Algebraic General Topology. Vol 1: Paperback / E-book || Axiomatic Theory of Formulas: Paperback / E-book

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Algebraic General Topology. Volume 2 partial
draft
Victor Porton
Email address: porton@narod.ru
URL: http://www.mathematics21.org
2000 Mathematics Subject Classification. 54J05, 54A05, 54D99, 54E05, 54E15,
54E17, 54E99
Key words and phrases. algebraic general topology, quasi-uniform spaces,
generalizations of proximity spaces, generalizations of nearness spaces,
generalizations of uniform spaces
Abstract. Partial rough draft of volume 2 of Algebraic General Topology
book. This volume is meant to contain materials which refer to more advanced
prerequisites than plain ZFC (such as category theory and classical pointfree
topology). This is a very rough draft.
Contents
Chapter 1. Introduction 6
Chapter 2. Products in dagger categories with complete ordered Hom-sets 8
1. General product in partially ordered dagger category 8
2. On duality 11
3. General coproduct in partially ordered dagger category 11
4. Applying this to the theory of funcoids and reloids 13
5. Initial and terminal objects 14
6. Canonical product and subatomic product 14
7. Further plans 15
8. Cartesian closedness 15
9. Is category Rld cartesian closed? 18
Chapter 3. Equalizers and co-Equalizers in Certain Categories 19
1. Equalizers 19
2. Co-equalizers 19
3. Rest 20
Chapter 4. Categories of filters 21
Chapter 5. Power of filters 23
1. Germs of functions 23
2. Power of filters 24
Chapter 6. Matters related to tensor product 25
Chapter 7. Mappings between endofuncoids and topological spaces 27
Chapter 8. Funcoids as closed sets 30
Chapter 9. Categories related with funcoids 31
1. Draft status 31
2. Topic of this article 31
3. Category of continuous morphisms 31
4. Definition of the categories 32
5. Isomorphisms 32
6. Direct products 33
Chapter 10. Product of funcoids over a filter 35
1. More on product of reloids 36
Chapter 11. Compact funcoids 37
1. The rest 37
Chapter 12. Pointfree funcoids as a generalization of frames 41
1. Definitions 41
2. Postface 42
3
CONTENTS 4
Chapter 13. Singularities 43
1. Singularities funcoids: some special cases 43
2. Using plain funcoids 43
3. Singularities funcoids: special cases proof attempts 44
Bibliography 46
CONTENTS 5
This file is a rough draft.
It is a continuation of [5].
CHAPTER 1
Introduction
I remind some definitions from volume 1 [5].
I denote a set definition like
n
xA
P (x)
o
instead of customary {x A | P (x)} (in
order to reduce formulas size).
I denote partial order as v. I denote lattice operations as
d
,
d
, u, t.
The following generalizes monovalued morphisms in category Rel.
Let Hom-sets be complete lattices.
Definition 1955. A morphism f of a partially ordered category is metamono-
valued when (
d
G) f =
d
gG
(g f) whenever G is a set of morphisms with a
suitable domain and image.
Definition 1956. A morphism f of a partially ordered category is metainjec-
tive when f (
d
G) =
d
gG
(f g) whenever G is a set of morphisms with a suitable
domain and image.
Obvious 1957. Metamonovaluedness and metainjectivity are dual to each
other.
Definition 1958. A morphism f of a partially ordered category is metacom-
plete when f (
d
G) =
d
gG
(f g) whenever G is a set of morphisms with a
suitable domain and image.
Definition 1959. A morphism f of a partially ordered category is co-
metacomplete when (
d
G) f =
d
gG
(g f) whenever G is a set of morphisms
with a suitable domain and image.
Let now Hom-sets be meet-semilattices.
Definition 1960. A morphism f of a partially ordered category is weakly
metamonovalued when (g uh)f = (g f)u(hf ) whenever g and h are morphisms
with a suitable domain and image.
Definition 1961. A morphism f of a partially ordered category is weakly
metainjective when f (g u h) = (f g) u (f h) whenever g and h are morphisms
with a suitable domain and image.
Let now Hom-sets be join-semilattices.
Definition 1962. A morphism f of a partially ordered category is weakly
metacomplete when f (g t h) = (f g) t (f h) whenever g and h are morphisms
with a suitable domain and image.
Definition 1963. A morphism f of a partially ordered category is weakly co-
metacomplete when (g t h) f = (g f) t (h f ) whenever g and h are morphisms
with a suitable domain and image.
Obvious 1964.
1
. Metamonovalued morphisms are weakly metamonovalued.
2
. Metainjective morphisms are weakly metainjective.
6
1. INTRODUCTION 7
3
. Metacomplete morphisms are weakly metacomplete.
4
. Co-metacomplete morphisms are weakly co-metacomplete.
Definition 1965. For a partially ordered dagger category I will call monoval-
ued morphism such a morphism f that f f
v 1
Dst f
.
Definition 1966. For a partially ordered dagger category I will call entirely
defined morphism such a morphism f that f
f w 1
Src f
.
Definition 1967. For a partially ordered dagger category I will call injective
morphism such a morphism f that f
f v 1
Src f
.
Definition 1968. For a partially ordered dagger category I will call surjective
morphism such a morphism f that f f
w 1
Dst f
.
Remark 1969. It is easy to show that this is a generalization of monovalued,
entirely defined, injective, and surjective functions as morphisms of the category
Rel.
Obvious 1970. “Injective morphism” is a dual of “monovalued morphism” and
“surjective morphism” is a dual of “entirely defined morphism”.
CHAPTER 2
Products in dagger categories with complete
ordered Hom-sets
FiXme: This is a rough draft. It is not yet checked for errors.
Note 1971. What I previously denoted
Q
F is now denoted
Q
(L)
F (and like-
wise for
`
). The other draft chapters referring to this chapter may be not yet
updated.
Proposition 1972. FiXme: Should we move this to volume 1?
1
. Every entirely defined monovalued morphism is metamonovalued and
metacomplete.
2
. Every surjective injective morphism is metainjective and co-
metacomplete.
Proof. Let’s prove the first (the second follows from duality):
Let f be an entirely defined monovalued morphism.
(
d
G) f v
d
gG
(g f) by monotonicity of composition.
Using the fact that f is monovalued and entirely defined:
d
gG
(g f)
f
v
d
gG
(g f f
) v
d
G;
d
gG
(g f) v
d
gG
(g f)
f
f v (
d
G) f.
So (
d
G) f =
d
gG
(g f).
Let f be a entirely defined monovalued morphism.
f (
d
G) w
d
gG
(f g) by monotonicity of composition.
Using the fact that f is entirely defined and monovalued:
f
d
gG
(f g)
w
d
gG
(f
f g) w
d
G;
d
gG
(f g) w f f
d
gG
(f g) w f (
d
G).
So f (
d
G) =
d
gG
(f g).
1. General product in partially ordered dagger category
To understand the below better, you can restrict your imagination to the case
when C is the category Rel.
1.1. Infimum product. Let C be a dagger category, each Hom-set of which
is a complete lattice (having order agreed with the dagger).
We will designate some morphisms as principal and require that principal mor-
phisms are both metacomplete and co-metacomplete. (For a particular example of
the category Rel, all morphisms are considered principal.)
Let
Q
(Q)
X be an object for each indexed family X of objects.
Let π be a partial function mapping elements X dom π (which consists of
small indexed families of objects of C) to indexed families
Q
(Q)
X X
i
of principal
morphisms (called projections) for every i dom X.
We will denote particular morphisms as π
X
i
.
