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8.7. COMPLETE RELOIDS AND COMPLETION OF RELOIDS

192

Conjecture

1038

.

A reloid

f

is monovalued iff

g

RLD

(Src

f,

Dst

f

) : (

g

v

f

⇒ ∃A ∈

F

(Src

f

) :

g

=

f

|

A

)

.

8.7. Complete reloids and completion of reloids

Definition

1039

.

A

complete

reloid is a reloid representable as a join of reloidal

products

A

{

α

} ×

RLD

b

where

α

A

and

b

is an ultrafilter on

B

for some sets

A

and

B

.

Definition

1040

.

A

co-complete

reloid is a reloid representable as a join of

reloidal products

a

×

RLD

A

{

β

}

where

β

B

and

a

is an ultrafilter on

A

for some

sets

A

and

B

.

I will denote the sets of complete and co-complete reloids from a set

A

to a

set

B

as

ComplRLD

(

A, B

) and

CoComplRLD

(

A, B

) correspondingly and set of all

(co-)complete reloids (for small sets) as

ComplRLD

and

CoComplRLD

.

Obvious

1041

.

Complete and co-complete are dual.

Theorem

1042

.

G

7→

d

n

A

{

α

RLD

G

(

α

)

α

A

o

is an order isomorphism from the set

of functions

G

F

(

B

)

A

to the set

ComplRLD

(

A, B

).

The inverse isomorphism is described by the formula

G

(

α

) = im(

f

|

↑{

α

}

) where

f

is a complete reloid.

Proof.

d

n

A

{

α

RLD

G

(

α

)

α

A

o

is complete because

G

(

α

) =

d

atoms

G

(

α

) and

thus

l

A

{

α

} ×

RLD

G

(

α

)

α

A

=

l

A

{

α

} ×

RLD

b

α

A, b

atoms

G

(

α

)

is complete. So

G

7→

d

n

A

{

α

RLD

G

(

α

)

α

A

o

is a function from

G

F

(

B

)

A

to

ComplRLD

(

A, B

).

Let

f

be complete. Then take

G

(

α

) =

l

(

b

atoms

F

(Dst

f

)

A

{

α

} ×

RLD

b

v

f

)

and we have

f

=

d

n

A

{

α

RLD

G

(

α

)

α

A

o

obviously. So

G

7→

d

n

A

{

α

RLD

G

(

α

)

α

A

o

is

surjection onto

ComplRLD

(

A, B

).

Let now prove that it is an injection:

Let

f

=

l

A

{

α

} ×

RLD

F

(

α

)

α

A

=

l

A

{

α

} ×

RLD

G

(

α

)

α

A

for some

F, G

F

(

B

)

A

. We need to prove

F

=

G

. Let

β

Src

f

.

f

u

(

A

{

β

} ×

RLD

>

F

(

B

)

) = (theorem

610

)

l

(

A

{

α

} ×

RLD

F

(

α

))

u

(

A

{

β

} ×

RLD

>

F

(

B

)

)

α

A

=

A

{

β

} ×

RLD

F

(

β

)

.

Similarly

f

u

(

A

{

β

RLD

>

F

(

B

)

) =

A

{

β

RLD

G

(

β

). Thus

A

{

β

RLD

F

(

β

) =

A

{

β

} ×

RLD

G

(

β

) and so

F

(

β

) =

G

(

β

).

We have proved that it is a bijection. To show that it is monotone is trivial.