Database Reference
In-Depth Information
1
Notice that an empty arrow
⊥
={
q
⊥
:⊥→⊥}
between two databases, with
⊥
0
1
, is equivalent to the absence of the schema mapping between these
two databases. In fact, because of that, we are always able to introduce the empty
arrows between databases, as it was used in Lemma
8
for the construction of canon-
ical complex morphisms in Sect.
3.2
. In fact, an empty arrow is not a ptp arrow,
so that the presence or absence of them in a given complex arrow does not change
such an arrow (from Definition
23
two complex arrows are equal iff they have the
same set of ptp arrows and the empty arrows are on ptp arrows). Moreover, note
that each empty arrow
=⊥
={⊥}
1
0
⊥
:
A
j
→
B
i
is not monic if
A
j
=⊥
(to be monic it must
satisfy
TA
j
= ⊥
0
) and is not epic if
B
i
=⊥
0
1
=⊥
(to be epic it must satisfy
TB
i
=⊥
0
), and is not an isomorphism if
A
j
=⊥
0
0
.
1
=⊥
or
B
i
=⊥
0
is isomorphic to
A
, we can replace any
original complex-object
B
by a
strictly complex
object
B
obtained by elimination of
the empty objects
Notice also that, from the fact that
A
⊥
0
from the disjoint union. Thus, any complex arrow
g
:
A
→
B
can be substituted by an equivalent arrow
f
⊥
A
→
B
between two strictly-complex
:
objects
A
and
B
such that
f
g
=
so that
f
can always be well-defined from the set
g
by using the canonical representation of
g
between strictly complex objects
A
and
B
(by Lemma
8
). Consequently, any proof for the morphisms can be done by
using its canonical morphism between the strictly-complex objects obtained from
the original source and target objects (as we did in the last part of the proof of
Proposition
8
).
Lemma 15
Each complex arrow f
:
A
→
B between the strictly-complex objects
=
1
≤
j
≤
m
A
j
,B
=
1
≤
i
≤
k
B
i
,
with m,k
A
≥
2,
is an isomorphism iff it is both
monic and epic and composed of m
=
k point-to-point arrows
.
Proof
From Proposition
6
and the fact that
f
is monic, for each
A
j
there must exist
a ptp arrow
f
ji
∈
f
f
|≥
with
dom(f
ji
)
=
A
j
,
|
m
. Moreover,
f
is epic and hence
f
f
|≥
for each
B
i
there must exist a ptp arrow
f
li
∈
with
cod(f
li
)
=
B
i
, so that
|
k
.
f
Thus the fact that
|=
m
=
k
is consistent with the facts that
f
is monic and epic,
and for each
A
j
we have only one outgoing ptp arrow from it while for each
B
i
we
have only one ingoing arrow into it.
Moreover, from the fact that
f
is both monic and epic, each of its ptp arrows
(f
ji
:
A
j
→
|
f
must be epic and monic (by Proposition
6
), and from the fact that it
is a simple arrow, it must be an isomorphism with
f
ji
=
B
i
)
∈
TA
j
=
TB
i
(i.e.,
A
j
B
i
),
A
j
. Consequently, from Definition
22
,
f
◦
f
OP
with its inverse
f
OP
ji
:
B
i
→
={
f
ji
◦
f
OP
f
f
(
from the fact that
f
is epic, for each
B
i
there
must exist a ptp arrow
f
ji
with
cod(f
ji
)
=
B
i
)
={
ji
|
f
ji
∈
}={
id
B
i
|
f
ji
∈
}=
id
B
id
B
i
|
1
≤
i
≤
m
}=
. Thus,
f
OP
◦
f
from Definition
23
, we obtain
f
◦
f
OP
={
f
OP
=
id
B
. Analogously,
ji
◦
f
}={
f
}=
f
ji
|
(
from the fact that
f
is monic, for each
A
j
there
must exist a ptp arrow
f
ji
with
dom(f
ji
)
=
A
j
)
={
f
ji
∈
id
A
j
|
f
ji
∈
id
A
id
A
j
|
1
≤
i
≤
m
}=
. Thus,
