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4. EQUIVALENT MORPHISMS

9

Proof.

ι

A,B

f

=

E

Dst

f,B

C

f

◦ E

A,

Src

f

C

= (by definition of IM

f

and DOM

f

) =

E

Dst

f,B

C

◦ E

Y,

Dst

f

C

◦ E

Dst

f,Y

C

f

◦ E

X,

Src

f

C

◦ E

Src

f,X

C

◦ E

A,

Src

f

C

=

E

Y,B

C

◦ E

Dst

f,Y

C

f

E

X,

Src

f

C

◦ E

A,X

C

=

ι

A,B

ι

X,Y

f

because

E

Dst

f,B

◦ E

Y,

Dst

f

◦ E

Dst

f,Y

= id

C

(Dst

f,B

)

Y

u

Dst

f

u

B

= id

C

(

Y,B

)

Y

u

B

id

C

(Dst

f,Y

)

Y

u

Dst

f

=

E

Y,B

◦ E

Dst

f,Y

and thus

E

Dst

f,B

C

◦ E

Y,

Dst

f

C

◦ E

Dst

f,Y

C

=

E

Y,B

C

◦ E

Dst

f,Y

C

and similarly

for

E

X,

Src

f

C

◦ E

Src

f,X

C

◦ E

A,

Src

f

C

.

Definition

2084

.

I call two morphisms

f

∈ C

(

A

0

, B

0

) and

g

∈ C

(

A

1

, B

1

) of a

category with restricted morphisms

equivalent

(and denote

f

g

) when

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

,B

0

t

B

1

g.

Proposition

2085

.

f

g

iff

ι

A,B

f

=

ι

A,B

g

for some

A

DOM

f

DOM

g

,

B

IM

f

IM

g

.

Proof.

Both

ι

A,B

f

=

ι

A,B

g

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

,B

0

t

B

1

g

and

ι

A,B

f

=

ι

A,B

g

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

,B

0

t

B

1

g

follow from proposition

2083

.

Theorem

2086

.

Let

f

:

A

0

B

0

and

g

:

A

1

B

1

(for a partially ordered

category with restricted identities). The following are pairwise equivalent:

1

.

f

g

;

2

.

ι

A

1

,B

1

f

=

g

and

ι

A

0

,B

0

g

=

f

;

3

.

ι

A

1

,B

1

f

w

g

and

ι

A

0

,B

0

g

w

f

.

Proof.

1

2

.

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

,B

0

t

B

1

g

;

ι

A

1

,B

1

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

1

,B

1

ι

A

0

t

A

1

,B

0

t

B

1

g

;

ι

A

1

,B

1

f

=

ι

A

1

,B

1

g

;

ι

A

1

,B

1

f

=

g

.

ι

A

0

,B

0

g

=

f

is similar.

3

1

Let

ι

A

1

,B

1

f

w

g

and

ι

A

0

,B

0

g

w

f

.

ι

A

1

,B

1

ι

A

0

,B

0

g

w

g

;

E

B

0

,B

1

◦ E

B

1

,B

0

g

◦ E

A

0

,A

1

◦ E

A

1

,A

0

w

g

;

id

C

(

B

1

,B

1

)

[

B

0

]

u

[

B

1

]

g

id

C

(

A

1

,A

1

)

[

A

0

]

u

[

A

1

]

w

g

; id

C

(

B

1

,B

1

)

[

B

0

]

u

[

B

1

]

g

w

g

; id

C

(

B

1

,B

1

)

[

B

0

]

u

[

B

1

]

g

=

g

;

id

C

(

B

0

u

B

1

,B

1

)

[

B

0

]

u

[

B

1

]

id

C

(

B

1

,B

0

u

B

1

)

[

B

0

]

u

[

B

1

]

g

=

g

;

E

B

0

u

B

1

,B

1

◦ E

B

1

,B

0

u

B

1

g

=

g

.

Thus

B

0

u

B

1

Im

g

. Similarly

A

0

u

A

1

Dom

g

.

So

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

,B

0

t

B

1

ι

A

0

,B

0

g

=

ι

A

0

t

A

1

,B

0

t

B

1

g

.

2

3

Obvious.

Proposition

2087

.

Above defined equivalence of morphisms (for a small cat-

egory) is an equivalence relation.

Proof.

Reflexivity. Obvious.
Symmetry. Obvious.
Transitivity. Let

f

g

and

g

h

for

f

:

A

0

B

0

,

g

:

A

1

B

1

,

h

:

A

2

B

2

. Then

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

,B

0

t

B

1

g

and

ι

A

1

t

A

2

,B

1

t

B

2

g

=

ι

A

1

t

A

2

,B

1

t

B

2

h

.

Thus

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

ι

A

0

t

A

1

,B

0

t

B

1

f

=

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

ι

A

0

t

A

1

,B

0

t

B

1

g