Database Reference
In-Depth Information
TAwith its inverse is
−
A
:
TA
→
A such that is
A
=
is
−
A
=
id
A
=
TA
.
Thus
,
A
TA
.
2. (
Commutativity
)
A
B
B
A
.
3. (
Associativity
)
A
(B
C)
A
B
C
(A
B)
C
.
4. (
Composition with zero object
)
C
1.
For any object A
,
we have the isomorphism is
A
:
A
→
0
0
⊥
C
⊥
C
.
Proof
Claim 1. Let us consider a simple object (instance-database)
A
. The SOtgd
Φ
A
of the schema mappings, corresponding to the morphisms
id
A
:
A
→
A
,
is
A
:
TA
and
is
−
A
:
A
→
TA
→
A
, is equal to the
InverseOperads(
{
1
r
i
∈
O(r
i
,r
i
)
|
r
i
∈
)
, i.e., to the logical formula
{∀
S
A
}∪{
1
r
∅
}
x
i
(r
i
(
x
i
)
⇒
r
i
(
x
i
))
|
r
i
∈
S
A
}
where
A
⊆
A
α
∗
(
A
α
∗
(
α
∗
(
A
=
A
)
of a schema
A
and
(thus,
A
=
A
)
⊆
)
=
TA
). Thus,
α
∗
(MakeOperads(
α
∗
(MakeOperads(
id
A
=
{
Φ
A
}
))
:
A
→
A
,
is
A
=
{
Φ
A
}
))
:
A
→
TA
, and
is
−
A
=
α
∗
(MakeOperads(
A
,are
atomic
morphisms (Def-
inition
18
). That is, these three morphisms are based on the same SOtgd
Φ
A
and, consequently, for each object
A
in
DB
(
id
A
=
{
Φ
}
A
))
:
TA
→
TA
is demonstrated by The-
=
1
≤
j
≤
m
A
j
,
orem
1
) that
is
A
=
is
−
A
=
id
A
=
TA
. For a complex object
A
2,
is
A
=
1
≤
j
≤
m
is
A
j
,
is
−
A
=
1
≤
j
≤
m
is
−
1
is
A
=
1
≤
i
≤
k
is
−
1
A
j
: In fact,
is
−
A
◦
m
≥
A
i
◦
1
≤
i
≤
k
id
A
i
is
A
i
=
1
≤
i
≤
k
id
A
i
=
id
A
id
A
(i.e.,
=
={
id
Ai
:
A
i
→
A
i
|
1
≤
i
≤
k
}
).
A
i
=
1
≤
i
≤
k
id
TA
i
=
id
TA
. Hence
is
A
=
is
−
A
=
id
A
=
is
−
1
Analogously,
is
A
i
◦
1
≤
j
≤
m
TA
j
=
TA
.
Claim 2. (Commutativity) For any two objects in
DB
there is a commutativity
isomorphism
A
B
B
A
.
Let us show that this isomorphism is represented by the complex morphism
i
1
=[⊥
1
, id
B
]
1
1
, id
B
]:
,
[
id
A
,
⊥
] :
A
B
→
B
A
(where
[⊥
A
B
→
B
1
i
OP
1
1
, id
B
]
OP
,
and
[
id
A
,
⊥
]:
A
B
→
A
) and its inverse by
i
2
=
=[[⊥
[
id
A
,
1
OP
1
, id
B
1
⊥
]
]=[⊥
,
id
A
,
⊥
]:
B
A
→
A
B
.
In fact,
i
2
=
⊥
1
, id
B
,
id
A
,
1
◦
⊥
1
, id
B
,
id
A
,
1
i
1
◦
⊥
⊥
=
⊥
1
, id
B
◦
⊥
1
, id
B
,
⊥
1
, id
B
◦
id
A
,
1
,
⊥
id
A
,
1
◦
⊥
1
, id
B
,
id
A
,
1
◦
id
A
,
1
⊥
⊥
⊥
=
,
⊥
id
A
=
id
B
,
1
,
⊥
1
, id
A
.
1
1
id
B
,
⊥
,
⊥
Thus,
i
1
◦
i
2
id
B
id
A
={
id
A
, id
B
}=
, and hence, by Definition
23
,
i
1
◦
i
2
=
id
B
id
A
=
(
point 2 of Example
19
)
=
id
B
A
:
B
A
→
B
A.
Analogously,
i
1
=
⊥
1
, id
B
,
id
A
,
1
◦
⊥
1
, id
B
,
id
A
,
1
i
2
◦
⊥
⊥
=
⊥
1
, id
B
◦
⊥
1
, id
B
,
id
A
,
1
◦
id
A
,
1
⊥
⊥
