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Equalizers and co-Equa lizers in Certain Categories
by Victor Porton
Email: porton@narod.ru
Web: http://www.mathematics21.org
February 9, 2014
1 Draft status
It is a rough dr aft. Errors are possible. Subscribe to my blog for further results.
http://math.stackexchange.com/questions/54 0220/right-adjoint-of-forgetful-functor-from-top
[TODO: Change notation
Q
Q
(L)
.]
2 Categories with embeddings
Note 1. This section in not used below, it is just to feed your intuition.
The following generalizes the well known concept of embedding function A
B for from a set
A to a set B where A B.
I will set that the unique morphism from an object A to an object B of a thin category is equal
to the pair (A; B).
Definition 2. A category with embeddings of objects is a dagger category with a preorder of the
set of objects together with a functor
(we will denote applying this functor to the object (A; B)
as A
B.) such that:
is an identity on objects.
Every A
B is a mono morphism.
(A
B)
(A
B) = 1
A
.
Obvious 3. A
B is defined when (A; B) is a morphism of the preorder that is when A B.
Obvious 4. A
B: A B when A B.
Proposition 5. A
A = 1
A
.
Proof. Because (A; A) is a n identity morphis m and
preserves identities.
Proposition 6. (B
C) (A
B) = A
C whenever A B C.
Proof. (B
C) (A
B) =
(B; C)
(A; B) =
((B; C) (A; B)) =
(A; C) = A
C.
3 Categories under Rel
Definition 7. The Rel-morphism A B (restriction-embedding) is defined by the fo rmula:
A B = (A; B; id
AB
).
Obvious 8. If A B then A B is an embedding A
B = (A; B; id
A
).
Obvious 9. If A B then A B = (A; B; id
B
).
1
Obvious 10. A A = 1
Rel
A
.
Obvious 11. (A B)
1
= B A.
Definition 12 . Dagger functor between two dagger categories is a functor between these cate-
gories, which commutes with the daggers.
[TODO: Clea rer wording.]
Definition 13. Category under Rel is a pair (C; ) where C is a category whose objects are small
sets and is an i de ntity-on-objects functor Rel C. I call up-arrow functor. [TODO: A B
A B for s ets.]
Definition 14. Dagger category unde r Rel is a pair (C; ) where C is a dagger ca tegory whose
objects are small sets and is a dagger identity-on-objects functor Rel C.
Definition 15. A
C
B = (A B). In other words,
C
= .
Proposition 16. A
C
A = 1
C
A
.
Proof. A
C
A = (A A) = 1
Rel
= 1
C
A
.
Proposition 17. If f : X Y is a Rel-morphis m and im f = A Y then
(A Y ) (Y A) f = f .
Proof. (A Y ) (Y A) f = id
A
f = f .
Corollary 18. If f : X Y is a morphism of a categor y under Rel and im f = A Y , then
(A
C
Y ) (Y
C
A) f = f .
Proposition 19.
1. If A B then A
C
B is a monomorphism.
2. If A B then A
C
B is a epimorphism.
Proof. We’ll prove only the first as the second is dual.
Let (A
C
B) f = (A
C
B) g. Then (B
C
A) (A
C
B) f = (B
C
A) (A
C
B) g;
1
A
f = 1
A
g; f = g.
Proposition 20. (B C) (A B) = A C iff B A C (for every sets A, B, C).
Proof. (B C) (A B) = A C is e quivalent to:
(B; C; id
BC
) (A; B; id
AB
) = (A; C; id
AC
);
(A; C; id
AB C
) = (A; C; id
AC
);
A B C = A C;
B A C.
Corollary 21. (B
C
C) (A
C
B) = (A
C
C) if B A C (f or every sets A, B, C).
Definition 22. Partially ordered dagger category under Rel is a category which is both a partially
ordered dagger category and a category under Rel such that f
1
= ( f )
and A B
A B.
Proposition 23. (A
C
B)
= B
C
A for a dagger category under Rel.
Proof. (A
C
B)
= ((A B))
= (A B)
1
= (B A) = B
C
A.
Proposition 24. For a partially ordered dagger category C unde r Rel we have A
C
B is:
1. m onovalued;
2 Section 3
2. in jective;
3. entirely defined if A B;
4. surjective if B A.
Proof.
1. (A B) (B A) 1
B
Rel
; (A B) (A B)
1
1
B
Rel
; (A
C
B) (A
C
B)
1
B
C
.
2. (B A) (A B) 1
A
Rel
; (A B)
1
(A B) 1
A
Rel
; (A
C
B)
(A
C
B) 1
A
C
.
3. (B A) (A B) 1
A
Rel
; (A B)
1
(A B) 1
A
Rel
; (A
C
B)
(A
C
B) 1
A
C
.