8
1. GENERAL PRODUCT IN PARTIALLY ORDERED DAGGER CATEGORY 9
Remark 1973. In some important examples the function π is entire, that is
dom π is the set of all small indexed families of objects of C. However there are also
some important examples where it is partial.
Definition 1974. Infimum product
Q
F (such that π is defined at λj n :
Src F
j
and λj n : Dst F
j
) is defined by the formula
(L)
Y
F =
l
idom F
((π
λjn:Dst F
j
i
)
F
i
π
λjn:Src F
j
i
).
This formula can be (over)simplified to:
(L)
Y
F =
l
idom F
((π
Dst F
i
)
F
i
π
Src F
i
).
Remark 1975. (π
λjn:Dst F
j
i
)
F
i
π
λjn:Src F
j
i
Hom
Q
(Q)
jn
Src F
j
,
Q
(Q)
jn
Dst F
j
are properly defined and have the same sources
and destination (whenever i dom F is), thus the meet in the formulas is properly
defined.
Remark 1976. Thus
F
0
×
(L)
F
1
= ((π
(Dst F
0
,Dst F
1
)
0
)
F
0
π
(Src F
0
,Src F
1
)
0
)u((π
(Dst F
0
,Dst F
1
)
1
)
F
1
π
(Src F
0
,Src F
1
)
1
)
that is product is defined by a pure algebraic formula.
Proposition 1977.
Q
(L)
F = max
(
ΦHom
Q
(Q)
jn
Src F
j
,
Q
(Q)
jn
Dst F
j
inv(π
λjn:Dst F
j
i
)
F
i
π
λjn:Src F
j
i
)
.
Proof. By definition of meet on a complete lattice.
Corollary 1978.
Q
(L)
F =
d
(
ΦHom
Q
(Q)
jn
Src F
j
,
Q
(Q)
jn
Dst F
j
inv(π
λjn:Dst F
j
i
)
F
i
π
λjn:Src F
j
i
)
.
Theorem 1979. Let π
X
i
be metamonovalued morphisms. If S
P(Hom(A
0
, B
0
) × Hom(A
1
, B
1
)) for some sets A
0
, B
0
, A
1
, B
1
then
l
a ×
(L)
b
(a, b) S
=
l
dom S ×
(L)
l
im S.
Proof.
l
a × b
(a, b) S
=
l
(
((π
(Dst a,Dst b)
0
)
a π
(Src a,Src b)
0
) u ((π
(Dst a,Dst b)
1
)
b π
(Src a,Src b)
1
)
(a, b) S
)
=
l
(
(π
(Dst a,Dst b)
0
)
a π
(Src a,Src b)
0
a dom S
)
u
l
(
(π
(Dst a,Dst b)
1
)
b π
(Src a,Src b)
1
b im S
)
=
(π
(Dst a,Dst b)
0
)
l
n
a
a dom S
o
π
(Src a,Src b)
0
u
(π
(Dst a,Dst b)
1
)
l
b
b im S
π
(Src a,Src b)
1
=
(π
(Dst a,Dst b)
0
)
l
dom S
π
(Src a,Src b)
0
u
(π
(Dst a,Dst b)
1
)
l
im S
π
(Src a,Src b)
1
=
l
dom S ×
l
im S.
Corollary 1980. (a
0
×
(L)
b
0
) u (a
1
×
(L)
b
1
) = (a
0
u a
1
) ×
(L)
(b
0
u b
1
).
Corollary 1981. a
0
×
(L)
b
0
6 a
1
×
(L)
b
1
a
0
6 a
1
b
0
6 b
1
.
1. GENERAL PRODUCT IN PARTIALLY ORDERED DAGGER CATEGORY 10
1.2. Infimum product for endomorphisms. Let F is an indexed family of
endomorphisms of C.
I will denote Ob f the object (source and destination) of an endomorphism f.
Let also π
X
i
be a monovalued entirely defined morphism (for each i dom F).
Then
Q
(L)
F =
d
idom F
((π
λjn:Ob F
j
i
)
F
i
π
λjn:Ob F
j
i
) (if π is defined at
λj n : Ob F
j
).
Abbreviate π
i
= π
λjn:Ob F
j
i
.
So
Q
(L)
F =
d
idom F
((π
i
)
F
i
π
i
).
Q
(L)
F = max
(
ΦEnd
Q
(Q)
jn
Ob F
j
inv(π
i
)
F
i
π
i
)
.
Taking into account that π
i
is a monovalued entirely defined morphism, we get:
Obvious 1982.
Q
(L)
F = max
(
ΦEnd
Q
(Q)
jn
Ob F
j
in:π
i
C(Φ,F
i
)
)
.
Remark 1983. The above formula may allow to define the product for non-
dagger categories (but only for endomorphisms). In this writing I don’t introduce
a notation for this, however.
Corollary 1984. π
i
C
Q
(L)
F, F
i
for every i dom F .
1.3. Category of continuous morphisms. Let π
i
= π
X
i
(for i dom F) be
entirely defined monovalued morphisms (we suppose it is defined at X).
Let
N
of an indexed family of morphisms is a morphism; π
i
N
f = f
i
;
N
in
(π
i
f ) = f .
Definition 1985. The category cont(C) is defined as follows:
Objects are endomorphisms of the category C.
Morphisms are triples (f, a, b) where a and b are objects and f : Ob a
Ob b is an entirely defined monovalue principal morphism of the category
C such that f C(a, b) (in other words, f a v b f).
Composition of morphisms is defined by the formula (g, b, c) (f, a, b) =
(g f, a, c).
Identity morphisms are (a, a, 1
C
a
).
It is really a category:
Proof. We need to prove that: composition of morphisms is a morphism,
composition is associative, and identity morphisms can be canceled on the left and
on the right.
That composition of morphisms is a morphism by properties of generalized
continuity.
That composition is associative is obvious.
That identity morphisms can be canceled on the left and on the right is obvious.
Remark 1986. The “physical” meaning of this category is:
Objects (endomorphisms of C) are spaces.
Morphisms are continuous functions between spaces.
f a v b f intuitively means that f combined with an infinitely small is
less than infinitely small combined with f (that is f is continuous).
Definition 1987. π
cont(C)
i
=
Q
(L)
F, F
i
, π
i
.
Proposition 1988. π
i
are continuous, that is π
cont
(C)
i
are morphisms.
3. GENERAL COPRODUCT IN PARTIALLY ORDERED DAGGER CATEGORY 11
Proof. We need to prove π
i
C
Q
(L)
F, F
i
but that was proved above.
Lemma 1989. f Hom
cont(C)
Y,
Q
(L)
F
is continuous iff all π
i
f are con-
tinuous.
Proof.
. Let f Hom
cont(C)
Y,
Q
(L)
F
. Then f Y v
Q
(L)
F
f ; π
i
f Y v
π
i
Q
(L)
F
f; π
i
f Y v
Q
(L)
F
π
i
f. Thus π
i
f is continuous.
. Let all π
i
f be continuous. Then π
cont(C)
i
f Hom
cont(C)
(Y, F
i
); π
cont(C)
i
f Y v F
i
π
cont(C)
i
f. We need to prove Y v f
Q
(L)
F
f that is
Y v f
l
in
((π
i
)
F
i
π
i
) f
for what is enough (because f is metamonovalued)
Y v
l
in
(f
(π
i
)
F
i
π
i
f )
what follows from Y v
d
in
(f
(π
i
)
π
i
f Y ) what is obvious.
Theorem 1990.
Q
(L)
together with
N
is a (partial) product in the category
cont(C).
