13 . Suppose range(α) = {0, 1, 2, 3, 7}. (See Figure 3.2 on the following
page.) With X j , a j , and b j defined as usual, let d2 = b2 ∧ a3 , d1 = a3 ∧ b1 , d0 = b0 ∧ a1 , c7 = a7 ∨ b3
and e0 = a2 ∧ b0 .
24

b7

X7
c7
b3
a7

X3

b1

b2
a3

X2 d

X1 d1

2

b0
a2

a1
e0 d0
X0

a0

Figure 3.2: Diagram for range(α) = {0, 1, 2, 3, 7}.

25

Here we see that (d0 ≤ a1 ≤ d1 ≤ a3 ≤ b3 ≤ c7 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤
Y4 ≤ Y5 ≤ Y6 ) and that (e0 ≤ a2 ≤ d2 ≤ a3 ) is one instance of (Y7 ≤ Y8 ≤ Y9 ≤ Y4 ).
Let s = hd0 , a1 , d1 , a3 , b3 , c7 , e0 , a2 , d2 , i. Note that s ∈ Q013 by the above. Suppose also that
s ∈ Q13 then
α(s) = hα(d0 ), α(a1 ), α(d1 ), α(a3 ), α(b3 ), α(c7 ), α(e0 ), α(a2 ), α(d2 )i
= h0, 1, 1, 3, 3, 7, 0, 2, 2i.
But h0, 1, 1, 3, 3, 7, 0, 2, 2i ∈
/ Q13 , so α(s) ∈
/ Q13 , and thus s ∈
/ Q13 because α respects Q13 . Therefore there exists a coordinate, say i, with 1 ≤ i ≤ n for which
s(i) = hd0 (i), a1 (i), d1 (i), a3 (i), b3 (i), c7 (i), e0 (i), a2 (i), d2 (i)i ∈
/ Q13 .
Because
d0 ≤ a1 ≤ d1 ≤ a3 ≤ b3 ≤ c7 , and e0 ≤ a2 ≤ d2 ≤ a3
we have that
d0 (i) ≤ a1 (i) ≤ d1 (i) ≤ a3 (i) ≤ b3 (i) ≤ c7 (i), and e0 (i) ≤ a2 (i) ≤ d2 (i) ≤ a3 (i).
So s(i) ∈ Q013 and s(i) ∈
/ Q13 , and there exists an f ∈ Clou (M) with f (s(i)) = h0, 1, 1, 3, 3, 7, 0, 2, 2i
but this forces f = id, as the only term operation of M with 7 in the range is the identity operation.
Thus we have s(i) = h0, 1, 1, 3, 3, 7, 0, 2, 2i.
Now we show that α is the restriction of the term function πi , that is, that α satisfies the
interpolation condition for this case, and that this means α(x) = x(i) for all x ∈ X. By Lemma 3.3.1,
this reduces to showing that
j = a j (i) = b j (i)

for j ∈ range(α) = {0, 1, 2, 3, 7}.

Since c7 (i) = b3 (i) ∨ a7 (i), and ∨ is a homomorphism, we have 7 = 3 ∨ a7 (i), so a7 (i) = 7. As
b7 ≥ a7 we have b7 (i) ≥ a7 (i) = 7, so we also have b7 (i) = 7.
26

We know that a2 (i) = 2. Again since d2 (i) = b2 (i) ∧ a3 (i), and ∧ is a homomorphism, we have
2 = b2 (i) ∧ 3, so b2 (i) = 2.
Similarly, we know that a1 (i) = 1, and since d1 (i) = b1 (i) ∧ a3 (i) we have 1 = b1 (i) ∧ 3, so
b1 (i) = 1.
Likewise, since
0 = e0 (i) ∨ d0 (i)
= (a2 (i) ∧ b0 (i)) ∨ (a1 (i) ∧ b0 (i))
= (2 ∧ b0 (i)) ∨ (1 ∧ b0 (i))
we must have b0 (i) = 0. As a0 ≤ b0 we have a0 (i) ≤ b0 (i) = 0, so we also have a0 (i) = 0.
We already know that a3 (i) = b3 (i) = 3 from the co-ordinates of s.
Hence, by Lemma 3.3.1 we have α = πi X , a term function of M.

We omit the detailed explanations in the following cases because all are very similar to
the above case.
Lemma 3.3.4. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 1, 2, 3} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q12 . Suppose range(α) = {0, 1, 2, 3}. (See Figure 3.3 on the next page.) With
X j , a j and b j defined as usual, let d2 = b2 ∧ a3 , d1 = a3 ∧ b1 , d0 = a1 ∧ b0 and e0 = a2 ∧ b0 . We
know that α(a j ) = j, essentially by definition (and α respects finite meets), and similarly we know
that α(b j ) = j. We get α(e0 ) = 0 and α(d j ) = j from definitions and the fact that α respects
meets.

27

b3

X3

b2

b1
a3

X2 d

X1 d1

2

b0
a2

a1
e0 d0
X0

a0

Figure 3.3: Diagram for range(α) = {0, 1, 2, 3}.

28

Here we see that (d0 ≤ a1 ≤ d1 ≤ a3 ≤ b3 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤
Y5 ) and that (e0 ≤ a2 ≤ d2 ≤ a3 ) is one instance of (Y6 ≤ Y7 ≤ Y8 ≤ Y4 ).
Let s = hd0 , a1 , d1 , a3 , b3 , e0 , a2 , d2 i. Note that s ∈ Q012 by the above. Then the tuple α(s) =
h0, 1, 1, 3, 3, 0, 2, 2i, and so α(s) ∈
/ Q12 . Thus s ∈
/ Q12 and there exists a coordinate i for which
s(i) ∈
/ Q12 . From this we have s(i) = h0, 1, 1, 3, 3, 0, 2, 2i. We now show that for all x ∈ X we have
x(i) = α(x). By Lemma 3.3.1, it suffices to show that for j ∈ range(α) we have a j (i) = b j (i) = j.
For j = 0 we know that
0 = e0 (i) ∨ d0 (i)
= (a2 (i) ∧ b0 (i)) ∨ (a1 (i) ∧ b0 (i))
= (2 ∧ b0 (i)) ∨ (1 ∧ b0 (i))
we must have b0 (i) = 0. As a0 ≤ b0 we have a0 (i) ≤ b0 (i) = 0, so a0 (i) = 0.
For j = 1 we argue as follows. From the value of s we have a1 (i) = 1, d1 (i) = 1, and a3 (i) = 3.
Now 1 = d1 (i) = b1 (i) ∧ a3 (i) = b1 (i) ∧ 3, so b1 (i) = 1.
For j = 2 we argue similarly. From the value of s we have a2 (i) = 2, d2 (i) = 2, and a3 (i) = 3.
Now 2 = d2 (i) = b2 (i) ∧ a3 (i) = b2 (i) ∧ 3, so b2 (i) = 2.
For j = 3 we have a3 (i) = b3 (i) = 3 from the co-ordinates of s.
Hence, by Lemma 3.3.1 we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.5. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 2, 3, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q11 . Suppose range(α) = {0, 2, 3, 7} (See Figure 3.4 on the following page.)
29

With X j , a j , and b j defined as usual, let d0 = b0 ∧ a2 , d2 = a3 ∧ b2 and c7 = a7 ∨ b3 .
Here we see that (d0 ≤ a2 ≤ d2 ≤ a3 ≤ b3 ≤ c7 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤
Y4 ≤ Y5 ≤ Y6 ) and that (d0 ≤ b0 ) is one instant of (Y1 ≤ Y7 ).
Let s = hd0 , a2 , d2 , a3 , b3 , c7 , b0 i. Note that s ∈ Q011 by the above. Then α(s) = h0, 2, 2, 3, 3, 7, 0i,
and so α(s) ∈
/ Q11 . Thus s ∈
/ Q11 and there is some coordinate i for which s(i) ∈
/ Q11 . From this
we have s(i) = h0, 2, 2, 3, 3, 7, 0i. Here we use that fact that the unique solution of
2 = d2 (i) = b2 (i) ∧ a3 (i) = b2 (i) ∧ 3

is

b2 (i) = 2,

is

a7 (i) = 7,

and the unique solution of
7 = c7 (i) = b3 (i) ∨ a7 (i) = 3 ∨ a7 (i)
and the unique solution of
7 = a7 (i) ≤ c7 (i) ≤ b7 (i)

is

b7 (i) = 7.

Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.6. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 1, 3, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q10 . Suppose range(α) = {0, 1, 3, 7}. (See Figure 3.5 on the next page.) With
X j , a j , and b j defined as usual, let d0 = b0 ∧ a1 , d1 = b1 ∧ a3 and c7 = b3 ∨ a7 .
Here we see that (d0 ≤ a1 ≤ d1 ≤ a3 ≤ b3 ≤ c7 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤
Y4 ≤ Y5 ≤ Y6 ) and that (d0 ≤ b0 ) is one instance of (Y1 ≤ Y7 ) .
Let s = hd0 , a1 , d1 , a3 , b3 , c7 , b0 i. Note that s ∈ Q010 by the above. Then α(s) = h0, 1, 1, 3, 3, 7, 0i,
and so α(s) ∈
/ Q10 . Thus s ∈
/ Q10 and hence there exists a coordinate i for which s(i) ∈
/ Q10 . From
30

b7

X7
c7
b3
a7

X3

b2
a3

X2 d

2

b0
a2
d0
X0

a0

Figure 3.4: Diagram for range(α) = {0, 2, 3, 7}.

31

b7

X7
c7
b3
a7

X3

b1
a3

X1 d1

b0
a1
d0
X0

a0

Figure 3.5: Diagram for range(α) = {0, 1, 3, 7}.

32

this we have s(i) = h0, 1, 1, 3, 3, 7, 0i. Here we use that fact that the unique solution of
7 = c7 (i) = b3 (i) ∨ a7 (i) = 3 ∨ a7 (i)

is

a7 (i) = 7,

is

b1 (i) = 1.

and the unique solution of
1 = d1 (i) = b1 (i) ∧ a3 (i) = b1 (i) ∧ 3

Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.7. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 2, 3} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q9 . Suppose range(α) = {0, 2, 3}. (See Figure 3.6.) With X j , a j , and b j
b3

b2
X3
b0

X2
a3
d2

X0

a2
d0
a0

Figure 3.6: Diagram for range(α) = {0, 2, 3}.
33

defined as usual, let d0 = b0 ∧ a2 and d2 = b2 ∧ a3 .
Here we see that (d0 ≤ a2 ≤ d2 ≤ a3 ≤ b3 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤
Y5 ) and that (d0 ≤ b0 ) is one instance of (Y1 ≤ Y6 ).
Let s = hd0 , a2 , d2 , a3 , b3 , b0 i. Note that s ∈ Q09 by the above. Then α(s) = h0, 2, 2, 3, 3, 0i, and
so α(s) ∈
/ Q9 . Thus s ∈
/ Q9 and hence there is some coordinate i for which s(i) ∈
/ Q9 . From this we
have s(i) = h0, 2, 2, 3, 3, 0i. We now show that for all x ∈ X we have x(i) = α(x). By Lemma 3.3.1
it suffices to show that for j ∈ range(α) we have a j (i) = b j (i) = j.
For j = 0 we have b0 (i) = 0. As a0 ≤ b0 we have a0 (i) ≤ b0 (i) = 0.
For j = 2 we argue as follows. From the value of s we have a2 (i) = 2, d2 (i) = 2, and a3 (i) = 3.
Now 2 = d2 (i) = b2 (i) ∧ a3 (i) = b2 (i) ∧ 3, so b2 (i) = 2.
For j = 3 we have a3 (i) = b3 (i) = 3 from the co-ordinates of s.
Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.8. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 1, 3} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q8 . Suppose range(α) = {0, 1, 3}. (See Figure 3.7 on the next page.) With
X j , a j , and b j defined as usual, let d0 = b0 ∧ a1 and d1 = b1 ∧ a3 .
Here we see that (d0 ≤ a1 ≤ d1 ≤ a3 ≤ b3 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤ Y4 ≤
Y5 ) and that (d0 ≤ b0 ) is one instance of (Y1 ≤ Y6 ).
Let s = hd0 , a1 , d1 , a3 , b3 , b0 i. Note that s ∈ Q08 by the above. Then α(s) = h0, 1, 1, 3, 3, 0i, and

34

b3

b1
X3
b0

X1
a3
d1

X0

a1
d0
a0

Figure 3.7: Diagram for range(α) = {0, 1, 3}.
so α(s) ∈
/ Q8 . Thus s ∈
/ Q8 and hence there is some coordinate i for which s(i) ∈
/ Q8 . From this we
have s(i) = h0, 1, 1, 3, 3, 0i. Here we use that fact that the unique solution of
1 = d1 (i) = b1 (i) ∧ a3 (i) = b1 (i) ∧ 3

is

b1 (i) = 1.

Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.9. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 3, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q7 . We have range(α) = {0, 3, 7}. (See Figure 3.8 on the following page.)
With X j , a j , and b j defined as usual, let d0 = b0 ∧ a3 and d3 = b3 ∧ a7 .
Here we see that (d0 ≤ a3 ≤ d3 ≤ a7 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤ Y4 ) and
35

b7

b3
X7
b0

X3
a7
d3

X0

a3
d0
a0

Figure 3.8: Diagram for range(α) = {0, 3, 7}.
(d0 ≤ b0 ) is one instance of (Y1 ≤ Y5 ).
Let s = hd0 , a3 , d3 , a7 , b0 i. Note that s ∈ Q07 by the above. Then α(s) = h0, 3, 3, 7, 0i, and so
α(s) ∈
/ Q7 . Thus s ∈
/ Q7 and we get that there is a coordinate i for which s(i) ∈
/ Q7 . From this we
have s(i) = h0, 3, 3, 7, 0i. Here we use that fact that the unique solution of
3 = d3 (i) = b3 (i) ∧ a7 (i) = b3 (i) ∧ 7

is

b3 (i) = 3.

Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.10. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 2, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q6 . Suppose range(α) = {0, 2, 7}. (See Figure 3.9 on the next page.) With
36

b7

b2
X7
b0

X2
a7
d2

X0

a2
d0
a0

Figure 3.9: Diagram for range(α) = {0, 2, 7}.
X j , a j , and b j defined as usual, let d0 = b0 ∧ a2 and d2 = b2 ∧ a7 .
Here we see that (d0 ≤ a2 ≤ d2 ≤ a7 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤ Y4 ) and
that (d0 ≤ b0 ) is one instance of (Y1 ≤ Y5 ).
Let s = hd0 , a2 , d2 , a7 , b0 i. Note that s ∈ Q06 by the above. Then α(s) = h0, 2, 2, 7, 0i, and so
α(s) ∈
/ Q6 . Thus s ∈
/ Q6 and we get that there exists i for which s(i) ∈
/ Q6 . From this we have
s(i) = h0, 2, 2, 7, 0i. Here we use that fact that the unique solution of
2 = d2 (i) = b2 (i) ∧ a7 (i) = b2 (i) ∧ 7

is

b2 (i) = 2.

