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can be expressed in the Set category by the following (total) function from the sets
TA to TB such that for any relation R
∈ f ,
∈
TA , f T (R)
=
F sk (f )(R)
=
R if R
⊥
otherwise. An analogous property is valid for its reversed equivalent morphism
f OP
T
TA in DB , too.
Let us demonstrate that the functor F sk :
:
TB
−→
DB sk −→
Set is well defined (we recall
that any object A in DB sk is closed, i.e., A
TA ):
1. For any identity morphism in DB , we have the following two possible cases:
1.1. m = k =
=
1 (case 1 of the definition of F sk ). Thus, for f =
id A : TA →
id A
and id A =
TA ( A
=
P (T A) (thus, S ⊆ TA ), F sk (id A )(S) ={ R | R ∈ S ∩ f l ,f l : A → B ∈
id A
=
TA ), with
={
id A }
TA , for any S
∈ P
(A)
∩ id A }={
}={
R
|
R
∈
S
R
|
R
∈
S
∩
TA
}={
R
|
R
∈
S
}=
S and hence
F sk (id A )
=
id
P (A) is the identity function for the powerset set
P
(A) in Set .
1.2. m
=
k
≥
2 (case 4 of the definition of F sk ). Thus, for
with id TA j = TA j
f =
id A =
id TA j : TA j → TA j
1
≤
j
≤
m
( 1 ≤ j ≤ m A j ) ,
id A
(hence
={
id A 1 ,...,id A m }
), for any S
∈ P
(i,R)
id A
∈ f l ,f l :
=
|
∈
A j →
B i ∈
F sk (id A )(S)
(j,R)
S,R
(j,R) | (j,R) ∈ S,R ∈ id A j , id A j : A j → A j ∈
id A
=
= (j,R) | (j,R) ∈ S,R ∈ TA j = A j = S,
and hence F sk (id A )
=
id
P (A) is the identity function for the powerset set
(A) in Set .
2. For any g
P
= 1 ≤ l ≤ n C l , n
1, and a composition
g ◦ f : A → C , let us show that F sk (g ◦ f )(S) = F sk (g)(F sk (f )(S)) for any
S ∈ P (A) . In the case when n =
:
B
→
C in DB sk , with C
≥
1, it can be directly reduced to one of the cases
1 and 2 of the definition of F sk (f ) above. Thus, let us show this property for two
most complex cases:
2.1. m
=
1 ,k,n
≥
2. Let
(i,R) | R ∈ S ∩ f 1 i ,(f 1 i : A 1 → B i ) ∈
.
f
S 1 = F sk (f )(S) =
=
={
|
∈
∈
g il ,(g il :
Then, F sk (g)(F sk (f )(S))
F sk (g)(S 1 )
(l,R)
(i,R)
S 1 ,R
f
}={ (l,R) | R ∈ S ∩ f 1 i ,(f 1 i : A 1 → B i ) ∈
f
B i → C l ) ∈
,R ∈ g il ,(g il :
f
}={
g
◦
f
}=
∩ f 1 i ∩
B i →
C l )
∈
(l,R)
|
R
∈
S
g il ,(g il ◦
f 1 i :
A 1 →
C l )
∈
{ (l,R) | R ∈ S ∩ g il ◦ f ii ,(g il ◦ f 1 i : A 1 → C l ) ∈
g
◦
f
}= F sk (g ◦ f )(S) .
2.2. m,k,n
≥
2. Let
(i,R)
.
f
∈ f ji ,(f ji :
S 1 =
F sk (f )(S)
=
|
(j,R)
∈
S,R
A j →
B i )
∈
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