6. OPERATIONS ON THE SET OF UNFIXED MORPHISMS

13

Proof.

Take arbitrary id

C

(

A,B

0

)

X

id

X

and id

C

(

B

1

,C

)

Y

id

Y

.

Obviously, id

C

(

A,B

0

t

B

1

)

X

id

X

and id

C

(

B

0

t

B

1

,C

)

Y

id

Y

.

Thus id

Y

id

X

=

[id

C

(

B

0

t

B

1

,C

)

Y

]

[id

C

(

A,B

0

t

B

1

)

X

] = [id

C

(

A,C

)

X

u

Y

] = id

X

u

Y

.

6.3. Poset of unfixed morphisms.

Lemma

2103

.

f

v

g

ι

A,B

f

v

ι

A,B

g

for every morphisms

f

and

g

such that

Src

f

= Src

g

and Dst

f

= Dst

g

.

Proof.

ι

A,B

f

v

ι

A,B

g

⇔ E

Dst

f,B

f

◦ E

A,

Src

f

v E

Dst

g,B

g

◦ E

A,

Src

g

id

C

(Dst

f,B

)

[

B

]

u

[Dst

f

]

f

id

C

(

A,

Src

f

)

[

A

]

u

[Src

f

]

v

id

C

(Dst

g,B

)

[

B

]

u

[Dst

g

]

g

id

C

(

A,

Src

g

)

[

A

]

u

[Src

g

]

f

v

g

because

id

C

(Dst

f,B

)

[

B

]

u

[Dst

f

]

= id

C

(Dst

g,B

)

[

B

]

u

[Dst

g

]

and id

C

(

A,

Src

f

)

[

A

]

u

[Src

f

]

= id

C

(

A,

Src

g

)

[

A

]

u

[Src

g

]

.

Corollary

2104

.

1

.

f

0

v

g

0

f

0

f

1

g

0

g

1

f

1

v

g

1

whenever Src

f

0

= Src

g

0

and

Dst

f

0

= Dst

g

0

and Src

f

1

= Src

g

1

and Dst

f

1

= Dst

g

1

.

2

.

f

0

v

g

0

f

1

v

g

1

whenever Src

f

0

= Src

g

0

and Dst

f

0

= Dst

g

0

and

Src

f

1

= Src

g

1

and Dst

f

1

= Dst

g

1

and

f

0

f

1

g

0

g

1

.

Proof.

1

Because

f

1

=

ι

Src

f

1

,

Dst

f

1

f

0

and

g

1

=

ι

Src

g

1

,

Dst

g

1

f

0

.

2

A consequence of the previous.

The above corollary warrants validity of the following definition:

Definition

2105

.

The order on the set of unfixed morphisms is defined by the

formula [

f

]

v

[

g

]

f

v

g

whenever Src

f

= Src

g

Dst

f

= Dst

g

.

It is really an order:

Proof.

Reflexivity. Obvious.
Transitivity. Obvious.
Antisymmetry. Let [

f

]

v

[

g

] and [

g

]

v

[

f

] and Src

f

= Src

g

Dst

f

= Dst

g

. Then

f

v

g

and

g

v

f

and thus

f

=

g

so having [

f

] = [

g

].

Obvious

2106

.

f

7→

[

f

] is an order embedding from the set

C

(

A, B

) to unfixed

morphisms, for every objects

A

,

B

.

Proposition

2107

.

If

S

is a set of parallel morphisms of a partially ordered

category with an equivalence relation respecting the order, then

1

.

d

X

S

[

X

] exists and

d

X

S

[

X

] = [

d

S

];

2

.

d

X

S

[

X

] exists and

d

X

S

[

X

] = [

d

S

].

Proof.

1

[

d

S

]

v

[

X

] for every

X

S

because

d

S

v

X

.

Let now

L

v

[

X

] for every

X

S

for an equivalence class

L

. Then

L

v

[

d

S

]

because

l

v

d

S

for

l

L

because

l

v

X

for every

X

S

.

Thus [

d

S

] is the greatest lower bound of

n

[

X

]

X

S

o

.

2

By duality.

Proposition

2108

.