Proof. Obvious.
Check http://math.stackexchange.com/questions/102632/
how-to-check-whether-it-is-a-direct-product/102677#102677
2. On duality
We will consider duality where both the category C and orders on Mor-sets are
replaced with their dual. I will denote A
dual
B when two formulas A and B are
dual with this duality.
Proposition 1991. f C(µ, ν)
dual
f
C(ν
, µ
).
Proof. f C(µ, ν) f µ v ν f
dual
µ
f
w f
ν
1
f
C(ν
, µ
).
f is entirely defined f
f w 1
Src f
dual
f
f v 1
Src f
f is injective
f
is monovalued.
f is monovalued f f
v 1
Dst f
dual
f f
w 1
Dst f
f is surjective
f
is entirely defined.
3. General coproduct in partially ordered dagger category
The below is the dual of the above, proofs are omitted as they are dual.
Let ι
i
FiXme: What is ι? are entirely defined monovalued morphisms to an
object Z.
Let ι
i
dual
π
i
that is ι
i
= (π
i
)
. We have the above equivalent to π
i
being
monovalued and entirely defined.
3. GENERAL COPRODUCT IN PARTIALLY ORDERED DAGGER CATEGORY 12
3.1. Supremum coproduct. Let C be a dagger category, each Hom-set of
which is a complete lattice (having order agreed with the dagger).
We will designate some morphisms as principal and require that principal mor-
phisms are both metacomplete and co-metacomplete. (For a particular example of
the category Rel, all morphisms are considered principal.)
Let
`
(Q)
X be an object for each indexed family X of objects.
Let ι be a partial function mapping elements X dom ι (which consists of
small indexed families of objects of C) to indexed families X
i
`
(Q)
X of principal
morphisms (called injections) for every i dom X.
Definition 1992. Supremum coproduct
`
(L)
F (such that ι is defined at λj
n : Dst F
j
and λj n : Src F
j
) is defined by the formula
(L)
a
F =
l
idom F
(ι
λjn:Src F
j
i
F
i
(ι
λjn:Dst F
j
i
)
).
This formula can be (over)simplified to:
(L)
a
F =
l
idom F
(ι
Src F
i
F
i
(ι
Dst F
i
)
).
Remark 1993. ι
λjn:Src F
j
i
F
i
(ι
λjn:Dst F
j
i
)
Hom
`
(Q)
jn
Src F
j
,
`
(Q)
jn
Dst F
j
are properly defined and have the same sources
and destination (whenever i dom F is), thus the meet in the formulas is properly
defined.
Remark 1994. Thus
F
0
q
(L)
F
1
= (ι
(Src F
0
,Src F
1
)
0
F
0
(ι
(Dst F
0
,Dst F
1
)
0
)
)t(ι
(Src F
0
,Src F
1
)
1
F
1
(ι
(Dst F
0
,Dst F
1
)
1
)
)
that is coproduct is defined by a pure algebraic formula.
Proposition 1995.
`
(L)
F = min
(
ΦEnd
`
(Q)
jn
Ob F
j
inwι
λjn:Src F
j
i
F
i
(ι
λjn:Dst F
j
i
)
)
.
Proof. By definition of meet on a complete lattice.
Corollary 1996.
`
(L)
F =
d
(
ΦEnd
`
(Q)
jn
Ob F
j
inwι
λjn:Src F
j
i
F
i
(ι
λjn:Dst F
j
i
)
)
.
Theorem 1997. Let π
X
i
be metainjective morphisms. If S
P(Hom(A
0
, B
0
) × Hom(A
1
, B
1
)) for some sets A
0
, B
0
, A
1
, B
1
then
l
a ×
(L)
b
(a, b) S
=
l
dom S ×
(L)
l
im S.
Corollary 1998. (a
0
q
(L)
b
0
) t (a
1
q
(L)
b
1
) = (a
0
u a
1
) q
(L)
(b
0
u b
1
).
Corollary 1999. a
0
q
(L)
b
0
a
1
q
(L)
b
1
a
0
a
1
b
0
b
1
.
3.2. Supremum coproduct for endomorphisms. Let F be an indexed
family of endomorphisms of C.
I will denote Ob f the object (source and destination) of an endomorphism f.
Let also ι
i
be a monovalued entirely defined morphism (for each i dom F).
Definition 2000.
`
(L)
F =
d
idom F
(ι
λjn:Ob F
j
i
F
i
(ι
λjn:Ob F
j
i
)
) (if ι is
defined at λj n : Ob F
j
). (I call it supremum coproduct).
4. APPLYING THIS TO THE THEORY OF FUNCOIDS AND RELOIDS 13
Abbreviate ι
i
= ι
λjn:Ob F
j
i
.
So
`
F =
d
idom F
(ι
i
F
i
(ι
i
)
).
`
F = min
(
ΦEnd
`
(Q)
jn
Ob F
j
inwι
i
F
i
(ι
i
)
)
.
Taking into account that ι
i
is a monovalued entirely defined morphism, we get:
Obvious 2001.
`
(L)
= min
(
ΦEnd
`
(Q)
jn
Ob F
j
in:ι
i
C(F
i
,Φ)
)
.
Corollary 2002. ι
i
C
F
i
,
`
(L)
F
for every i dom F .
3.3. Category of continuous morphisms. Let ι
i
(for i dom F ) be en-
tirely defined monovalued and metacomplete morphisms.
Let
L
of an indexed family of morphisms is a morphism; (
L
f) ι
i
= f
i
;
L
in
(f ι
i
) = f (a dual of the above).
Let F
i
End
`
(Q)
jn
Ob F
j
for all i n (where n is some index set) (a self-dual
of the above).
Definition 2003. ι
cont(C)
i
=
`
(L)
F, F
i
, ι
i
.
Proposition 2004. ι
i
are continuous, that is ι
cont(C)
i
are morphisms.
Lemma 2005. f Hom
cont(C)
`
(L)
F, Y
FiXme: What is Y ? is continuous
iff all f ι
cont(C)
are continuous.
Theorem 2006.
`
(L)
together with
L
is a (partial) coproduct in the category
cont(C).
4. Applying this to the theory of funcoids and reloids
4.1. Funcoids.
Definition 2007. Fcd
def
= cont FCD.
Let F be a family of endofuncoids.
The cartesian product
Q
(Q)
X
def
=
Q
X.
I define π
i
= π
X
i
FCD(
Q
X, X
i
) as the principal funcoid corresponding to the
i-th projection. (Here π is entirely defined.)
The disjoint union
`
(Q)
X
def
=
`
X.
I define ι
i
= ι
X
i
FCD(X
i
,
`
X) as the principal funcoid corresponding to the
i-th canonical injection. (Here ι is entirely defined.)
Let
N
and
L
be defined in the same way as in category Set.
Obvious 2008. π
i
N
f = f
i
;
N
in
(π
i
f ) = f .
Obvious 2009. (
L
f) ι
i
= f
i
;
L
in
(f ι
i
) = f.
It is easy to show that π
i
is entirely defined monovalued, and ι
i
is metacomplete
and co-metacomplete.
Thus we are under conditions for both canonical products and canonical co-
products and thus both
Q
(L)
F and
`
(L)
F are defined.
6. CANONICAL PRODUCT AND SUBATOMIC PRODUCT 14
4.2. Reloids.
Definition 2010. Rld
def
= cont RLD.
Let F be a family of endoreloids.
The cartesian product
Q
(Q)
X
def
=
Q
X.