4. (A B) (B A) 1
A
Rel
; (A B) (A B)
1
1
A
Rel
; (A
C
B) (A
C
B)
1
A
C
.
4 Rectangular embedding-rest rictio n
Definition 25. ι
B
0
,B
1
f = (A
1
C
B
1
) f (B
0
C
A
0
) for f Mor
C
(A
0
; A
1
).
For brevity ι
B
f = ι
B,B
f.
Proposition 26. ι
Src f ,Dst f
f = f .
Proof. ι
Src f ,Dst f
f = (Dst f
C
Dst f ) f (Src f
C
Src f ) = 1
C
Dst f
f 1
C
Src f
= f .
Proposition 27. The function ι
B
0
,B
1
|
f Mor
C
(A
0
;A
1
)
is injective, if A
0
B
0
A
1
B
1
.
Proof. Because A
1
C
B
1
is a monomorphism and A
0
C
B
0
is a n epimorphism.
Proposition 28. ι
C
0
,C
1
ι
B
0
,B
1
f = ι
C
0
,C
1
f for B
0
A
0
C
0
, B
1
A
1
C
1
and f : A
0
A
1
.
Proof. ι
C
0
,C
1
ι
B
0
,B
1
f = (B
1
C
C
1
) (A
1
C
B
1
) f (B
0
C
A
0
) (C
0
C
B
0
) = (A
1
C
C
1
) f
(C
0
C
A
0
) = ι
C
0
,C
1
f.
Proposition 29. Let f : A
0
A
1
and g: A
1
A
2
and A
1
B
1
. Then ι
B
0
,B
2
(g f )= ι
B
1
,B
1
g ι
B
0
,B
1
f.
Proof. ι
B
0
,B
2
(g f ) = (A
2
C
B
2
) (g f ) (B
0
C
A
0
) = (A
2
C
B
2
) g id
A
1
f (B
0
C
A
0
) =
(A
2
C
B
2
) g (B
1
A
1
) (A
1
B
1
) f (B
0
C
A
0
) = ι
B
1
,B
1
g ι
B
0
,B
1
f.
5 Examples of partially ordered dagger categories under Rel
5.1 Generalize d rebase of filters
In [2] I define d rebase A ÷ A for a set-theoretic lter A and a set X such that X A: X A.
Now define a generalized rebase for every set-theoretic filter A and every set A:
Definition 30. A ÷ A =
d
{↑
A
(X A) | X A}.
Proposition 31. In the case of X A: X A these two definitions coincide.
Proof. Let X A: X A. Then as it is proved in [2] {X PA | Y A: Y X } is a filter.
If P {X PA | Y A: Y X } then P PA and Y P for some Y A. Thus
P Y A
d
{↑
B
(Y A) | Y A}.
If P
d
{↑
B
(X A) | X A} then by properties of generalized filter bases, there exists X A
such that P X A. Also P PA. Thus P X. Thus P {X PA | Y A: Y X }.
[TODO: Clea r this proof: wording, consistent use of letters.]
Examples of partially ordered dagger categories under Rel 3
Proposition 32. (X ÷ A) ÷ B = X ÷ B if B A.
Proof. (X ÷ A) ÷ B =
d
{↑
B
(Y B) | Y
d
{↑
A
(X A) | X X }} =
d
{↑
B
(X A) | X
X }
B
B =
d
{↑
B
(X A B) | X X } = X ÷ (A B) = X ÷ B.
5.2 Category Rel
Category Rel with the identity up-arrow functor to itself and “reverse relation” as the dagger is
an obvious example of a partially ordered dagger category under Rel.
Definition 33. f ÷ (A × B) = (A; B; (GR f ) ÷ (A × B)) for every Rel-morphism f .
Proposition 34. ι
A,B
f = (A; B; GR f (A × B)).
Proof. ι
A,B
f = (Dst f B) f (A Src f ) = (A; B; GR f (A × B)).
5.3 Category FCD
Category FCD with the u p-arrow functor
FCD
and “reverse funcoid” as the dagger is a partially
ordered dagg er category under Rel.
Proposition 35. A
FCD
B = (A; B; λ X F(A): X ÷ B; λ Y F(B): Y ÷ A) for objects A B of
FCD.
Proof. hA
FCD
B iX =
d
{hA
FCD
B i
X | X X } =
d
{↑
B
hA BiX | X X } =
d
{↑
B
(X A B) | X X } =
d
{↑
B
(X B) | X X } = X ÷ B.
Rest follows from symmetry.
Proposition 36.
1. hA
FCD
B i
X =
B
X for every X PA if A B.
2. h(B
FCD
A)i
Y =
A
(Y A) for every Y PB if A B.
