4. EQUIVALENT MORPHISMS

10

and

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

ι

A

1

t

A

2

,B

1

t

B

2

g

=

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

ι

A

1

t

A

2

,B

1

t

B

2

h

that is (proposition

2083

)

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

f

=

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

g

and

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

g

=

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

h.

Combining,

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

f

=

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

h

and thus

ι

A

0

t

A

2

,B

0

t

B

2

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

f

=

ι

A

0

t

A

2

,B

0

t

B

2

ι

A

0

t

A

1

t

A

2

,B

0

t

B

1

t

B

2

h

;

(again proposition

2083

)

ι

A

0

t

A

2

,B

0

t

B

2

f

=

ι

A

0

t

A

2

,B

0

t

B

2

h

that is

f

h

.

Proposition

2088

.

[

f

] =

n

ι

A,B

f

A

DOM

f,B

IM

f

o

.

Proof.

If

A

DOM

f

,

B

IM

f

then

ι

A

t

Src

f,B

t

Dst

f

ι

A,B

f

=

ι

A

t

Src

f,B

t

Dst

f

f.

Thus

ι

A,B

f

f

that is

ι

A,B

f

[

f

].

Let now

g

[

f

] that is

f

g

;

ι

Src

f

t

Src

g,

Dst

f

t

Dst

g

f

=

ι

Src

f

t

Src

g,

Dst

f

t

Dst

g

g.

Take

A

= Src

g

,

B

= Dst

g

. We have

ι

A,B

ι

Src

f

t

Src

g,

Dst

f

t

Dst

g

f

=

ι

A,B

ι

Src

f

t

Src

g,

Dst

f

t

Dst

g

g

;

ι

A,B

f

=

ι

A,B

g

=

g.

Proposition

2089

.

1

. IM

f

=

Y

Z

E

Dst

f,Y

C

f

f

;

2

. DOM

f

=

X

Z

f

◦E

X,

Src

f

C

f

.

Proof.

E

Dst

f,Y

C

f

f

ι

Src

f,Y

t

Dst

f

(

E

Dst

f,Y

C

f

) =

ι

Src

f,Y

t

Dst

f

f

E

Y,Y

t

Dst

f

◦ E

Dst

f,Y

f

◦ E

Src

f,

Src

f

=

E

Dst

f,Y

t

Dst

f

f

◦ E

Src

f,

Src

f

E

Y,Y

t

Dst

f

◦ E

Dst

f,Y

f

=

E

Dst

f,Y

t

Dst

f

f

(proposition

2059

)

⇔ E

Y

t

Dst

f,

Dst

f

◦ E

Y,Y

t

Dst

f

◦ E

Dst

f,Y

f

=

E

Y

t

Dst

f,

Dst

f

◦ E

Dst

f,Y

t

Dst

f

f

E

Y,

Dst

f

◦ E

Dst

f,Y

f

=

f.

From this our thesis follows obviously.

Proposition

2090

.

ι

A

1

,B

1

ι

A

0

,B

0

f

v

ι

A

1

,B

1

f

.

Proof.

ι

A

1

,B

1

ι

A

0

,B

0

f

=

E

B

0

,B

1

◦ E

Dst

f,B

0

f

◦ E

A

0

,

Src

f

◦ E

A

1

,A

0

=

id

C

(

B

0

,B

1

)

[

B

0

]

u

[

B

1

]

id

C

(Dst

f,B

0

)

[Dst

f

]

u

[

B

0

]

f

id

C

(

A

0

,

Src

f

)

[

A

0

]

u

[Src

f

]

id

C

(

A

1

,A

0

)

[

A

1

]

u

[

A

0

]

=

id

C

(Dst

f,B

1

)

[Dst

f

]

u

[

B

0

]

u

[

B

1

]

f

id

C

(

A

1

,

Src

f

)

[

A

0

]

u

[

A

1

]

u

[Src

f

]

v

id

C

(Dst

f,B

1

)

[Dst

f

]

u

[

B

1

]

f

id

C

(

A

1

,

Src

f

)

[

A

1

]

u

[Src

f

]

=

ι

A

1

,B

1

f

.