I define π
i
= π
X
i
RLD(
Q
X, X
i
) as the principal reloid corresponding to the
i-th projection. (Here π is entirely defined.)
The disjoint union
`
(Q)
X
def
=
`
X.
I define ι
i
= ι
X
RLD(X
i
,
`
X) as the principal reloid corresponding to the
i-th canonical injection. (Here ι is entirely defined.)
Let
N
and
L
are defined in the same way as in category Set.
Obvious 2011. π
i
N
f = f
i
;
N
in
(π
i
f ) = f .
Obvious 2012. (
L
f) ι
i
= f
i
;
L
in
(f ι
i
) = f.
It is easy to show that π
i
is entirely defined monovalued, and ι
i
is metacomplete
and co-metacomplete.
Thus we are under conditions for both canonical products and canonical co-
products and thus both
Q
(L)
F and
`
(L)
F are defined.
It is trivial that for uniform spaces infimum product of reloids coincides with
product uniformilty.
5. Initial and terminal objects
Initial object of Fcd is the endofuncoid
FCD(,)
. It is initial because it has
precisely one morphism o (the empty set considered as a function) to any object
Y . o is a morphism because o
FCD(,)
v Y o.
Proposition 2013. Terminal objects of Fcd are exactly
F
{∗}×
FCD
F
{∗} =
FCD
{(, )} where is an arbitrary point.
Proof. In order for a function f : X →↑
FCD
{(, )} be a morphism, it is
required exactly f X v↑
FCD
{(, )} f
f X v (f
1
FCD
{(, )})
1
; f X v ({∗} ×
FCD
hf
1
i{∗})
1
; f X v
hf
1
i{∗} ×
FCD
{∗} what true exactly when f is a constant function with the value
.
If n = then Z = {∅};
Q
(L)
= max FCD(Z, Z) =
F
{∅}×
FCD
F
{∅} =
FCD
{(, )}.
FiXme: Initial and terminal objects of Rld.
6. Canonical product and subatomic product
FiXme: Confusion between filters on products and multireloids.
Proposition 2014. Pr
RLD
i
|
F(Z)
= hπ
i
i for every index i of a cartesian product
Z.
Proof. If X F(Z) then (Pr
RLD
i
|
F(Z)
)X = Pr
RLD
i
X =
d
F
hPr
i
i
X =
d
hπ
i
i up X = hπ
i
iX .
Proposition 2015.
Q
(A)
F =
d
in
π
FCD
Q
in
Dst F
i
1
F
i
π
FCD
Q
in
Src F
i
!
.
8. CARTESIAN CLOSEDNESS 15
Proof. a
h
Q
(A)
F
i
b i dom F : Pr
RLD
i
a [F
i
] Pr
RLD
i
b
i dom F :
*
π
FCD
Q
in
Dst F
i
1
+
[F
i
]
π
FCD
Q
in
Src F
i
i dom F : a
"
π
FCD
Q
in
Dst F
i
1
F
i
π
FCD
Q
in
Src F
i
#
b
a
"
d
in
π
FCD
Q
in
Dst F
i
1
F
i
π
FCD
Q
in
Src F
i
!#
b for ultrafilters a and
b.
Corollary 2016.
Q
(L)
F =
Q
(A)
F is F is a small indexed family of funcoids.
7. Further plans
Does the formula
Q
(L)
in
(g
i
f
i
) =
Q
(L)
g
Q
(L)
f hold?
Coordinate-wise continuity.
8. Cartesian closedness
We are not only to prove (or maybe disprove) that our categories are cartesian
closed, but also to find (if any) explicit formulas for exponential transpose and
evaluation.
”Definition” A category is //cartesian closed// iff:
1
. It has finite products.
2
. For each objects A, B is given an object MOR(A, B) (//exponentiation//)
and a morphism ε
Dig
A,B
: MOR(A, B) × A B.
3
. For each morphism f : Z × A B there is given a morphism (//expo-
nential transpose//) f : Z MOR(A, B).
4
. ε
B,C
( f × 1
A
) = f for f : A B × C.
5
. (ε
B,C
(g × 1
A
)) = g for g : A MOR(B, C).
We will also denote f 7→ (f) the reverse of the bijection f 7→ ( f).
Our purpose is to prove (or disprove) that categories Dig, Fcd, and Rld are
cartesian closed. Note that they have finite (and even infinite) products is already
proved.
Alternative way to prove: you can prove that the functor × B is left adjoint
to the exponentiation
B
where the counit is given by the evaluation map.
8.1. Definitions. Categories Dig, Fcd, and Rld are respectively categories
of:
1
. discretely continuous maps between digraphs;
2
. (proximally) continuous maps between endofuncoids;
3
. (uniformly) continuous maps between endoreloids.
”Definition” //Digraph// is an endomorphism of the category Rel.
For a digraph A we denote Ob A the set of vertexes or A and GR A the set of
edges or A.
”Definition” Category Dig of digraphs is the category whose objects are di-
graphs and morphisms are discretely continuous maps between digraphs. That is
morphisms from a digraph µ to a digraph ν are functions (or more precisely mor-
phisms of Set) f such that f µ v ν f (or equivalently µ v f
1
ν f or
equivalently f µ f
1
v ν).
”Remark” Category of digraphs is sometimes defined in an other (non equiva-
lent) way, allowing multiple edges between two given vertices.
8. CARTESIAN CLOSEDNESS 16
8.2. Conjectures.
Conjecture 2017. The categories Fcd and Rld are cartesian closed (actually
two conjectures).
http://mathoverflow.net/questions/141615/how-to-prove-that-there-are-no-exponential-object-in-a-category
suggests to investigate colimits to prove that there are no exponential object.
Our purpose is to prove (or disprove) that categories Dig, Fcd, and Rld are
cartesian closed. Note that they have finite (and even infinite) products is already
proved.
Alternative way to prove: you can prove that the functor × B is left adjoint
to the exponentiation
B
where the counit is given by the evaluation map.
See http://www.springer.com/us/book/9780387977102 for another way to
prove Cartesian closedness.
8.3. Category Dig is cartesian closed. Category of digraphs is the sim-
plest of our three categories and it is easy to demonstrate that it is cartesian closed.
I demonstrate cartesian closedness of Dig mainly with the purpose to show a pat-
tern similarly to which we may probably demonstrate our two other categories are
cartesian closed.
Let G and H be graphs:
Ob MOR(G, H) = (Ob H)
Ob G
;
(f, g) GR MOR(G, H) (v, w) GR G : (f(v), g(w)) GR H for
every f, g Ob MOR(G, H) = (Ob H)
Ob G
;
GR 1
MOR(B,C)
= id
Ob MOR(B,C)
= id
(Ob H)
Ob G
Equivalently
(f, g) GR MOR(G, H) (v, w) GR G : g {(v, w)} f
1
GR H
(f, g) GR MOR(G, H) g (GR G) f
1
GR H
(f, g) GR MOR(G, H) hf ×
(C)
gi GR G GR H
The transposition (the isomorphism) is uncurrying.
f = λa Zλy A : f(a, y) that is ( f)(a)(y) = f(a, y).
(f)(a, y) = f(a)(y)
If f : A × B C then f : A MOR(B, C)
”Proposition” Transposition and its inverse are morphisms of Dig.
”Proof” It follows from the equivalence f : A MOR(B, C) x, y :
(xAy ( f)x(MOR(B, C))( f )y)
x, y : (xAy (v, w) B : (( f)xv, ( f)yw) C)
x, y, v, w : (xAy vBw (( f)xv, ( f )yw) C)
x, y, v, w : ((x, v)(A × B)(y, w) (f (x, v), f(y, w)) C) f : A × B C.
