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Proposition 39.

For a bijective

Set

-morphism

f

:

S

up

A →

S

up

B

the following are equivalent:

1. up

B

=

f

∗A

.

2.

C

S

up

B

:

C

up

B ⇔

f

1

C

up

A

.

3.

C

S

up

B

: (

h

f

i

C

up

B ⇔

C

up

A

)

.

4.

h

f

i|

up

A

is a bijection from up

A

to up

B

.

5.

h

f

i|

up

A

is a function onto up

B

.

6.

B

=

h

f

iA

.

7.

f

Mor

G reFunc

2

(

A

;

B

)

.

8.

f

Mor

FuncBij

(

A

;

B

)

.

Proof.

(1)

(2).

up

B

=

f

∗A ⇔

up

B

=

C

P

S

up

B |

f

1

C

up

A

 

⇔ ∀

C

S

up

B

:

C

up

B ⇔

f

1

C

up

A

.

(2)

(3).

Because

f

is a bijection.

(2)

(5).

For every

C

up

B

we have

f

1

C

up

A

and thus

h

f

i|

up

A

f

1

C

=

h

f

i

f

1

C

=

C

up

B

. Thus

h

f

i|

up

A

is onto up

B

.

(4)

(5).

Obvious.

(5)

(4).

We need to prove only that

h

f

i|

up

A

is an injection. But this follows from the fact

that

f

is a bijection.

(4)

(3).

We have

C

S

up

B

: ((

h

f

i|

up

A

)

C

up

B ⇔

C

up

A

)

and consequently

C

S

up

B

: (

h

f

i

C

up

B ⇔

C

up

A

)

.

(6)

(1).

From the last corollary.

(1)

(7).

Obvious.

(7)

(8).

Obvious.

Corollary 40.

The following are equivalent for every f.o.

A

and

B

:

1.

A

is directly isomorphic to a f.o.

B

.

2. There are a bijective

Set

-morphism

f

:

S

up

A →

S

up

B

such that for every

C

P

S

up

B

C

up

B ⇔

f

1

C

up

A

3. There are a bijective

Set

-morphism

f

:

S

up

A →

S

up

B

such that for every

C

P

S

up

B

h

f

i

C

up

B ⇔

C

up

A

.

4. There are a bijective

Set

-morphism

f

:

S

up

A →

S

up

B

such that

h

f

i|

up

A

is a bijection

from up

A

to up

B

.

5. There are a bijective

Set

-morphism

f

:

S

up

A →

S

up

B

such that

h

f

i|

up

A

is a function

onto up

B

.

6. There are a bijective

Set

-morphism

f

:

S

up

A →

S

up

B

such that

B

=

h

f

iA

.

7. There are a bijective morphism

f

Mor

G reFunc

2

(

A

;

B

)

.

Proposition 41. GreFunc

1

and

GreFunc

2

with function composition are categories.

Proof.

Let

f

:

A → B

and

g

:

B → C

are morphisms of

GreFunc

1

. Then

B ⊆ h

f

iA

and

C ⊆ h

g

iB

.

So

h

g

f

iA

=

h

g

ih

f

iA ⊇ h

g

iB ⊇ C

. Thus

g

f

is a morphism of

GreFunc

1

. Associativity law is

evident. Id

S

up

A

is the identity morphism of

GreFunc

1

for every f.o.

A

.

6

Section 3