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