Database Reference
In-Depth Information
simple morphisms in
DB
and only after that the general properties of (also complex)
morphisms.
Corollary 9
For each simple arrow f
:
A
→
B the following are valid
:
1.
f is monic iff f
=
TA
.
2.
f is epic iff f
TB
.
3.
f is an isomorphism iff it is monic and epic
(
i
.
e
.,
when f
=
=
TA
=
TB
).
Proof
Claim 1. (From right to left) Let us show that if
f
=
TA
then
f
:
A
→
B
is monic. A simple arrow
f
:
A
−→
B
is monic iff for any (possibly complex) two
:
1
≤
i
≤
n
X
i
−→
◦
=
◦
=
arrows
h,g
A
,
f
h
f
g
implies
h
g
. Let us show that it is
satisfied when
f
=
TA
. We have (from Proposition
6
) that for each point-to-point
g
simple arrow
(g
i
:
X
i
→
A)
∈
,
g
i
⊆
TA
(analogously for each
(h
i
:
X
i
→
A)
∈
h
,
h
i
⊆
TA
). Thus, if
f
◦
h
=
f
◦
g
then, from Definition
22
,
f
◦
g
f
◦
h
, i.e., they
have the same ptp arrows, that is,
f
◦
g
i
=
f
◦
h
i
:
X
i
→
B
, so that they have the
same fluxes, i.e.,
f
=
g
i
=
f
h
i
. But we have that
f
g
i
=
f
◦
◦
◦
∩
g
i
=
TA
∩
g
i
=
g
i
and
f
h
i
=
f
∩
h
i
=
∩
h
i
=
h
i
, so that
h
i
=
g
i
, i.e., they are equal ptp
arrows
h
i
=
g
i
:
X
i
→
A
. It holds for all ptp arrows in
g
and
h
, so that
g
=
h
, and,
consequently,
f
:
A
−→
B
is a monomorphism.
(From left to right) If a simple arrow
f
:
A
→
B
is monic then for any (possibly
complex) two arrows
h,g
◦
TA
:
1
≤
i
≤
n
X
i
−→
A
,
f
◦
h
=[
f
◦
h
1
,...,f
◦
h
n
]=
f
◦
g
n
]:
1
≤
i
≤
n
X
i
−→
g
=[
f
◦
g
1
,...,f
◦
B
, implies
h
=
g
, that is,
∀
1
≤
l
≤
n(f
◦
A)
. Let us suppose that
f
g
l
=
◦
h
l
:
X
l
→
B
implies
g
l
=
h
l
:
X
l
→
⊂
f
TA
,
g
l
=
f
=
h
l
, that is,
g
l
=
so that we can have, for example,
=
TA
h
l
but with
f
g
l
=
f
=
f
◦
◦
h
l
, i.e.,
f
◦
g
l
=
f
◦
h
l
, which is a contradiction. Thus, it must
hold that
f
=
TA
.
Claim 2. The proof is analogous to that of Claim 1.
Claim 3. (From left to right) It holds for every category.
(From right to left) If
f
:
A
→
B
is both monic and epic then, from the first
two points,
f
=
TA
=
TB
. Let us show that
f
=
is
−
B
◦
is
−
B
◦
is
A
. In fact,
is
A
=
is
−
B
∩
is
A
=
(from Lemma
9
)
=
TB
∩
TA
=
TA
=
TB
, so that, from Definition
23
,
is
−
B
◦
=
is
A
:
→
f
B
, that is,
f
is a composition of two isomorphisms, and hence
it is an isomorphism as well.
A
Now we will also analyze the properties of complex arrows.
Notice that the complex morphism
f
C
has to be for-
mally represented by 'indexing by position' formalism, that is,
f
=[
id
C
, id
C
]:
C
C
→
=[
id
C
, id
C
]:
C
, so that
f
A
1
A
2
→
B
1
where
A
1
=
A
2
=
B
1
=
={
f
11
=
id
C
:
A
1
→
B
1
,f
21
=
id
C
:
A
2
→
B
1
}
,thatis,by
two
mutually independent mapping-morphisms
id
C
:
1
1
, id
C
]:
C
→
C
. Thus, the morphisms
g
=[
id
C
,
⊥
]:
C
C
→
C
and
h
=[⊥
g
={
f
h
C
C
→
C
have
g
11
=
id
C
:
A
1
→
B
1
}={
id
C
:
C
→
C
}=
,
={
h
21
=
f
.But
g
h
id
C
:
A
2
→
B
1
}={
id
C
:
C
→
C
}=
={
id
C
:
C
→
C
}=
so that, from
1
1
, id
C
]
Definition
23
,
g
=
h
. Thus,
[
id
C
,
⊥
]=
g
=
h
=[⊥
.