Hence, by Lemma 3.3.1, we have α(x) = x(i) for each x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.11. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
37

If range(α) = {0, 1, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q5 . Suppose range(α) = {0, 1, 7}. (See Figure 3.10.) With X j , a j , and b j
b7

b1
X7
b0

X1
a7
d1

X0

a1
d0
a0

Figure 3.10: Diagram for range(α) = {0, 1, 7}.

defined as usual, let d0 = b0 ∧ a1 , and d1 = b1 ∧ a7 .
Here we see that (d0 ≤ a1 ≤ d1 ≤ a7 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ≤ Y4 ) and
that (d0 ≤ b0 ) is one instance of (Y1 ≤ Y5 ).
Let s = hd0 , a1 , d1 , a7 , b0 i. Note that s ∈ Q05 by the above. Then α(s) = h0, 1, 1, 7, 0i, and so
α(s) ∈
/ Q5 . Thus s ∈
/ Q5 and hence there is some coordinate i for which s(i) ∈
/ Q5 . From this we
have s(i) = h0, 1, 1, 7, 0i. Here we use that fact that the unique solution of
1 = d1 (i) = b1 (i) ∧ a7 (i) = b1 (i) ∧ 7
38

is

b1 (i) = 1.

Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.12. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 3} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q4 . Suppose range(α) = {0, 3}. (See Figure 3.11.) With X j , a j , and b j defined
b3

b0
X3

X0
a3
d0
a0

Figure 3.11: Diagram for range(α) = {0, 3}.

as usual, let d0 = b0 ∧ a3 .
Here we see that (d0 ≤ a3 ≤ b3 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ) and that (d0 ≤ b0 )
is one instance of (Y1 ≤ Y4 ).
Let s = hd0 , a3 , b3 , b0 i. Note that s ∈ Q04 by the above. Then α(s) = h0, 3, 3, 0i, and so α(s) ∈
/
Q4 . Thus s ∈
/ Q4 and hence there is some coordinate i for which s(i) ∈
/ Q4 . From this we have
s(i) = h0, 3, 3, 0i.

39

Hence, by Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.
Lemma 3.3.13. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 2} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q3 . Suppose range(α) = {0, 2}. (See Figure 3.12.) With X j , a j , and b j defined
b2

b0
X2

X0
a2
d0
a0

Figure 3.12: Diagram for range(α) = {0, 2}.

as usual, let d0 = b0 ∧ a2 .
Here we see that (d0 ≤ a2 ≤ b2 ) is one instance of the relation (Y1 ≤ Y2 ≤ Y3 ), and that (d0 ≤ b0 )
is one instance of (Y1 ≤ Y4 ).
Let s = hd0 , a2 , b2 , b0 i. Note that s ∈ Q03 by the above. Then α(s) = h0, 2, 2, 0i, and so α(s) ∈
/
Q3 . Thus s ∈
/ Q3 and hence there is some coordinate i for which s(i) ∈
/ Q3 . From this we have
s(i) = h0, 2, 2, 0i.

40

Hence, by the Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term
function of M.

Lemma 3.3.2 and the next lemma are the cases where α might not be the restriction of a
projection.
Lemma 3.3.14. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 1} then α = f (πi )X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q2 . Suppose range(α) = {0, 1}. (See Figure 3.13.) With X j , a j , and b j defined
b1

b0

X1

X0
a1
d0
a0

Figure 3.13: Diagram for range(α) = {0, 1}.

as usual, let d0 = b0 ∧ a1 .

Let U be the subset of M 4 defined by hY1 ,Y2 ,Y3 ,Y4 i ∈ M 4 : Y1 ≤ Y2 ≤ Y3 & Y1 ≤ Y4 . Let S be
the subset of U defined by S = {u ∈ U : h0, 1, 1, 0i ∈ SgM4 ({u})}. We note that if t ∈ U \ S then
h0, 1, 1, 0i ∈
/ SgM4 ({t}).
41

Note that Q2 = U \ S, and Q02 = U ∪ S. Let s = hd0 , a1 , b1 , b0 i. Note that s ∈ Q02 by the above.
Then α(s) = h0, 1, 1, 0i ∈
/ Q2 , and therefore s ∈
/ Q2 and hence there is some coordinate i for which
s(i) ∈
/ Q2 . However d0 ≤ a1 ≤ b1 and d0 ≤ b0 , so s(i) ∈ U. Thus s(i) ∈ S, and there exists an f ∈
Clou (M), such that f (s(i)) = h0, 1, 1, 0i. We now show that for all x ∈ X we have f (x(i)) = α(x).
By Lemma 3.3.1 it suffices to show that for j ∈ range(α) we have f (a j (i)) = f (b j (i)) = j.
For j = 0 we have f (b0 (i)) = 0. As a0 ≤ b0 , and each possible f is order preserving, we have
f (a0 (i)) ≤ f (b0 (i)) = 0, so f (a0 (i)) = 0.
For j = 1 we have f (a1 (i)) = f (b1 (i)) = 1 from the co-ordinates of s.
Hence, by Lemma 3.3.1, we have α(x) = f (x(i)) for every x ∈ X, and so α = f (πi )X , the
restriction of a term function of M.
Lemma 3.3.15. Suppose α : X → M is a morphism, where X ≤ Mn , and M := hM, ∧, ∨, 0, Q, T i.
If range(α) = {0, 7} then α = πi X , for some i where 1 ≤ i ≤ n.

Proof. Here we use Q1 . Suppose range(α) = {0, 7}. (See Figure 3.14 on the following page.)
With X j , a j , and b j defined as usual, let d0 = b0 ∧ a7 .
Here we see that (d0 ≤ a7 ) is an instance of the relation (Y1 ≤ Y2 ).
Let s = hd0 , a7 i. Note that s ∈ Q01 by the above. Then α(s) = h0, 7i, and so α(s) ∈
/ Q1 . Thus
s∈
/ Q1 and there is some coordinate i for which s(i) ∈
/ Q1 . From this we have s(i) = h0, 7i. Here
we use that fact that the unique solution of
0 = d0 (i) = b0 (i) ∧ a7 (i) = b0 (i) ∧ 7

is

b0 (i) = 0,

and the unique solution of
7 = a7 (i) ≤ b7 (i)
42

is

b7 (i) = 7.

b7

b0
X7

X0
a7
d0
a0

Figure 3.14: Diagram for range(α) = {0, 7}.
Hence, the Lemma 3.3.1, we have α(x) = x(i) for every x ∈ X, and so α = πi X , a term function
of M.

Now we prove the following Theorem by using Interpolation Condition.
Theorem 3.3.1. The algebra M is dualized by the alter ego hM, ∧, ∨, 0, Q, T i, where the set
Q = {Q1 , . . . , Q13 } as given in Table 3.2 on page 21 .

Proof. For range(α) = {0}, we have shown that α = 0̄ ◦ π1 X ( Lemma 3.3.2).
Moreover, for range(α) = {0, 1}, we have proved that
α = f (πi )X
is a term function of the algebra M ( Lemma 3.3.14).
Furthermore, we have shown (Lemmas 3.3.3, 3.3.4, 3.3.5, 3.3.6, 3.3.7, 3.3.8, 3.3.9,
3.3.10, 3.3.11, 3.3.12, 3.3.13, and 3.3.15) that for any range(α) (except range(α) = {0} and
43

range(α) = {0, 1}), for every x ∈ X, and 1 ≤ i ≤ n we have
α = πi X ,
a term function of M. This concludes that the alter ego hM, ∧, ∨, 0, Q, T i satisfies the interpolation
condition relative to the algebra M, and hence M dualizes M; that means M is dualizable.

The previous Theorem 3.3.1 gives the following corollary.
Corollary 3.3.1. The alter ego hM, ∧, ∨, 0, R9 , T i yields a duality on ISP(M).

Proof. This follows from Theorem 3.3.1 because Q ⊆ R9 .

By Theorem 2.6.3, the alter ego hM, ∧, ∨, 0, R10 , T i yields a duality on ISP(M). From Theorem 3.3.1, the algebra M is dualized by the alter ego hM, ∧, ∨, 0, Q, T i. From Corollary 3.3.1,
the alter ego hM, ∧, ∨, 0, R9 , T i yields a duality on ISP(M). In the alter ego M, a set of algebraic
relations Q is smaller than both R10 and R9 . We now are going to show whether or not the alter ego
M is a minimal dualizing structure.

