266

ϕ

=

is obvious.

ϕ

(

I t J

) =

d

F

F

up

Γ(Src

f,

Dst

f

)

f

(

h

F

iI t h

F

iJ

) =

d

F

F

up

Γ(Src

f,

Dst

f

)

f

h

F

iI t

d

F

F

up

Γ(Src

f,

Dst

f

)

f

h

F

iJ

=

ϕ

I t

ϕ

J

. If

S

is a general-

ized filter base of Src

f

, then

ϕ

F

l

S

=

F

l

F

up

Γ(Src

f,

Dst

f

)

f

h

F

i

F

l

S

=

F

l

F

up

Γ(Src

f,

Dst

f

)

f

F

l

hh

F

ii

S

=

F

l

F

up

Γ(Src

f,

Dst

f

)

f

F

l

X ∈

S

h

F

iX

=

F

l

X ∈

S

F

l

F

up

Γ(Src

f,

Dst

f

)

f

h

F

iX

=

F

l

X ∈

S

ϕ

X

=

F

l

h

ϕ

i

S.

So

ϕ

is a component of a funcoid.

Definition

1399

.

f

=

d

RLD

up

Γ(Src

f,

Dst

f

)

f

for reloid

f

.

Conjecture

1400

.

f

= (

RLD

)

in

(

FCD

)

f

for every reloid

f

.

Obvious

1401

.

f

w

f

for every reloid

f

.

Example

1402

.

(

RLD

)

in

f

6

=

(

RLD

)

out

f

for some funcoid

f

.

Proof.

Take

f

= id

FCD

Ω(

N

)

. Then, as it was shown above, (

RLD

)

out

f

=

and thus

(

RLD

)

out

f

=

. But (

RLD

)

in

f

w

(

RLD

)

in

f

6

=

. So (

RLD

)

in

f

6

=

(

RLD

)

out

f

.

Another proof of the theorem “dom(

RLD

)

in

f

= dom

f

and im(

RLD

)

in

f

= im

f

for every funcoid

f

.”:

Proof.

We have for every filter

X ∈

F

(Src

f

):

X w

dom(

RLD

)

in

f

⇔ X ×

RLD

> w

(

RLD

)

in

f

a

F

(Src

f

)

, b

F

(Dst

f

) : (

a

×

FCD

b

v

f

a

×

RLD

b

v X ×

RLD

>

)

a

F

(Src

f

)

, b

F

(Dst

f

) : (

a

×

FCD

b

v

f

a

v X

)

and

X w

dom

f

⇔ X ×

FCD

> w

f

a

F

(Src

f

)

, b

F

(Dst

f

) : (

a

×

FCD

b

v

f

a

×

FCD

b

v X ×

FCD

>

)

a

F

(Src

f

)

, b

F

(Dst

f

) : (

a

×

FCD

b

v

f

a

v X

)

.

Thus dom(

RLD

)

in

f

= dom

f

. The rest follows from symmetry.

Another proof that dom(

RLD

)

in

f

= dom

f

and im(

RLD

)

in

f

= im

f

for every

funcoid

f

:

Proof.

dom(

RLD

)

in

f

w

dom

f

and im(

RLD

)

in

f

w

im

f

because (

RLD

)

in

f

w

(

RLD

)

in

and dom(

RLD

)

in

f

= dom

f

and im(

RLD

)

in

f

= im

f

.

It remains to prove (as the rest follows from symmetry) that dom(

RLD

)

in

f

v

dom

f

.

Really,

dom(

RLD

)

in

f

v

F

l

X

up dom

f

X

× > ∈

up

f

=

F

l

X

up dom

f

X

up dom

f

=

F

l

up dom

f

= dom

f.