Proof. By definition of principal funcoid.
5.4 Category RLD
Category R LD with the up-arrow functor
RLD
and “reverse reloid” as the dagger is a partially
ordered dagg er category under Rel.
Obvious 37. A
RLD
B =
RLD(A;B)
id
AB
.
Definition 38. f ÷ (A × B) = (A; B; (GR f ) ÷ (A × B)) for every reloid f .
Proposition 39. ι
A,B
f = f ÷ (A × B).
Proof. ι
A,B
f = (Dst f
RLD
B) f (A
RLD
Src f) =
d
{↑
RLD
((Dst f B) F (A Src f )) | F
xyGR f } =
d
{↑
RLD
(F (A × B)) | F xyGR f } = f ÷ (A × B).
[TODO: Filters on cartesian
products vs reloids.]
6 Equa lizers
Categories cont(C) are define d in [1].
I will denote W th e f orgetful functor from cont(C) to C.
In the definition of the category cont(C) take values of as principal morphisms. [TODO:
Wording.]
4 Section 6
Lemma 40. Let f: X Y be a morphism of the category cont(C) where C is a concrete category
(so Wf = ϕ for a Rel-morphism ϕ because f is principal) and im ϕ = A Ob Y . Factor it
ϕ = (A Ob Y ) u where u: Ob X A using properties of Set. The n u is a morphism of cont(C)
(that is a continuous function X ι
A
Y ).
Proof. (A Ob Y )
1
ϕ = (A Ob Y )
1
(A Ob Y ) u;
(A
C
Ob Y )
1
ϕ = (A
C
Ob Y )
1
(A
C
Ob Y ) u;
(A
C
Ob Y )
1
ϕ = u;
X (u)
1
π
A
Y u X ( ϕ)
1
(A
C
Ob Y ) π
A
Y (A
C
Ob Y )
1
ϕ
X ( ϕ)
1
(A
C
Ob Y ) (A
C
Ob Y )
1
Y (A
C
Ob Y ) (A
C
Ob Y )
1
ϕ
X ( ϕ)
1
Y ϕ X (Wf )
1
Y Wf what is true by definition of continuity.
Equational definition of equalizers:
http://nforum.mathforge.org/comments.php?DiscussionID=5328/
Theorem 41. T he 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 Ob X Ob A considered as a morphism X A .
Proof. Denote 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 o f 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 c ont(C)-morphism because o f the lemma above. It is unique by
properties of Set.
We can (over)simplify the above theorem by the obvious below:
Obvious 42. {x Ob A | fx = gx} = dom(f g).
7 Co-e qualizers
http://math.stackexchange.c om/questions/539717/how-to-constru ct-co-equalizers-in-mathbftop
Let be an equivalence relation. Let’s denote π its cano nical projection.
Definition 43. f /=π f π
1
for every morphism f .
Obvious 44. Ob(f/) = (Ob f )/r.
Obvious 45. f /=
FCD
π ×
(C)
FCD
π
f for every morphism f.
To define co-equalizers of morphisms f and g let b e is the smallest equivalence relation such
that fx = gx.
Lemma 46. Let f: X Y be a morphism of the category cont(C) where C is a concrete category
(so Wf = ϕ for a Rel-morphism ϕ because f is prin cipal) 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
Y ; u π X π
1
u
1
Y ; u C(π X π
1
; Y ) = C(X/; Y ).
Theorem 47. The following is a co-equalizer of parallel morphisms f , g: A B of category
cont(C):
the object Y = f /;
Co-equalizers 5
the morphism π considered a s a morphism B Y .
Proof. Let z f = z g for some morphism z.
Let’s prove u π = z for some u: Y Dstz. 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).
8 Rest
[TODO: Specify what is C.]
Theorem 48. The c ategories cont(C) (f or example in Fcd and Rld) are complete.
Proof. They h ave products [1] and equalizers.
Theorem 49. The c ategories cont(C) (f or example in Fcd and Rld) are co-complete.
Proof. They h ave co-products [1] and co-equalizers.
Definition 50. I call morphisms f and g of a category with embeddings equivalent (f g) when
there exist a morphism p such that Src p Src f, Src p Src g, Dst p Dst f , Dst p Dst g and
ι
Src f ,Dst f
p = f and ι
Src g,Dst g
p = g.
Problem 51. Find under which conditions:
1. Equivale nce of morphisms is an equivalence relatio n.
2. Equivale nce of morphisms is a congruenc e for our category.
Bib liography
[1] Victor Porton. Products in dagger categories with complete ordered mor-sets. At http://www.mathe-
matics21.org/binaries/product.pdf.
[2] Victor Porton. Algebraic General Topology. Volume 1. 2013.
6 Section