44

Chapter 4

Minimality

4.1

Non-evaluation morphisms

We now want to show that the set Q defined on page 22 is minimal. We must show that if Q0 is
a proper subset of Q containing Q12 and Q13 then the algebra M is not dualized by the alter ego
hM, ∧, ∨, 0, Q0 , T i. Note that if for some L ∈ ISP(M), we have that L is not isomorphic to ED(L)
because of a non-evaluation morphism, then hM, ∧, ∨, 0, Q0 , T i does not yield a duality on M.
We first want to prove that the dual of L with respect to hM, {∧, ∨, 0}, T i has non-evaluation
morphisms.
Lemma 4.1.1. [17] Let L be a k-ary relation on M. Let β be an element of M k \ L, where
L0 ⊇ L ∪ {β } and L ≤ L0 ≤ Mk are all algebras.
If there is an embedding ι of Hom(L, M) into Hom(L0 , M) such that

• ι(h1 ∧ h2 ) = ι(h1 ) ∧ ι(h2 ),
45

• ι(h1 ∨ h2 ) = ι(h1 ) ∨ ι(h2 ),
• ι(0) = 0, and
• [ι(π j )](β ) = π j (β );

then the dual of L with respect to hM, {∧, ∨, 0}, T i has a non-evaluation morphism.

Proof. Let ι be the embedding mentioned above. Define α : Hom(L, M) → M by
α(h) = [ιh](β ).
Note that for a ∈ L we have
hea (πk )ik = hπk (a)ik = hak ik = a 6= β ;
whereas
hα(πk )ik = β .
Thus α is not an evaluation morphism. On the other hand, for h ∈ D(L), α(h) is well defined.
Because there is an embedding ι of Hom(L, M) into Hom(L0 , M) and α : Hom(L, M) → M we
have
α(h1 ∧ h2 ) = [ι(h1 ∧ h2 )](β )
= [ι(h1 ) ∧ ι(h2 )](β )
= [ιh1 ](β ) ∧ [ιh2 ](β )
= α(h1 ) ∧ α(h2 ),
and similarly, α(h1 ∨ h2 ) = α(h1 ) ∨ α(h2 ), and α(0) = [ι(0)](β ) = 0(β ) = 0, hence α respects ∧,
∨ and 0. Thus D(L) contains a non-evaluation morphism with respect to hM, {∧, ∨, 0}, T i.

46

Any alter ego hM, ∧, ∨, 0, Q0 , T i that dualizes M must contain relations that the morphisms
provided by Lemma 4.1.1 do not respect.
We want to prove that Hom(Qi , M) embeds into Hom(Q0i , M). We prove this in two separate
lemmas. Firstly, we prove that Hom(Qi , M) embeds into Hom(Q0i , M) for 1 ≤ i ≤ 13 (i 6= 2)
and secondly, we prove that Hom(Q2 , M) embeds into Hom(Q02 , M). The general argument in
Lemma 4.1.2 does not work for Q2 as |Q02 \ Q2 | > 1, so we need a second lemma.
Lemma 4.1.2. Let Qi , Q0i and βi be defined as in Chapter 3. For 1 ≤ i ≤ 13 (i 6= 2), there is an
embedding ι of Hom(Qi , M) into Hom(Q0i , M) satisfying the properties of Lemma 4.1.1.

Proof. In general, note that we have p(x) ≤ q(x) and p(x) ≤ r(x). Consulting Table 4.1 on the
following page and Table 4.2 on page 49, note that in each case we have bi , ci , di , ui , and vi are all
elements of Qi and that
bi = p(βi ) = p(ui ) = p(vi ),
ci = q(βi ) = q(vi ), and
di = r(βi ) = r(ui ).

Let h ∈ Hom(Qi , M). We know that hh(bi ), h(ci ), h(di )i ∈ {0, 1}3 . We can establish that h(bi ) ≤
h(ci ) and h(bi ) ≤ h(di ) by noting that
h(bi ) = h(p(vi )) = p(h(vi )) ≤ q(h(vi )) = h(q(vi )) = h(ci ),
h(bi ) = h(p(ui )) = p(h(ui )) ≤ r(h(ui )) = h(r(ui )) = h(di ).
Thus, there exists an mi ∈ M such that hh(bi ), h(ci ), h(di )i = hp(mi ), q(mi ), r(mi )i. By Lemma 3.2.1,
extend h to h0 : Q0i → M by setting h0 (βi ) = mi . We have
h0 (p(βi )) = h0 (bi ) = h(bi ) = p(mi ) = p(h0 (βi )),
47

i

βi

ui

vi

1

h0, 7i

h2, 7i

h1, 7i

3

h0, 2, 2, 0i

h0, 0, 2, 2i

h0, 2, 3, 0i

4

h0, 3, 3, 0i

h0, 1, 3, 0i

h0, 2, 3, 1i

5

h0, 1, 1, 7, 0i

h0, 1, 3, 7, 0i

h0, 0, 1, 7, 0i

6

h0, 2, 2, 7, 0i

h0, 0, 0, 7, 0i

h0, 2, 3, 7, 0i

7

h0, 3, 3, 7, 0i

h0, 1, 3, 7, 2i

h0, 2, 2, 7, 1i

8

h0, 1, 1, 3, 3, 0i

h0, 3, 3, 3, 3, 0i

h0, 0, 0, 3, 3, 1i

9

h0, 2, 2, 3, 3, 0i

h0, 0, 0, 3, 3, 2i

h0, 2, 2, 2, 3, 0i

10

h0, 1, 1, 3, 3, 7, 0i

h0, 3, 3, 3, 3, 7, 0i

h0, 1, 1, 2, 3, 7, 0i

11

h0, 2, 2, 3, 3, 7, 0i

h0, 0, 0, 3, 3, 7, 0i

h0, 3, 3, 3, 3, 7, 1i

12

h0, 1, 1, 3, 3, 0, 2, 2i

h0, 1, 1, 1, 3, 0, 0, 0i

h0, 0, 0, 2, 3, 0, 2, 2i

13 h0, 1, 1, 3, 3, 7, 0, 2, 2i h0, 3, 3, 3, 3, 7, 0, 2, 2i h0, 0, 0, 3, 3, 7, 0, 2, 2i
Table 4.1: Elements βi , ui , vi for proof of the Lemma 4.1.2

48

i

bi

ci

di

1

h0, 1i

h0, 1i

h0, 1i

3

h0, 0, 0, 0i

h0, 1, 1, 0i

h0, 0, 0, 0i

4

h0, 0, 0, 0i

h0, 1, 1, 0i

h0, 1, 1, 0i

5

h0, 0, 0, 1, 0i

h0, 0, 0, 1, 0i

h0, 1, 1, 1, 0i

6

h0, 0, 0, 1, 0i

h0, 1, 1, 1, 0i

h0, 0, 0, 1, 0i

7

h0, 0, 0, 1, 0i

h0, 1, 1, 1, 0i

h0, 1, 1, 1, 0i

8

h0, 0, 0, 0, 0, 0i

h0, 0, 0, 1, 1, 0i

h0, 1, 1, 1, 1, 0i

9

h0, 0, 0, 0, 0, 0i

h0, 1, 1, 1, 1, 0i

h0, 0, 0, 1, 1, 0i

10

h0, 0, 0, 0, 0, 1, 0i

h0, 0, 0, 1, 1, 1, 0i

h0, 1, 1, 1, 1, 1, 0i

11

h0, 0, 0, 0, 0, 1, 0i

h0, 1, 1, 1, 1, 1, 0i

h0, 0, 0, 1, 1, 1, 0i

12

h0, 0, 0, 0, 0, 0, 0, 0i

h0, 0, 0, 1, 1, 0, 1, 1i

h0, 1, 1, 1, 1, 0, 0, 0i

13 h0, 0, 0, 0, 0, 1, 0, 0, 0i h0, 0, 0, 1, 1, 1, 0, 1, 1i h0, 1, 1, 1, 1, 1, 0, 0, 0i
Table 4.2: Elements bi , ci , di for proof of the Lemma 4.1.2

49

h0 (q(βi )) = h0 (ci ) = h(ci ) = q(mi ) = q(h0 (βi )),
h0 (r(βi )) = h0 (di ) = h(di ) = r(mi ) = r(h0 (βi )).
Thus h0 is a homomorphism, and every homomorphisms in Hom(Qi , M) extends to Hom(Q0i , M).
The properties of the Lemma 4.1.1 hold because ∧, ∨ and 0 are homomorphism; all Qi ’s are
subalgebras and are closed under p, q, r and 0̄; Qi is a k-ary relation on M; β be an element of
M k \ Qi , where Q0i ⊇ Qi ∪ {β } and Qi ≤ Q0i ≤ Mk are all algebras; and there is an embedding ι of
Hom(Qi , M) into Hom(Q0i , M).
Therefore, for 1 ≤ i ≤ 13 (i 6= 2), Hom(Qi , M) embeds into Hom(Q0i , M) in a way that satisfies
the Lemma 4.1.1.