Evaluation ε : MOR(G, H) × G H is defined by the formula:
Then evaluation is ε
B,C
= (1
MOR(B,C)
).
So ε
B,C
(p, q) = ((1
MOR(B,C)
))(p, q) = (1
MOR(B,C)
)(p)(q) = p(q).
”Proposition” Evaluation is a morphism of Dig.
”Proof” Because ε
B,C
(p, q) = (1
MOR(B,C)
).
It remains to prove: * ε
B,C
( f × 1
A
) = f for f : A B × C; *
(ε
B,C
(g × 1
A
)) = g for g : A MOR(B, C).
”Proof” ε
B,C
( f × 1
A
)(a, p) = ε
B,C
(( f )a, p) = ( f )ap = f(a, p). So
ε
B,C
( f × 1
A
) = f.
(ε
B,C
(g × 1
A
))(p)(q) = (ε
B,C
(g × 1
A
))(p, q) = ε
B,C
(g × 1
A
)(p, q) =
ε
B,C
(gp, q) = g(p)(q). So (ε
B,C
(g × 1
A
)) = g.
8.4. By analogy with the proof that Dig is cartesian closed. The most
obvious way for proof attempt that Fcd is cartesian closed is an analogy with the
proof that Dig is cartesian closed.
8. CARTESIAN CLOSEDNESS 17
Consider the long formula above. The proof would arise if we replace x and y in
this formula with filters and operations and relations on set element with operations
and relations on filters.
This proof could be simplified in either of two ways:
replace x and y with ultrafilters, see [[Proof for Fcd using ultrafilters]];
replace x and y with sets (principal filter), see [[Proof for Fcd using sets]].
This is not quite easy however, because we need to calculate uncurrying for
a entirely defined monovalued principal funcoid (what is essentially the same as a
function of a Set-morphisms) taking either ultrafilters or principal filters as argu-
ments. Such (generalized) uncurrying is not quite easy.
To sum what we need to prove:
Transposition is a morphism.
Evaluation is a morphism.
ε
B,C
( f × 1
A
) = f for f : A B × C.
(ε
B,C
(g × 1
A
)) = g for g : A MOR(B, C).
8.5. Attempt to describe exponentials in Fcd.
Exponential object HOM(A, B) is the following endofuncoid:
Object Ob HOM(A, B) = (Ob B)
Ob A
;
Graph is GR HOM(A, B) =
FCD
n
(f,g)
f,gHom
Set
(Ob A,Ob B)∧↑
FCD
gA◦↑
FCD
f
1
vB
o
.
Transposition is uncurrying.
Evaluation is ε
A,B
x = hPr
(A)
0
xi Pr
(A)
1
x.
We need to prove that the above defined are really an exponential and an
evaluation.
Possible ways to prove that Fcd is cartesian closed follow:
NEW IDEA: Prove GR HOM(A, B) =
FCD
(Pr
(A)
0
p,Pr
(A)
1
p)
p??∧hpiAvB
(what’s about
other kinds of projections?)
8.6. Proof for Fcd using sets. Currying for sets is hfi(X × Y ) =
S
hh∼
fiXiY (as it’s easy to prove). This simple formula gives hope, but...
It does not work with sets because an analogy for sets of the last equality of
the above mentioned long formula would be:
X, Y, V, W P Ob A : (X × V [A × B]
Y × W hf i(X × V ) [C]
hfi(Y × W ))
f : A × B C
but this implication seems false.
The most obvious way for proof attempt that Fcd is cartesian closed is an
analogy with the proof that Dig is cartesian closed.
Use the exponential object, transposition, and evaluation as defined in [[this
page|Is category Fcd cartesian closed?]]
8.7. Reducing to the fact that Dig is cartesian closed. It is probably a
simpler way to prove that Fcd is cartesian closed by embedding it into Dig (which
is [[already known to be cartesian closed|Category Dig is cartesian closed]]).
Fcd can be embedded into Dig by the formulas:
A 7→ hAi;
f 7→ hfi.
That this really maps a morphism of Fcd into a morphism of Dig follows from
the fact that hg fi = hgi hfi.
Obviously this embedding (denote it T ) is an injective (both on objects and
morphisms) functor.
9. IS CATEGORY RLD CARTESIAN CLOSED? 18
We will define:
ε
Fcd
A,B
= T
1
ε
Dig
T A,T B
;
Fcd
f = T
1
Dig
T f.
Due to functoriality and injectivity of T it is enough to prove that above defined
ε
Fcd
A,B
and
Fcd
f exist and are morphisms of Fcd.
ε
Dig
T A,T B
6= T ε
Fcd
A,B
because ε
Dig
T A,T B
accepts ordered pairs as the argument and
T ε
Fcd
A,B
accepts sets as the argument. So this is a dead end. Can the proof idea be
salvaged?
9. Is category Rld cartesian closed?
We may attempt to prove that Rld is cartesian closed by embedding it into
supposedly cartesian closed category Fcd by the function ρ:
hρfix = f x and hρf
1
iy = f
1
y.
TODO: More to write on this topic.
CHAPTER 3
Equalizers and co-Equalizers in Certain Categories
It is a rough draft. Errors are possible.
FiXme: Change notation
Q
Q
(L)
.
1. Equalizers
Categories cont(C) are defined above.
I will denote W the forgetful functor from cont(C) to C.
In the definition of the category cont(C) take values of as principal morphisms.
FiXme: Wording.
Lemma 2018. Let f : X Y be a morphism of the category cont(C) where C is
a concrete category (so W f = ϕ for a Rel-morphism ϕ because f is principal) and
im ϕ = A Ob Y . Factor it ϕ = E
Ob Y
u where u : Ob X A using properties of
Set. Then u is a morphism of cont(C) (that is a continuous function X ι
A
Y ).
Proof. (E
Ob Y
)
1
ϕ = (E
Ob Y
)
1
E
Ob Y
u;
(E
Ob Y
C
)
1
ϕ = (E
Ob Y
C
)
1
E
Ob Y
C
u;
(E
Ob Y
C
)
1
ϕ = u;
X v ( u)
1
π
A
Y u X v ( ϕ)
1
E
Ob Y
C
π
A
Y (E
Ob Y
C
)
1
ϕ
X v ( ϕ)
1
E
Ob Y
C
(E
Ob Y
C
)
1
Y E
Ob Y
C
(E
Ob Y
C
)
1
ϕ X v ( ϕ)
1
Y
ϕ X v (W f)
1
Y W f what is true by definition of continuity.
Equational definition of equalizers:
http://nforum.mathforge.org/comments.php?DiscussionID=5328/
Theorem 2019. The following is an equalizer of parallel morphisms f, g : A
B of category cont(C):
the object X = ι
{
xOb A
f x=gx
}
A;
the morphism E
Ob X,Ob A
considered as a morphism X A.
Proof. Denote e = E
Ob X,Ob A
.
Let f z = g z for some morphism z.
Let’s prove e u = z for some u : Src z X. Really, as a morphism of Set it
exists and is unique.
Consider z as as a generalized element.
f(z) = g(z). So z X (that is Dst z X). Thus z = e u for some u (by
properties of Set). The generalized element u is a cont(C)-morphism because of
the lemma above. It is unique by properties of Set.
We can (over)simplify the above theorem by the obvious below:
Obvious 2020.
n
xOb A
fx=gx
o
= dom(f g).
