Lemma 40.

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

= (

A

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.

(

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.

The following is an equalizer of parallel morphisms

f , g

:

A

B

of category

cont

(

C

)

:

the object

X

=

ι

{

x

Ob

A

|

f x

=

g x

}

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 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 42.

{

x

Ob

A

|

fx

=

gx

}

=

dom

(

f

g

)

.

7 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.

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

be 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

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

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