As Q02 is strictly larger Q2 ∪ {β2 }, a slightly different argument is needed for Q2 .
Lemma 4.1.3. Let Q2 , Q02 and β2 be defined as in Chapter 3. There is an embedding ι of
Hom(Q2 , M) into Hom(Q02 , M) satisfying the properties of Lemma 4.1.1.

Proof. First we determine that Hom(Q2 , M) = {0̄, π1 , π2 , π3 ∧ π4 , π3 , π3 ∨ π4 , π4 }, illustrated in
Table 4.3 on the following page.
We can do this by direct calculation (or by using Lemma 4.2.1 and Lemma 4.2.2). It is clear
that each of these extends to D(Q02 ). Thus every homomorphism in Hom(Q2 , M) embeds into
Hom(Q02 , M) in a way that satisfies the Lemma 4.1.1.

Now we define the maps αi : D(Qi ) → M. Recall that Qi and Q0i are subuniverses of Mk as
defined in Chapter 3. Qi values are given in the fourth column of the Table 3.2 on page 21 and
βi is given in the fifth column of the Table 3.2 on page 21. Recall that for h in Hom(Qi , M) there

50

0̄ π1 π2 π3 π4 π3 ∧ π4 π3 ∨ π4
0000 0

0

0

0

0

0

0

0001 0

0

0

0

1

0

1

0010 0

0

0

1

0

0

1

0011 0

0

0

1

1

1

1

0111 0

0

1

1

1

1

1

1111 0

1

1

1

1

1

1

Table 4.3: Homomorphisms of Q2 on {0, 1}-elements.
is a unique h0 in Hom(Q0i , M) such that h = h0 Qi . From Lemma 4.1.1 we see the non-evaluation
morphism αi with respect to hM, ∧, ∨, 0, T i for Qi ≤ Q0i is given by:
αi (h) = eβi (h0 )

for

1 ≤ i ≤ 13.

Note that:

• αi is “evaluate at βi ”, an element in Q0i \ Qi ;
• αi does not respect Qi ;
• αi respects ∧, ∨, 0.
Lemma 4.1.4. For 1 ≤ i ≤ 13, we have range αi ⊇ {π j (βi ) | 1 ≤ j ≤ n}.

Proof. We have αi (π j ) = π j (βi ), and βi ∈ range(αi )k , where βi is k-ary and we are done.

51

4.2

When does the Map αi Respect Q j for j 6= i?

In this section, we verify whether or not the map αi : D(Qi ) → M respects all of the relations
Q j ∈ Q \ {Qi }. First recall that the morphism αi : D(Qi ) → M respects the relation Q j of arity k,
D(Qi )

means if (h1 , . . . , hk ) ∈ Qki is such that (h1 , . . . , hk ) ∈ Q j

then (αi (h1 ), . . . , αi (hk )) ∈ QM
j . Thus

to check that αi preserves a k-ary relation R, we need to know Hom(Qi , M) = D(Qi ). Second recall
that a subset S of a partial ordered set is an up-set if x ∈ S and y ≥ x implies y ∈ S. We define
S01 (X) := X ∩ {0, 1}s

for

X ⊆ Ms.

Lemma 4.2.1. Let A ≤ Ms be an algebra with the property that for all s, and t in S01 (A) where
s < t there is an element u ∈ A such that s = p(u) and t = q(u). Then for h ∈ D(A) we have
S01 (h−1 (1)) is an up-set of S01 (A).

Proof. Let s ∈ S01 (h−1 (1)) and t ∈ S01 (A) be elements such that s < t. Pick u as in the hypothesis. We have 1 = h(s) = h(p(u)) = p(h(u)) ≤ q(h(u)) = h(q(u)) = h(t), so t ∈ S01 (h−1 (1)), and
S01 (h−1 (1)) is an up-set.
Corollary 4.2.1. For A ∈ Q ∪ {Q0i | 1 ≤ i ≤ 13}, and h ∈ D(A) we have S01 (h−1 (1)) is an up-set
of S01 (A).

Proof. This follows from Lemma 4.2.1 because if π j (s) ≤ πk (s) and π j (t) ≤ πk (t) then π j (u) ≤
πk (u) for s ∈ S01 (h−1 (1)), t ∈ S01 (A) and u ∈ A.
−1
Lemma 4.2.2. If h1 , h2 ∈ Hom(Qi , M) and S01 (h−1
1 (1)) = S01 (h2 (1)) then h1 = h2 .

Proof. For a ∈ Qi , we have hp(a), q(a), r(a)i ∈ {0, 1}s . Since h1 , h2 ∈ Hom(Qi , M) and
−1
S01 (h−1
1 (1)) = S01 (h2 (1)) we have h2 (p(a)) = h1 (p(a)), h2 (q(a)) = h1 (q(a)), and h2 (r(a)) =

52

h1 (r(a)) or, we have p(h2 (a)) = p(h1 (a)), q(h2 (a)) = q(h1 (a)), and r(h2 (a)) = r(h1 (a)). From
this we have h2 (a) = h1 (a) (because rows of the Table 3.1 on page 20 are unique); that means
h1 = h2 .

The previous lemmas allow us to compute Hom(Qi , M) by up-sets of S01 (Qi ) := Qi ∩ {0, 1}s .
Before launching into the details of the proof, let us examine the logical structure of what
we wish to show. Recall that αi is a map from D(Qi ) to M. To say that αi respects Q j , where
Q j is k-ary, means that for all ĥ = hh1 , . . . , hk i ∈ D(Qi )k we have that ĥ in Q j implies α(ĥ) :=
hα(h1 ), . . . , α(hk )i in Q j , or contrapositively if α(ĥ) is not in Q j then ĥ is not in Q j . There are two
ways to have α(ĥ) not in Q j :

1. α(ĥ) ∈ M k \ Q0j ,
2. α(ĥ) ∈ Q0j \ Q j .

For j 6= 2, α(ĥ) ∈ Q0j \ Q j means α(ĥ) = β j . We first deal with α(ĥ) ∈ M k \ Q0j . For the rest of this
chapter we look at α(ĥ) ∈ Q0j \ Q j .
Lemma 4.2.3. For i 6= j, and Q j a k-ary relation, we have that for all hh1 , . . . , hk i ∈ D(Qi )k
if hαi (h1 ), . . . , αi (hk )i ∈ M k \ Q0j then hh1 , . . . , hk i is not in Q j .

Proof. We prove the contrapositive. Suppose that hh1 , . . . , hk i in Q j . Clearly hh1 , . . . , hk i in Q0j .
Then, because ι is order-preserving, h(ιh1 ), . . . , (ιhk )i in Q0j , and, in particular,
h(ιh1 )(βi ), . . . , (ιhk )(βi )i = hαi (h1 ), . . . , αi (hk )i ∈ Q0j .