2. Co-equalizers
http://math.stackexchange.com/questions/539717/
how-to-construct-co-equalizers-in-mathbftop
Let be an equivalence relation. Let’s denote π its canonical projection.
19
3. REST 20
Definition 2021. f/ = π f π
1
for every morphism f.
Obvious 2022. Ob(f/ ) = (Ob f)/r.
Obvious 2023. f/ = h↑
FCD
π×
(C)
FCD
πif for every morphism f.
To define co-equalizers of morphisms f and g let be is the smallest equivalence
relation such that fx = gx.
Lemma 2024. Let f : X Y be a morphism of the category cont(C) where C
is a concrete category (so W f = ϕ for a Rel-morphism ϕ because f is principal)
such that ϕ respects . Factor it ϕ = u π where u : Ob(X/ ) Ob Y using
properties of Set. Then u is a morphism of cont(C) (that is a continuous function
X/ ∼→ Y ).
Proof. f X f
1
v Y ; u π X π
1
u
1
v Y ; u C( π X
π
1
, Y ) = C(X/ , Y ).
Theorem 2025. The following is a co-equalizer of parallel morphisms f, g :
A B of category cont(C):
the object Y = f/ ;
the morphism π considered as a morphism B Y .
Proof. Let z f = z g for some morphism z.
Let’s prove u π = z for some u : Y Dst z. Really, as a morphism of Set it
exists and is unique.
Src z Y . Thus z = u π for some u (by properties of Set). The function u is
a cont(C)-morphism because of the lemma above. It is unique by properties of Set
(π obviously respects equivalence classes).
3. Rest
Theorem 2026. The categories cont(C) (for example in Fcd and Rld) are
complete. FiXme: Note that small complete category is a preorder!
Proof. They have products and equalizers.
Theorem 2027. The categories cont(C) (for example in Fcd and Rld) are
co-complete.
Proof. They have co-products and co-equalizers.
Definition 2028. I call morphisms f and g of a category with embeddings
equivalent (f g) when there exist a morphism p such that Src p v Src f, Src p v
Src g, Dst p v Dst f, Dst p v Dst g and ι
Src f,Dst f
p = f and ι
Src g,Dst g
p = g.
Problem 2029. Find under which conditions:
1
. Equivalence of morphisms is an equivalence relation.
2
. Equivalence of morphisms is a congruence for our category.
CHAPTER 4
Categories of filters
In [1] two categories, whose objects are related with filters on sets, are defined
and researched.
Accordingly [1] infinite product is defined just in the first (denoted F there)
of these two categories. So we will for now consider the first category. (Usefulness
of the second category for our research is questionable.)
Let f : A B be a function, A be a filter on A.
Proposition 2030.
n
Y PB
hf
1
i
Y ∈A
o
is a filter.
Proof. That it is an upper set is obvious.
Let Y
0
, Y
1
n
Y PB
hf
1
i
Y ∈A
o
. Then
f
1
Y
0
A and
f
1
Y
1
A. We have
f
1
(Y
0
Y
1
) =
f
1
Y
0
f
1
Y
1
A
since f is monovalued. Thus Y
0
Y
1
n
Y PB
hf
1
i
Y ∈A
o
.
Theorem 2031. FiXme: Should be moved above in the book.
n
Y PB
hf
1
i
Y ∈A
o
is
equal to the filter generated by the filter base
hfi
A, for every filter A.
Proof. Denote B =
n
Y PB
hf
1
i
Y ∈A
o
, C =
hfi
A.
Let Y C. Then Y = hfi
A where A A. Then
f
1
hfi
A A and so
f
1
hfi
A A. This proves hf i
A B, that is Y B.
Let now Y B. Then hfi
f
1
Y Y . Since
f
1
Y A, we have that Y
is a supset of some set of the form hfi
A, so Y C.
Corollary 2032. uphfiA =
n
Y PB
hf
1
i
Y up A
o
.
Definition 2033. The category of filtered sets Filt is the category defined as
follows:
1
. Objects are pairs (A, A) where A is a (small) set and A is a filter on A.
2
. Morphisms from (A, A) to (B, B) are functions f : A B such that
hfiA v B.
3
. Identities are identity functions.
To verify that it is a category is straightforward.
It is the same category as F in [1], as follows from an above proposition.
We will prove that starred reloidal product is a categorical product in this
category. First we will prove the special case that binary reloidal product is a
categorical product in this category.
Theorem 2034. ×
RLD
(together with projections Pr
0
and Pr
1
) is a categorical
product in Filt.
Proof. Let our objects be A, B.
Denote p the left projection from Base(A) × Base(B) to Base(A).
21
4. CATEGORIES OF FILTERS 22
We need to check that p is a Filt-morphism that is p(A ×
RLD
B) v A what is
obvious.
Similarly for the right projection q.
It remains to check the universal property: Let C be a filter and f : C A,
g : C B. We need to prove that there are a unique u : C A ×
RLD
B such that
f = p u and g = q u. Denote h(z) = (f(z), g(z)).
h is the unique function Base(C) Base(A) × Base(B) such that f = p h and
g = qh, so it remains to check that h is a morphism of Filt that is hhiC v A×
RLD
B,
what obviously follows from hfiC v A and hgiC v B.
Theorem 2035.
Q
RLD
together with projections Pr
k
is a categorical product
in Filt.
Proof. Consider an indexed family A of objects.
Denote p
k
the k-th projection from
Q
idom A
Base(A
i
).
We need to check that p
k
s a Filt-morphism that is p
k
Q
RLD
A
v A
k
what
is obvious.
It remains to check the universal property: Let C be a filter and f
k
: C A
k
.
We need to prove that there are a unique u : C
Q
RLD
A such that f
k
= p
k
u.
Denote h(z) = λi dom A : f
i
z.
h is the unique function Base(C)
Q
idom A
Base(A
i
) such that f
k
= p
k
h,
so it remains to check that h is a morphism of Filt that is hhiC v
Q
RLD
A. It
follows from
RLD
Pr
i
hhiC =
l
hP r
i
i
hhi
up C =
l
hPr
i
hi
up C =
l
hf
i
i
up C = hf
i
iC v A
i
.
CHAPTER 5
Power of filters
1. Germs of functions
Definition 2036. Functions f, g Rel(Ob X , B) are of the same X -germ for
a filter object X iff there exists X up X such that f|
X
= g|
X
.
Proposition 2037. Being of the same germ is an equivalence relation.
Proof.
Reflexivity. Take arbitrary X up X .
Symmetry. Obvious.
Transitivity. Let f|
X
= g|
X
and g|
Y
= h|
Y
. Then f|
XY
= h|
XY
.
Definition 2038. A germ is an equivalence class of being the same germ.
Obvious 2039. Every germ is a filter on Set.
Theorem 2040. Let A, B be sets.
The following are mutually inverse bijections between monovalued reloids f :
A B with dom f = X and X -germs S of functions A B for X F A:
1
. f 7→ up
Set
f;
2
. S 7→ s|
X
if s S.
The second bijection can also be written as S 7→
d
RLD
S
|
X
or if card B 6= 1 as
S 7→
d
RLD
S.
Remark 2041. s|
X
is always defined because S is nonempty (it is an equiva-
lence class).
Proof. First prove that up
Set
f is an X -germ. Really, F up
Set
f F w
f F |
X
= f X up X : F |
X
w f; thus F, G up
Set
f X up X :
F |
X
w f Y up X : G|
Y
w f X up X : F |
XY
w f Y up X :
G|
XY
w f Z up X : (F |
Z
w f G|
Z
w f ) Z up X : (F u G)|
Z
w f
and F up
Set
f X up X : F |
X
= G|
X
F w f F |
X
= G|
X
G|
X
w f
G up
Set
f. We have proved that up
Set
f is an equivalence class of the suitable
equivalence relation, that is up
Set
f is an X -germ.
