Database Reference
In-Depth Information
(obtained from the sketch's arrow
M
AA
∅
:
A
→
A
∅
in Example
17
) and its dual
arrow (which is a monomorphism)
f
OP
1
0
A
, both with empty infor-
mation flux. It is easy to verify that each
empty database
(with all empty relations)
is isomorphic to the zero object
=⊥
:⊥
→
0
:
⊥
0
Lemma 10
The bottom object
⊥
={⊥}
is the zero
(
terminal and initial
)
closed
object in
DB
.
Each database A
={
R
1
,...,R
n
,
⊥}
with all empty relations R
i
=
0
.
That is
,
A
⊥
0
.
{}
,
i
=
1
,...,n
,
is isomorphic to the zero object
⊥
0
, we have from Proposition
7
that
Proof
For each simple object
A
and
f
:
A
→⊥
f
0
0
0
1
⊆
∩
⊥
=
∩⊥
=⊥
=⊥
TA
T
TA
(From point 1 of Theorem
1
), thus
f
0
, so that
0
is the unique arrow from
A
into
⊥
⊥
is a terminal object (analogously,
1
0
→
A
is the unique arrow, so that
0
⊥
:⊥
⊥
is an initial object as well).
1
,
1
[⊥
⊥
]:
Thus, for a complex object
A
B
, the unique arrow into zero object is
0
1
,
1
0
A
B
→⊥
while its opposite is
⊥
⊥
:⊥
→
A
B
.
For each database
A
={
R
1
,...,R
n
,
⊥}
where every
R
i
={}
,
k
=
ar(R
i
)
≥
1,
1
is a
k
-ary empty relation, for
i
=
1
,...,n
, we have the unique arrow
⊥
:
A
→
0
1
⊥
(into terminal object) with
⊥
={
q
⊥
:⊥→⊥}
and its unique inverted arrow
1
)
OP
0
1
)
OP
(
⊥
:⊥
→
A
(from initial object) with
(
⊥
={
q
⊥
:⊥→⊥}
, hence
⊥
1
OP
1
1
◦⊥
=⊥
:
A
→
A.
with
id
A
=
The identity arrow for
A
is
id
A
={
id
R
1
:
R
i
→
R
i
|
R
i
∈
A
}∪{
q
⊥
}
= ⊥
0
1
1
)
OP
1
⊥
so that, from Definition
23
,
(
⊥
◦⊥
=
id
A
and, analogously,
1
1
)
OP
1
0
0
.
⊥
◦
⊥
=
⊥
:
→⊥
⊥
(
id
0
. Thus,
A
is an isomorphism, that is,
A
⊥
3.2.3 Symmetry
The mapping
B
T
:
Ob
DB
, specified in the definition of the
DB
category
in Theorem
1
, is a fundamental concept for the categorial symmetry [
13
] introduced
in Definition
4
(Sect.
1.5.1
). Let us demonstrate the following equality for any mor-
phism in
DB
, which will be used in the proof of the next theorem.
Mor
DB
−→
f
OP
in
B
:
f
Lemma 11
For any morphism f
:
A
→
B
,
f
=
f
◦
◦
f
.
Let in
f
=
→
in
A
:
f
TB and in
f
OP
=
→
TAbe two monomorphisms defined by Proposition
7
.
in
OP
Tf
OP
◦
f
=
:
→
Then
,
in
TB
TA
.
f
OP
=
f
OP
(by duality), so that
Proof
If
f
is a simple arrow then
f
f
OP
f
◦
◦
f
=
∩
f
OP
f
=
f
and hence, from Definition
23
:
(i)
f
=
f
◦
f
OP
∩
f
◦
f
.
