Database Reference
In-Depth Information
Proposition 6
We have the following fundamental properties for the morphisms
in
DB
1.
If a morphism f
:
A
−→
B has a property P
∈{
isomorphic, monic, epic
}
then
f
each of its ptp arrows f
l
∈
has the same property P
;
=
1
≤
j
≤
m
A
j
,
m
2.
If f
:
A
→
B is monic and A
≥
1,
is strictly complex then for
f
each A
j
there must exist a ptp arrow f
n
∈
with dom(f
n
)
=
A
j
;
3.
If f
:
A
B is epic and B
=
1
≤
i
≤
k
B
i
,
k
≥
1,
k
≥
1,
is strictly complex then
f
for each B
i
there must exist a ptp arrow f
n
∈
with cod(f
n
)
=
B
i
.
=
1
≤
j
≤
m
C
j
and B
=
1
≤
i
≤
k
B
i
are strictly-complex objects and there is
4.
If C
an isomorphism is
:
C
B
,
then m
=
k
.
f
OP
={
f
OP
Proof
Claim 1.
Its inverted arrow is
ji
:
B
i
→
A
j
|
f
ji
:
A
j
→
:
1
≤
j
≤
m
A
j
→
1
≤
i
≤
k
B
i
be an isomor-
and
f
OP
f
ji
=
f
ji
}
B
i
in
.Let
f
f
◦
f
−
1
=
f
◦
f
OP
={
f
−
1
f
OP
phism, so that
f
◦
=
id
B
, thus
f
ji
◦
ji
=
id
i
:
B
i
→
f
OP
f
,f
OP
with
cod(f
OP
ji
B
i
|
f
ji
∈
ji
∈
)
=
dom(f
ji
)
=
A
j
and
cod(f
ji
)
=
B
i
}=
1
≤
i
≤
k
id
B
i
f
id
B
=
. Let us suppose that
f
ji
∈
is not an isomorphism. Thus,
f
ji
◦
f
OP
id
i
:
B
i
→
B
i
(with
cod(f
OP
ji
=
)
=
dom(f
ji
)
and
cod(f
ji
)
=
B
i
), so that
ji
f
−
1
f
id
B
and
f
would not be an isomorphism, which is a contradiction. Con-
sequently, each
f
ji
∈
◦
=
f
must be an isomorphism. The converse does not hold: let
us consider the morphism
f
=[
id
C
, id
C
]:
A
1
A
2
→
B
1
, with
A
1
=
A
2
=
B
1
=
C
f
where
where the point-to-point
arrows
id
C
are the identity arrows (thus isomorphisms as well) but
f
is not an iso-
morphism (from Proposition
5
). Let
f
:
A
→
B
be a monomorphism. Then, for
every two arrows
g,h
:
X
→
A
if
f
◦
g
=
f
◦
h
then we obtain
g
=
h
. Let us sup-
pose that there exists
(f
ji
:
A
j
→
B
i
)
∈
={
f
11
=
id
C
:
A
1
→
B
1
,f
21
=
id
C
:
A
2
→
B
1
}
f
which is not monic, that is, there exist
g
lj
,h
lj
such that
f
ji
◦
g
lj
=
f
ji
◦
h
lj
with
g
lj
=
h
lj
:
X
l
→
A
j
. However, from the
g
=
h
fact that
f
is a monomorphism it must hold that
g
=
h
and, by Definition
23
,
with
g
lj
=
A
j
, which is a contradiction. Thus, each point-to-point arrow
f
ji
of a monic arrow
f
must be monic as well.
The proof for an epimorphism
f
h
lj
:
X
l
→
B
is similar.
Claim 2. Let
A
be a strictly-complex object (with all
A
j
=⊥
:
A
0
,j
=
1
,...,m
) and
f
}
let us suppose that there exists
A
j
such that
A
j
/
∈{
dom(f
n
)
|
f
n
∈
.Let
X
=
X
1
be a simple object and
g,h
:
X
1
→
A
be two arrows such that
f
◦
g
=
f
◦
h
so
that from the fact that
f
is a monomorphism,
g
=
h
with two ptp arrows
(g
1
j
:
g
h
X
1
→
A
j
)
∈
and
(h
1
j
:
X
1
→
A
j
)
∈
with
g
1
j
=
h
1
j
. However, we can now
take any ptp arrow
g
1
j
:
A
j
such that
g
1
j
=
h
1
j
and change
g
into
g
by
X
1
→
substituting old
g
1
j
by
g
1
j
, so that still
f
g
=
g
=
◦
f
◦
h
is valid, that is,
f
◦
g
but not
g
=
f
g
and hence, we obtain that
f
is not monomorphic, which is
a contradiction. Consequently, for each
A
j
there must exist at least one ptp arrow
f
n
∈
◦
f
1
).
with
dom(f
n
)
=
A
j
if
f
is a monomorphism (note that
f
n
=⊥
Claim 3. The proof when
f
is epic is similar.