That
d
RLD
S is a monovalued reloid is obvious. Also im
d
RLD
S = X is obvious.
Now prove that our correspondences are mutually inverse.
Let f
0
: A B be a monovalued reloid and dom f = X . Let S = up
Set
f
0
and f
1
= s|
X
for an s S. We need to prove f
1
= f
0
. Really, f
1
= F |
X
for an
F up
Set
f
0
; thus f
1
= f
0
.
Let S
0
be an X -germ of functions A B. Let f = s|
X
for an s S
0
and
S
1
= up
Set
f. We need to prove S
1
= S
0
. Really,
S
1
= up
Set
(s|
X
) =
F Set
F w s|
X
=
F Set
X up X : F |
X
w s|
X
=
F Set
X up X : F |
X
= s|
X
= S
0
.
d
RLD
S
|
X
=
d
RLD
sS
s|
X
= s|
X
for every choice of s S.
23
2. POWER OF FILTERS 24
We can assume that B 6= because otherwise the theorem is obvious. Thus we
can assume card B > 1.
If X = X then obviously S has just one element F and im
d
RLD
S = im F =
X = X . Otherwise for every X up X there are elements F , G of S such that
dom(F u G) v X (using card B > 1).
By properties of generalized filter bases X × > w
d
RLD
S F, G S :
X × > w F u G X w X . Thus im
d
RLD
S = X .
2. Power of filters
Let’s define Y
X
for filters X , Y:
First define Y
X
for a set Y :
Y
X
=
f RLD(Ob X , Y )
dom f = X f is monovalued
.
Now Y
X
=
d
RLD
Y up Y
Y
X
.
[1] defines an isomorphic to this way to define “exponentiation” of filters.
TODO: Check Y
1
=
Y; Z
X ×
RLD
Y
=
(Z
X
)
Y
; Z
X qY
=
Z
X
×
RLD
Z
Y
; Y
2
=
Y ×
RLD
Y; Y
0
=
1; Y
N
=
Q
RLD
nN
Y. More formulas at https://en.wikipedia.org/
wiki/Cartesian_closed_category.
Andreas Blass says in a private email that it is not cartesian closed: “Unfor-
tunately, the two categories of filters in my paper are not cartesian closed. This is
mentioned in a parenthetical comment near the bottom of page 141. The operation
of cartesian product with the cofinite filter on the natural numbers has no right
adjoint, because it does not preserve infinite coproducts. about [1].
But it is probably a braided closed monoidal category?
See [1] for more categorical properties of filters.
CHAPTER 6
Matters related to tensor product
These consideration on (possibly infinite) indexed families of join-semilattices
is based on [7] (for the finite case).
Let A be an indexed family of join-semilattices with least elements. Let T also
be a join-semilattice.
Let F (X) mean free join-semilattice for a set X.
Definition 2042. SepJoin(
Q
A, T ) is the set of maps from
Q
A to T , pre-
serving joins in every argument i dom A.
Obvious 2043. The set of free join-semilattices F (X) is order-isomorphic to
the set of subsets X of a “universal” set f.
Let i :
Q
A F (
Q
A) be the universal embedding.
Let be defined as the smallest equivalence relation on F (
Q
A) that for every
k dom A, L
Q
i(dom A)\{k}
A
i
:
1
. i(L {(k, g t h)}) i(L {(k, g)}) t i(L {(k, h)});
2
. i(L {(k, )});
3
. x y x
0
y
0
x t x
0
y t y
0
for all x, y, x
0
, y
0
F (
Q
A).
Obvious 2044. Some function h : X Y induces a well defined map ψ :
X/E Y on equivalence classes, if E F where x F y hx = hy.
Lemma 2045. The set of join-homomorphisms ψ : F (
Q
A)/ ∼→ T is isomor-
phic to the set of maps φ :
Q
A T preserving finite joins in separate arguments.
Proof. The quotient map q : F (
Q
A) F (
Q
A)/ which takes an element
x to its equivalence class [x] map is well defined because
x y x
0
y
0
x t x
0
y t y
0
.
The map q preserves join. F (
Q
A)/ is associative, commutative, and idempotent
since it is so on F (
Q
A) and thus is a join-semilattice.
Let join-preserving map ψ : F (
Q
A)/ ∼→ T . It is easy to show that ψ q i
preserves joins in separate arguments.
Let now φ :
Q
A T preserves joins in separate arguments. There is a unique
join-preserving map
˜
φ : F (
Q
A) T such that
˜
φ i = φ. We must show that
this induces a well-defined join-preserving map ψ : F (
Q
A)/ ∼→ T such that
ψ(q(x)) =
˜
φ(x) for all x F (
Q
A) (clearly at most one function ψ can satisfy
this equation since q is surjective). This will show that ψ bijectively correspond
to
˜
φ and thus bijectively correspond to φ. (This will finish the proof as that this
bijection is monotone is obvious.)
Using the “obvious” above, it’s enough (taking into account that is the
minimal equivalence relation subject to the above formulas) to prove that:
1
.
˜
φ(i(L {(k, g t h)})) =
˜
φ(i(L {(k, g)}) t i(L {(k, h)}));
2
.
˜
φ() =
˜
φ(i(L {(k, )}));
3
.
˜
φ(x) =
˜
φ(y)
˜
φ(x
0
) =
˜
φ(y
0
)
˜
φ(x t x
0
) =
˜
φ(y t y
0
)
25
6. MATTERS RELATED TO TENSOR PRODUCT 26
The first easily follows from
˜
φ i = φ and the fact that
˜
φ preserves binary joins.
The second easily follows from
˜
φ i = φ and that φ preserves .
The third follows from the fact that
˜
φ preserves joins.
Corollary 2046. The poset of prestaroids preStrd(A) is isomorphic to an ideal
(on a join-semilattice), provided that A is an indexed family of join-semilattices.
Proof. preStrd(A)
=
SepJoin(A, 2)
=
F (
Q
A)/ ∼→ 2
=
I(F (
Q
A)/ ).
FiXme: Check below (especially posets vs dual posets) for errors.
Corollary 2047. preStrd is a complete lattice.
Proof. Corollary 515.
Corollary 2048. preStrd is a filtered filtrator.
Proof. Theorem 531.
FiXme: Try to prove that preStrd is atomic and moreover atomistic (under
certain conditions). Other properties?
CHAPTER 7
Mappings between endofuncoids and topological
spaces
Oreder topologies reversely to set-theoretic inclusion. That is for topologies t
and s we set t v s t s. (Intuitively: The less is the topology, the lesser are its
open sets.)
Let’s study mappings between topological spaces and endofuncoids.
Definition 2049. Let t be a topology.
1
. F
?
t =
d
xOb t
{x} ×
d
F
Et
xE
;
2
. (F
?
t)E =
T
n
Dt
ED
o
.
Proposition 2050. Let t be a topology.
1
. F
?
t is complete, reflexive, transitive funcoid.
2
. F
?
t is co-complete, reflexive, transitive funcoid.
3
. F
?
and F
?
are injections.
4
. F
?
t = (F
?
t)
1
.
Proof. By theorem 785.
Definition 2051. Let f be an endofuncoid.
T f =
E P Ob f
x E : hf i{x} v E
.
Proposition 2052. T f is a topology.
