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
A∩B
).
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
B∩C
) ◦ (A; B; id
A∩B
) = (A; C; id
A∩C
);
(A; C; id
A∩B ∩C
) = (A; C; id
A∩C
);
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 filter 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
A∩B
.
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
{x∈Ob 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