The following lemma is helpful to prove that the map αi respects Q j for j 6= i and j 6= 2.
53

Lemma 4.2.4. For i 6= j, j 6= 2, and Q j a k-ary relation, if range α j * range αi we have that for all
hh1 , . . . , hk i ∈ D(Qi )k
if hαi (h1 ), . . . , αi (hk )i = β j then hh1 , . . . , hk i is not in Q j .
That is, αi respects Q j .

Proof. This follows because for all hh1 , . . . , hk i ∈ D(Qi )k we have hαi (h1 ), . . . , αi (hk )i ∈ (range αi )k ,
but β j ∈
/ (range αi )k .

Now we start to verify whether or not the map αi respects all of the relations Q j for j 6= i.
Lemma 4.2.5. The map α1 : D(Q1 ) → M respects Q j for j 6= 1.

Proof. Note that Q1 is defined in Table 3.2 on page 21 as {hY1 ,Y2 i ∈ M 2 : Y1 ≤ Y2 } \ {h0, 7i}. We
now compute Hom(Q1 , M). To do this, we make an up-sets table. To get all up-sets we first draw
the lattice diagram of S01 (Q1 ). (See Figure 4.1.) From the lattice diagram we find the up-sets of
(2)

q1 := 11

(1)

q1 := 01

(0)

q1 := 00

Figure 4.1: The Lattice Diagram for S01 (Q1 ).

S01 (h−1 (1)), the corresponding h ∈ Hom(Q1 , M), and α1 (h) which are given by Table 4.4 on the
following page.
54

S01 (h−1 (1)) h ∈ Hom(Q1 , M) α1 (h)
0/

h1 = 0̄

(0)

0

{q1 }

(2)

h1 = π1

(1)

0

(2)

h1 = π2

(2)

7

(1)

{q1 , q1 }

Table 4.4: Up-sets table of S01 (Q1 )
(0)

(1)

(2)

(0)

(1)

(2)

We can calculate that Hom(Q1 , M) = {h1 , h1 , h1 }, where h1 = 0̄, h1 = π1 , and h1 = π2 ,
given in Table 4.5.
(0)

(1)

(2)

h1 = 0̄ h1 = π1 h1 = π2
00

0

0

0

01

0

0

1

11

0

1

1

Table 4.5: Homomorphism table of S01 (Q1 )

We have α1 : Hom(Q1 , M) → M is defined by
α1 (h) = (ιh)(β1 ), where β1 = h0, 7i.
Suppose that α1 does not respect R, a k-ary relation. Then there exists
{h1 , h2 , . . . , hk } ⊆ Hom(Q1 , M)
such that hh1 , h2 , . . . , hk i ∈ R, but hα1 (h1 ), α1 (h2 ), . . . , α1 (hk )i ∈
/ R. From the up-set table, for
h ∈ Hom(Q1 , M) we have α1 (h) ∈ {0, 0, 7}.
Firstly, we prove that α1 respects Q j for j 6= 1, 2. We have range α1 = {0, 7}, which by
Lemma 4.1.4 does not contain range α j for j ≥ 3, so by Lemma 4.2.4, the map α1 respects Q j
for j ≥ 3.
55

Secondly, we prove that α1 respects Q2 . Recall that Q2 is defined in Table 3.2 on page 21. Suppose hh1 , h2 , h3 , h4 i ∈ Hom(Q1 , M)4 where h1 ≤ h2 ≤ h3 and h1 ≤ h4 such that hα1 (h1 ), . . . , α1 (h4 )i
(0)

(1)

(2)

(2)

∈ Q02 \ Q2 , here (h1 , h4 ) ∈ {h1 , h1 }, h2 = h1 , and h3 = h1 . As range α1 = {0, 7} this means
hα1 (h1 ), . . . , α1 (h4 )i = h0, 7, 7, 0i. Thus hh1 , h2 , h3 , h4 i(0, 3) = h0, 3, 3, 0i ∈
/ Q2 .
So hh1 , h2 , h3 , h4 i ∈
/ Q2 . Thus α1 respects Q2 .
Hence the map α1 respects Q j for j 6= 1.

The above Lemma is looked at in detail. The following Lemmas are simplified.
Lemma 4.2.6. The map α2 : D(Q2 ) → M respects Q j for j 6= 2.

Proof. Note that Q2 is defined in Table 3.2 on page 21. By drawing the lattice diagram of S01 (Q2 )
(See Figure 4.2) we determine the up-sets of S01 (h−1 (1)), the corresponding h ∈ Hom(Q2 , M),
and α2 (h). See Table 4.6.
S01 (h−1 (1))

h ∈ Hom(Q2 , M) α2 (h)

0/

h2 = 0̄

(0)

0

{q2 }

(5)

h2 = π1

(1)

0

(5)

h2 = π2

(2)

1

(3)

0

(4)

{q2 , q2 }
(5)

(4)

(3)

h2 = π3 ∧ π4

{q2 , q2 , q2 }
(5)

(4)

(3)

(2)

h2 = π3

(4)

1

(5)

(4)

(3)

(1)

h2 = π4

(5)

0

(6)

1

{q2 , q2 , q2 , q2 }
{q2 , q2 , q2 , q2 }
(5)

(4)

(3)

(2)

(1)

{q2 , q2 , q2 , q2 , q2 }

h2 = π3 ∨ π4

Table 4.6: Up-sets table of S01 (Q2 )

(0)

(1)

(6)

(0)

(1)

(2)

We can calculate that Hom(Q2 , M) = {h2 , h2 , . . . , h2 }, where h2 = 0̄, h2 = π1 , h2 = π2 ,
56

(5)

q2 := 1111

(4)

q2 := 0111

(3)

q2 := 0011

(2)

(1)

q2 := 0010

q2 := 0001

(0)

q2 := 0000

Figure 4.2: The Lattice Diagram for S01 (Q2 ).

57

(3)

(4)

(5)

(6)

h2 = π3 ∧ π4 , h2 = π3 , h2 = π4 , and h2 = π3 ∨ π4 , defined in Table 4.7.
(0)

(1)

(2)

(3)

(4)

(5)

(6)

h2

h2

h2

h2

h2

h2

h2

0000

0

0

0

0

0

0

0

0001

0

0

0

0

0

1

1

0010

0

0

0

0

1

0

1

0011

0

0

0

1

1

1

1

0111

0

0

1

1

1

1

1

1111

0

1

1

1

1

1

1

Table 4.7: Homomorphism table of S01 (Q2 )

We have α2 : Hom(Q2 , M) → M is defined by
α2 (h) = (ιh)(β2 ), where β2 = h0, 1, 1, 0i.
From the up-set table, for h ∈ Hom(Q2 , M) we have α2 (h) ∈ {0, 0, 1, 0, 1, 0, 1}, and from Table 3.2
we get β j ∈
/ {0, 1}k for j 6= 2. By Lemma 4.2.4 the map α2 respects Q j for j 6= 2.
Lemma 4.2.7. The map α4 : D(Q4 ) → M respects Q j for j 6= 4.

Proof. Note that Q4 = {hY1 ,Y2 ,Y3 ,Y4 i ∈ M 4 : Y1 ≤ Y2 ≤ Y3 and Y1 ≤ Y4 } \ {h0, 3, 3, 0i}, defined in
Table 3.2. Now we draw the lattice diagram of S01 (Q4 ). (See Figure 4.3 on the following page.)
From the lattice diagram we find the up-sets of S01 (h−1 (1)), the corresponding h ∈ Hom(Q4 , M),
and α4 (h). See Table 4.8 on page 60.
(0)

(1)

(9)

(0)

(1)

(2)

We can determine that Hom(Q4 , M) = {h4 , h4 , . . . , h4 }, where h4 = 0̄, h4 = π1 , h4 =
(3)

(4)

(5)

(6)

(7)

(8)

π2 ∧ π4 , h4 = π2 , h4 = π3 ∧ π4 , h4 = π2 ∨ (π3 ∧ π4 ), h4 = π4 , h4 = π3 , h4 = π2 ∨ π4 , and
(9)

h4 = π3 ∨ π4 , defined in Table 4.9 on page 60.
58

(6)

q4 := 1111

(5)

q4 := 0111

q4 := 0110

(4)

q4 := 0011

(3)

(2)

q4 := 0001

(1)

q4 := 0010

(0)

q4 := 0000

Figure 4.3: The Lattice Diagram for S01 (Q4 ).