Proof. Union of open sets is open. S T f E Sx E : hfix v E
x
S
S : hfix v
S
S
Intersection of two open sets is open. Let X, Y T f. Then x X : hf ix v X
and x Y : hfix v Y . So if x X Y then hfix v X and hfix v Y ,
so hfix v X Y . So X Y T f.
Ob f is an open set. Obvious.
Obvious 2053. T f =
n
EP Ob f
hCompl fiEvE
o
.
In some reason when starting this research I assumed that the following funcoid
(for every endofuncoid f) is a Kuratowski closure:
1 t CoCompl f t (CoCompl f)
2
t . . . .
It is not true:
Example 2054. There exists such a co-complete endofuncoid f that 1 t f t
f
2
t . . . is not transitive that is
(1 t f t f
2
t . . .) (1 t f t f
2
t . . .) 6= 1 t f t f
2
t . . .
27
7. MAPPINGS BETWEEN ENDOFUNCOIDS AND TOPOLOGICAL SPACES 28
Proof. Take f = cl g where g is the principal funcoid which maps every real
number a into the closed interval
h
1−|a|
2
;
1+|a|
2
i
.
Take X =
1
2
;
1
2
. hf
n
i
X =
1 +
1
2
n+1
; 1
1
2
n+1
.
We have h1 t f t f
2
t . . .i
X =] 1; 1[;
h1 t f t f
2
t . . .i
h1 t f t f
2
t . . .i
X = [1; 1].
Thus follows our inequality.
That F
?
and F
?
are functors (if we map morphisms to themselves except of
changing the objects) follows from conjecture 1180.
Theorem 2055. T (if we map morphisms to themselves except of changing the
objects) is a functor.
Proof. Based on https://math.stackexchange.com/a/2792239/4876
Let f : µ ν that is f µ v ν f . We need to prove f : T µ T ν that is
E T ν hf
1
i
E T µ.
Suppose E T ν that is hνi
E v E. We will prove hµi
hf
1
i
E v hf
1
i
E.
FiXme: Can we use arbitrary filters rather than atoms?
Really, let atom y v hµi
hf
1
i
E. Then there exists atom x v hf
1
i
E such
that x [µ]
y.
x [f µ]
hfiy and thus x [ν f ]
hfiy, so hf ix [ν]
hfiy. But hfix v E, so
hfiy v hνi
E v E, that is hµi
hf
1
i
E v E.
Proposition 2056. f C(µ, ν) f C(µ
n
, ν
n
) for every endofuncoids µ
and ν and positive natural number n. FiXme: Move this proposition to the book.
Proof. f µ v ν f; f µ µ v ν f µ; f µ
2
v ν
2
f; f µ
3
v ν
3
f...
Proposition 2057. For every endofuncoid µ:
1
. F
?
T µ w Compl µ;
2
. F
?
T µ w Compl µ;
3
. F
?
T µ w CoCompl µ;
4
. F
?
T µ w CoCompl µ;
Proof. We will prove only the first two as the rest are dual.
hF
?
T µi
E =
T
n
DT µ
DE
o
=
T
n
DP Ob µ
hCompl µi
DvDDE
o
w
d
n
hCompl µi
D
DP Ob µ,hCompl µi
DvDDE
o
w hCompl µi
E.
hF
?
T µi
{x} =
d
F
n
ET µ
xE
o
=
d
F
n
EOb µ
xE,hCompl µi
EvE
o
w
d
F
n
hCompl µi
E
EOb µ,xE,hCompl µi
EvE
o
w hCompl µi
{x}.
Lemma 2058. For every endofuncoid µ:
1
. F
?
T µ v 1 t Compl µ t (Compl µ)
2
t . . .;
2
. F
?
T µ v 1 t CoCompl µ t (CoCompl µ)
2
t . . .
Proof. We will prove only the first as the second is dual.
h1 t Compl µ t (Compl µ)
2
t . . .i
E = E t hCompl µi
E t h(Compl µ)
2
i
E t . . .
Take D = E t hCompl µi
E t h(Compl µ)
2
i
E t . . . We have hCompl µi
D v
hCompl µi
E t h(Compl µ)
2
i
E t . . . v D. So
T
n
DP Ob µ
hCompl µi
DvDDE
o
D v h1 t Compl µ t (Compl µ)
2
t . . .i
E.
Theorem 2059. If we restrict the functor T only to complete endofuncoids
(= complete endoreloids), then T is a left adjoint of both F
?
and F
?
.
7. MAPPINGS BETWEEN ENDOFUNCOIDS AND TOPOLOGICAL SPACES 29
Proof. We will prove only for F
?
as the other is dual.
We will disprove f C(T µ, s) f C(µ, F
?
s) what is equivalent (because F
?
is full and faithful) to
f C(F
?
T µ, F
?
s) f C(µ, F
?
s);
F
?
T µ v f
1
F
?
s f µ v f
1
F
?
s f.
F
?
T µ v f
1
F
?
s f µ v f
1
F
?
s f because F
?
T µ w µ.
If µ v f
1
F
?
s f then µ
n
v f
1
(F
?
s)
n
f = f
1
F
?
s f. Also obviously
1 v f
1
F
?
s f. Thus
1 t µ t µ
2
t . . . v f
1
F
?
s f
and so 1 tCompl µt(Compl µ)
2
t. . . v f
1
F
?
s f . So F
?
T µ v f
1
F
?
s F .
FiXme: F and T are also a Galois connection, isn’t it?
Example 2060. T is a not left adjoint of both F
?
and F
?
, with bijection which
preserves the “function” part of the morphism.
Proof. We will disprove only from F
?
as the other is dual.
We will disprove f C(T µ, s) f C(µ, F
?
s) what is equivalent (because F
?
is full and faithful) to
f C(F
?
T µ, F
?
s) f C(µ, F
?
s);
F
?
T µ v f
1
F
?
s f µ v f
1
F
?
s f.
This equivalence does not hold: Take s the discrete space on R, f = id
R
, and
hµi
X = X for finite sets X and hµi
X = > for infinite X.
CHAPTER 8
Funcoids as closed sets
Idea [6] by Todd Trimble.
FiXme: https://ncatlab.org/toddtrimble/published/topogeny and https://
math.stackexchange.com/q/2681502/4876
FiXme: What about the infinite products?
Theorem 2061. The set of staroids PX
1
×· · ·×P X
n
2 is order isomorphic
to co-frame of closed subsets of topological product βX
1
× · · · × βX
n
.
Proof. PX
1
× · · · × PX
n
2 can be order-embedded to the frame of ideals
J(PX
1
×· · ·×P X
n
) what is dual (check!) to the frame of ideals of the distributive
lattice P X
1
· · · PX
n
. This by ?? is the coproduct
P
i
PX
i
in the category of
boolean algebras. By Stone duality it is isomorphic to the topology of it spectrum
βX
1
× · · · × βX
n
.
Elements of βX
1
× · · · × βX
n
are closed subsets. So every n-staroid one-to-one
corresponds to a closed set of βX
1
× · · · × βX
n
.
Note that βX
1
× · · · × βX
n
is a compact Hausdorff space (as a product of
compact Hausdorff spaces).
It seems that there is an easy way to describe the above order embedding in
both directions:
f 7→
(x
1
, . . . , x
n
)
x
1
, . . . , x
n
atoms
F
, x
1
×
FCD
· · · ×
FCD
x
n
v f
;
X 7→
l
p
1
×
FCD
· · · ×
FCD
p
n
p X
.
FiXme: Try to prove that composition of funcoids is isomorphic to composition
of relations βA × βB.
30