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6. OPERATIONS ON THE SET OF UNFIXED MORPHISMS

12

Note

2098

.

For example for below defined category of funcoids (with binary

product morphism), these filters are filters on filters on sets not filters of sets and
thus are not the same as im and dom.

6. Operations on the set of unfixed morphisms

6.1. Semigroup of unfixed morphisms.

Proposition

2099

.

Let

f

:

A

0

A

1

and

g

:

A

1

A

2

and

A

1

v

B

1

. Then

ι

B

0

,B

2

(

g

f

) =

ι

B

1

,B

2

g

ι

B

0

,B

1

f

.

Proof.

ι

B

0

,B

2

(

g

f

) =

E

A

2

,B

2

C

g

f

◦ E

B

0

,A

0

C

=

E

A

2

,B

2

C

g

1

A

1

f

◦ E

B

0

,A

0

C

=

E

A

2

,B

2

C

g

id

C

(Dst

f,

Src

g

)

A

1

f

◦ E

B

0

,A

0

C

=

E

A

2

,B

2

C

g

◦ E

B

1

,A

1

◦ E

A

1

,B

1

f

◦ E

B

0

,A

0

C

=

ι

B

1

,B

2

g

ι

B

0

,B

1

f

.

Definition

2100

.

We will turn the category

C

into a semigroup

U C

(

the semi-

group of unfixed morphisms

) by the formula [

g

]

[

f

] = [

g

f

] whenever

f

and

g

are

composable morphisms.

We need to prove that [

g

]

[

f

] does not depend on choice of

f

and

g

(provided

that

f

and

g

are composable). We also need to prove that [

g

]

[

f

] is always defined

for every morphisms (not necessarily composable)

f

and

g

. That the resulting

structure is a semigroup (that is,

is associative) is then obvious.

Proof.

That [

g

]

[

f

] is defined in at least one way for every morphisms

f

and

g

is simple to prove. Just consider the morphisms

f

0

=

ι

Src

f,

Dst

f

t

Src

g

f

f

and

g

0

=

ι

Dst

f

t

Src

g,

Dst

g

g

g

. Then we can take [

g

]

[

f

] = [

g

0

f

0

].

It remains to prove that [

g

]

[

f

] does not depend on choice of

f

and

g

. Really,

take arbitrary composable pairs of morphisms (

f

0

:

A

0

B

0

, g

0

:

B

0

C

0

) and

(

f

1

:

A

1

B

1

, g

1

:

B

1

C

1

) such that

f

0

f

1

and

g

0

g

1

. It remains to prove

that

g

0

f

0

g

1

f

1

. We have

ι

B

0

t

B

1

,C

0

t

C

1

g

0

ι

A

0

t

A

1

,B

0

t

B

1

f

0

= (proposition

2099

=

E

C

0

,C

0

t

C

1

C

g

0

f

0

◦ E

A

0

t

A

1

,B

0

C

=

ι

A

0

t

A

1

,C

0

t

C

1

(

g

0

f

0

)

.

Similarly

ι

B

0

t

B

1

,C

0

t

C

1

g

1

ι

A

0

t

A

1

,B

0

t

B

1

f

1

=

ι

A

0

t

A

1

,C

0

t

C

1

(

g

1

f

1

)

.

But

ι

B

0

t

B

1

,C

0

t

C

1

g

0

ι

A

0

t

A

1

,B

0

t

B

1

f

0

=

ι

B

0

t

B

1

,C

0

t

C

1

g

1

ι

A

0

t

A

1

,B

0

t

B

1

f

1

thus having

ι

A

0

t

A

1

,C

0

t

C

1

(

g

0

f

0

) =

ι

A

0

t

A

1

,C

0

t

C

1

(

g

1

f

1

) and so

g

0

f

0

g

1

f

1

.

6.2. Restricted identities.

Definition

2101

.

Restricted identity

for unfixed morphisms is defined as:

id

X

= [id

C

(

A,B

)

X

] for an

X

v

[

A

]

u

[

B

].

We need to prove that it does not depend on the choice of

A

and

B

.

Proof.

Let

A

3

X

v

[

A

0

]

u

[

B

0

] and

A

3

X

v

[

A

1

]

u

[

B

1

] for

A

0

, B

0

, A

1

, B

1

Z

. We need to prove id

C

(

A

0

,B

0

)

X

id

C

(

A

1

,B

1

)

X

.

Really,

ι

A

1

,B

1

id

C

(

A

0

,B

0

)

X

=

E

B

0

,B

1

id

C

(

A

0

,B

0

)

X

◦E

A

1

,A

0

=

id

C

(

B

0

,B

1

)

[

B

0

]

u

[

B

1

]

id

C

(

A

0

,B

0

)

X

id

C

(

A

1

,A

0

)

[

A

0

]

u

[

A

1

]

= id

C

(

A

1

,B

1

)

[

A

0

]

u

[

A

1

]

u

[

B

0

]

u

[

B

1

]

u

X

= id

C

(

A

1

,B

1

)

X

.

Sim-

ilarly

ι

A

0

,B

0

id

C

(

A

1

,B

1

)

X

= id

C

(

A

0

,B

0

)

X

.

So id

C

(

A

0

,B

0

)

X

id

C

(

A

1

,B

1

)

X

.

Proposition

2102

.

id

Y

id

X

= id

X

u

Y

for every

X, Y

A

.