59

S01 (h−1 (1))

h ∈ Hom(Q4 , M)

0/

h4 = 0̄

(0)

0

(6)

h4 = π 1

(1)

0

(2)

0

{q4 }
(6)

(5)

{q4 , q4 }

α4 (h)

h4 = π2 ∧ π4
(3)

(6)

(5)

(4)

h4 = π 2

3

(6)

(5)

(3)

h4 = π3 ∧ π4

(4)

0

{q4 , q4 , q4 }
{q4 , q4 , q4 }

(5)

(6)

(5)

(4)

(3)

h4 = π2 ∨ (π3 ∧ π4 )

{q4 , q4 , q4 , q4 }

(6)

(5)

(4)

(1)

h4 = π 4

(6)

0

(6)

(5)

(4)

(3)

(2)

h4 = π 3

(7)

3

{q4 , q4 , q4 , q4 , q4 }

(6)

(5)

(4)

(3)

(1)

h4 = π2 ∨ π4

(8)

3

(6)

h4 = π3 ∨ π4

(9)

3

{q4 , q4 , q4 , q4 }

{q4 , q4 , q4 , q4 , q4 }

(5)

(4)

(3)

(2)

(1)

{q4 , q4 , q4 , q4 , q4 , q4 }

3

Table 4.8: Up-sets table of S01 (Q4 )

(0)

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

h4

h4

h4

h4

h4

h4

h4

h4

h4

h4

0000

0

0

0

0

0

0

0

0

0

0

0001

0

0

0

0

0

0

1

0

1

1

0010

0

0

0

0

0

0

0

1

0

1

0011

0

0

0

0

1

1

1

1

1

1

0110

0

0

0

1

0

1

0

1

1

1

0111

0

0

1

1

1

1

1

1

1

1

1111

0

1

1

1

1

1

1

1

1

1

Table 4.9: Homomorphism table of S01 (Q4 )

60

We have α4 : Hom(Q4 , M) → M is defined by
α4 (h) = (ιh)(β4 ), where β4 = h0, 3, 3, 0i.
From the up-set table, we have α4 (h) ∈ {0, 0, 0, 3, 0, 3, 0, 3, 3, 3}.
Firstly, we prove that α4 respects Q j for j 6= 2, 4. We have range α4 = {0, 3}, which by
Lemma 4.1.4 does not contain range α j for j ≥ 3, so by Lemma 4.2.4, the map α4 respects Q j
for j ≥ 3.
Secondly, we prove that α4 respects Q2 . Recall that Q2 is defined in Table 3.2 on page 21. Suppose hh1 , h2 , h3 , h4 i ∈ Hom(Q4 , M)4 where h1 ≤ h2 ≤ h3 and h1 ≤ h4 such that hα4 (h1 ), . . . , α4 (h4 )i
(0)

(1)

(2)

(4)

(6)

(3)

(5)

(7)

(8)

(9)

∈ Q02 \ Q2 , here (h1 , h4 ) ∈ {h4 , h4 , h4 , h4 , h4 }, and (h2 , h3 ) ∈ {h4 , h4 , h4 , h4 , h4 }. As
range α3 = {0, 3} this means hα4 (h1 ), . . . , α4 (h4 )i = h0, 3, 3, 0i. Thus hh1 , h2 , h3 , h4 i(0, 7, 7, 0) =
h0, 7, 7, 0i ∈
/ Q2 . So hh1 , h2 , h3 , h4 i ∈
/ Q2 . Thus α4 respects Q2 .
Hence the map α4 : D(Q4 ) → M respects Q j for j 6= 4.
Lemma 4.2.8. If 7 ∈ range αi and 7 ∈
/ range α j and j 6= 2, then αi respects Q j .

Proof. Let a be an element of Qi gotten by replacing a 7 in βi with a 3. Suppose that η =

V

k∈K πk .

If η(βi ) ≤ 3 then η(βi ) = η(a) as 3 ∧ x = x for x ≤ 3. Now we know from the computation of
D(Qi ) that h ∈ D(Qi ) can be written as the join of meets of projections. If h(βi ) ≤ 3 then each of
the joinands is less than or equal to 3, and by the preceding argument h(βi ) = h(a).
As 7 ∈
/ range(α j ), if there is a k-tuple of homomorphisms in D(Qi ) such that hh1 , . . . , hk i(βi ) =
β j then hh1 , . . . , hk i(a) = β j , which proves that αi respects Q j .
Lemma 4.2.9. For i 6= 2, if there is a four tuple ĥ ∈ D(Qi )4 such that α(ĥ) ∈ Q02 \ Q2 then there is
an element a ∈ Sg({βi }) ∩ Qi such that ĥ(a) = h0, 1, 1, 0i.
61

Proof. Suppose that α(ĥ) ∈ Q02 \ Q2 . That means there is a term operation f such that f (α(ĥ)) =
h0, 1, 1, 0i. Set a = r( f (βi )). Clearly a ∈ Sg({βi }). As a 6= βi we have a ∈ Qi (Lemma 3.2.1). Now
ĥ(a) = ĥ(r( f (a)) = r( f (αi (ĥ))) = r(h0, 1, 1, 0i) = h0, 1, 1, 0i .
Corollary 4.2.2. For i 6= 2, the morphism αi respects the relation Q2 .

Now we can prove the following theorem. It implies that removing any Qi with i ≤ 11 results
in an alter ego that is not dualizing.
Theorem 4.2.1. The map αi : D(Qi ) → M respects Q j for j 6= i, with the following exceptions.

1. α12 does not respect Q8 or Q9 ; and
2. α13 does not respect Q10 or Q11 .

Proof. From Lemma 4.2.5, the map α1 respects Q j , for j 6= 1. From Lemma 4.2.6, the map α2
respects Q j , for j 6= 2. From Lemma 4.2.7, the map α4 respects Q j , for j 6= 4.
The remainder of the calculations are gathered in Tables 4.10 on page 64, 4.11 on page 65,
and 4.12 on page 66. In these tables, in row αi and column Q j a string of digits (for instance 00070
in α7 row and Q1 column) represents an element ai of Qi and means the following. Suppose that
Q j ≤ Ms , and that ĥ is an s-tuple of homomorphisms from D(αi ) such that αi (ĥ) = β j where β j is
the element excluded from Q j . Then ĥ(ai ) = β j also.
Such a claim is verified by computing D(Qi ), which is done by using Lemmas 4.2.1 and 4.2.2.
The lattice diagrams for the up-sets of S01 (Qi ) are shown in Figure 4.4 on page 67, Figure 4.5
on page 68, Figure 4.6 on page 69, Figure 4.7 on page 70, Figure 4.8 on page 71, Figure 4.9 on
page 72.

62

In the four cases where αi does not respect Q j (i 6= j), the calculations were verified by computer analysis by David Casperson.

Now we prove the following theorem:
Theorem 4.2.2. The alter ego hM, ∧, ∨, 0, Q, T i is a dualizing structure. The set Q is minimal
with respect to the the alter egos that extend hM, {∧, ∨, 0}, {Q12 , Q13 }, T i.

Proof. In Chapter 3 we showed that the algebra M is dualized by the alter-ego hM, ∧, ∨, 0, Q, T i.
In this Chapter we proved that each relation Q j in Q except Q12 and Q13 is a necessary part of a
dualizing structure for the five-element algebra by finding a specific morphism which is an nonevaluation morphism, when Q j is dropped. Thus the set Q is minimal with respect to the the alter
egos that extend hM, {∧, ∨, 0}, {Q12 , Q13 }, T i.

From Theorem 4.2.2, the set Q is minimal with respect to the the alter egos that extend
hM, {∧, ∨, 0}, {Q12 , Q13 }, T i.

63

Q1

Q2

Q3

Q4

Q5

α1

No, Lem 4.1.1

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

Lem 4.2.5

α2

Lem 4.2.6

No, Lem 4.1.1

Lem 4.2.6

Lem 4.2.6

Lem 4.2.6

α3

Lem 4.2.4

Lem 4.2.9

No, Lem 4.1.1

Lemm 4.2.4

Lem 4.2.4

α4

Lem 4.2.7

Lem 4.2.9

Lem 4.2.7

No, Lem 4.1.1

Lem 4.2.7

α5

00070

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

No, Lem 4.1.1

α6

00070

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

Lem 4.2.4

α7

00070

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

Lem 4.2.4

α8

Lem 4.2.4

Lem 4.2.9

Lem 4.2.4

000330

Lem 4.2.4

α9

Lem 4.2.4

Lem 4.2.9

022220

000330

Lem 4.2.4

α10

0000070

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

0111170

α11

0000070

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

Lem 4.2.4

α12

Lem 4.2.4

Lem 4.2.9

00022022

00033033

Lem 4.2.4

α13

000007000

Lem 4.2.9

Lem 4.2.8

Lem 4.2.8

011117000

See the proof of Theorem 4.2.1 for the meaning of the entries.
Table 4.10: The map αi respects Q j for j 6= i (Part I).

64

Q6

Q7

Q8

Q9

α1

Lem 4.2.5

Lem 4.2.5

Lem 4.2.8

Lem 4.2.8

α2

Lem 4.2.6

Lem 4.2.6

Lem 4.2.6

Lem 4.2.6

α3

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

α4

Lem 4.2.7

Lem 4.2.7

Lem 4.2.7

Lem 4.2.7

α5

Lem 4.2.4

Lem 4.2.4

Lem 4.2.8

Lem 4.2.8

α6

No, (Lem 4.1.1)

Lem 4.2.4

Lem 4.2.8

Lem 4.2.8

α7

Lem 4.2.4

No, (Lem 4.1.1)

Lem 4.2.8

Lem 4.2.8

α8

Lem 4.2.4

Lem 4.2.4

No, (Lem 4.1.1)

Lem 4.2.4

α9

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

No, (Lem 4.1.1)

α10

Lem 4.2.4

0003370

Lem 4.2.8

Lem 4.2.8

α11

0222270

0003370

Lem 4.2.8

Lem 4.2.8

α12

Lem 4.2.4

Lem 4.2.4

No! (comp)

No! (comp)

α13

000227022

000337033

Lem 4.2.8

Lem 4.2.8

See the proof of Theorem 4.2.1 for the meaning of the entries.
Table 4.11: Where αi respects Q j (Part II).

65

Q10

Q11

Q12

Q13

α1

Lem 4.2.5

Lem 4.2.5

Lem 4.2.8

Lem 4.2.5

α2

Lem 4.2.6

Lem 4.2.6

Lem 4.2.6

Lem 4.2.6

α3

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

α4

Lem 4.2.7

Lem 4.2.7

Lem 4.2.7

Lem 4.2.7

α5

Lem 4.2.4

Lem 4.2.4

Lem 4.2.8

Lem 4.2.4

α6

Lem 4.2.4

Lem 4.2.4

Lem 4.2.8

Lem 4.2.4

α7

Lem 4.2.4

Lem 4.2.4

Lem 4.2.8

Lem 4.2.4

α8

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

α9

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

Lem 4.2.4

α10

No, (Lem 4.1.1)

Lem 4.2.4

Lem 4.2.8

Lem 4.2.4

α11

Lem 4.2.4

No, (Lem 4.1.1)

Lem 4.2.8

Lem 4.2.4

α12

Lem 4.2.4

Lem 4.2.4

No, (Lem 4.1.1)

Lem 4.2.4

α13

No! (comp)

No! (comp)

Lem 4.2.8

No, (Lem 4.1.1)

See the proof of Theorem 4.2.1 for the meaning of the entries.
Table 4.12: Where αi respects Q j (Part III).

66

1111

0111

0110

0011

0010

0001

0000

Figure 4.4: The Lattice Diagram for S01 (Q3 ), and S01 (Q4 ).

67

11111

01111

01110

00111

00110

00011

00010

00001

00000

Figure 4.5: The Lattice Diagram for S01 (Q5 ), S01 (Q6 ), and S01 (Q7 ).

68

111111

011111

011110

001111

001110

000111

000110

000011

000010

000001

000000

Figure 4.6: The Lattice Diagram for S01 (Q8 ), and S01 (Q9 ).

69

1111111

0111111

0111110

0011111

0011110

0001111

0001110

0000111

0000110

0000011

0000010

0000001

0000000

Figure 4.7: The Lattice Diagram for S01 (Q10 ), and S01 (Q11 ).

70

11111111

11111011

11111001

11111000

01111111

01111011

01111001

01111000

00111111

00111011

00111001

00111000

00011111

00011011

00011001

00011000

00001000

00000000

Figure 4.8: The Lattice Diagram for S01 (Q12 ).

71

111111111

111111011

111111001

111111000

011111111

011111011

011111001

011111000

001111111

001111011

001111001

001111000

000111111

000111011

000111001

000111000

000011000

000001000

000000000

Figure 4.9: The Lattice Diagram for S01 (Q13 ).

72

Chapter 5

Conclusion

5.1

Summary

In this thesis we looked for when an alter ego is a minimal dualizing structure.
To summarize, we covered the required preliminary materials such as notations, definitions,
examples, and theorems for algebras, lattices, quasivarieties, topology, and dualizability in Chapter 2. In Chapter 3, we defined a five-element algebra, an alter ego and concluded that our defined
alter ego satisfies the interpolation condition relative to the algebra and thus dualizes the algebra.
That means, we proved the alter-ego M := hM, ∧, ∨, 0, Q, T i dualizes the algebra M. In Chapter
4, we showed that the set of relations is a necessary part of a dualizing structure including ∧, ∨,
and 0 by finding a specific morphism which is an non-evaluation morphism. We proved that the
set Q is minimal with respect to the the alter egos that extend hM, {∧, ∨, 0}, {Q12 , Q13 }, T i (see
Theorem 4.2.2).

73

5.2

Future Research

In this thesis we showed that our defined alter-ego dualizes the five-element algebra and a particular
relation of a dualizing structure is a necessary part for the alter ego. In this paper we did not show
that M was minimal in the way that we originally claimed it was, we have some obvious material
to include in future research.
For further research, with the current work of this paper, the following questions about unary
algebras and their dualizability are:
In this paper we used tables and 13 cases. Can we simplify the cases of Chapter 3 and Chapter
4 of this thesis to determine the dualizability results?
In this thesis we proved that the set Q is minimal with respect to the the alter egos that extend
hM, {∧, ∨, 0}, {Q12 , Q13 }, T i. Can we show that the set Q is a minimal dualizing structure?
Moreover, can we describe a minimal alter ego for any finite unary algebra with an underlying
lattice structure?
However, we did not look at fully dualizable or strongly dualizable. Can we find that our
defined alter-ego fully dualizes or strongly dualizes on the five-element algebra?
Furthermore, for {0, 1}-valued unary algebras with more than five elements, can we find nice
conditions for dualizability, full dualizability or not full dualizability, strong dualizability or not
strong dualizability?
Following the work of this paper, is it possible to find whether or not any {0, 1}-valued unary
algebras are dualizable or fully dualizable? Can we research {0, 1}-valued unary algebras that do
not have a constant zero-valued function?
74

Finally, with the technique of this thesis, can we find that there exists an algebra which is fully
dualizable but is not strongly dualizable?

75

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78