Database Reference
In-Depth Information
g
il
◦
g
il
◦
f
ji
=
h
il
◦
f
ji
:
A
j
→
X
l
and hence they have the same fluxes, i.e.,
f
ji
=
h
il
◦
g
il
◦
g
il
∩
f
ji
=
h
il
◦
f
ji
=
h
il
∩
f
ji
=
f
ji
. However,
f
ji
=
g
il
∩
TB
i
=
g
il
and
h
il
∩
TB
i
=
h
il
, so that
h
il
=
g
il
, i.e., are equal ptp arrows
h
il
=
g
il
:
B
i
→
X
l
.
It holds for all ptp arrows in
g
and
h
so that
g
=
h
and, consequently,
f
is an
epimorphism. If
B
is not strictly complex, then we have an isomorphism (which is
epic as well)
is
B
→
B
(thus,
TB
TB
) where
B
is strictly complex, so that
:
B
is an epimorphism above with
f
f
:
B
(where
f
:
B
→
f
=
is
◦
A
→
TB
TB
) is a composition of two epimorphism, with
f
f
, i.e.,
f
TB
and hence it
is epic as well.
Claim 3. For a simple arrow it holds from Corollary
9
. So let us consider
strictly complex objects
A
=
1
≤
j
≤
m
A
j
and
B
=
1
≤
i
≤
k
B
i
so that
m
=
k
≥
2
=
1
≤
j
≤
m
A
j
=
1
≤
i
≤
k
B
i
,wehaveabi-
and, from the isomorphism
TA
TB
jection (permutation)
σ
:{
1
,...,m
}→{
1
,...,k
}
such that
TA
j
=
TB
σ(j)
,for
j
=
1
,...,m
. Let us consider the most general case when
TA
j
=
TA
i
for all
(i.e., all
TA
j
are distinct). From
f
i
=
j
,
i,j
∈{
1
,...,m
}
TA
, the num-
f
={
ber of ptp arrows in
f
has to be equal to
m
and hence
f
1
,...,f
m
}
with
=
1
≤
i
≤
m
f
i
f
TA
, thus, with a permutation
σ
1
:{
1
,...,m
}→{
1
,...,k
}
such
that
f
i
=
TA
σ
1
(i)
=
TB
σ(σ
1
(i))
,for
i
=
1
,...,m
. Thus, from the fact that all
TA
j
are distinct values, each ptp arrow is (a)
f
i
:
A
σ
1
(i)
→
B
σ(σ
1
(i))
.
g
={
Let us define a morphism
g
:
TA
→
TB
such that
g
jσ(j)
:
TA
j
→
TB
σ(j)
|
1
≤
j
≤
m
}
where each ptp arrow
g
jσ(j)
is the identity arrow
id
TA
j
(from
g
−
1
g
−
1
the fact that
TA
j
=
TB
σ(j)
). From
={
jσ(j)
:
TB
σ(j)
→
TA
j
|
1
≤
j
≤
m
}
,
g
−
1
id
TB
and
g
−
1
we obtain that
g
◦
=
◦
g
=
id
TA
and hence
g
is an isomorphism.
where
is
A
=
1
≤
j
≤
m
is
A
j
:
is
−
B
◦
g
◦
is
A
f
It is easy to show that
=
A
j
→
TA
j
and
is
−
B
=
1
≤
i
≤
m
is
−
1
is
−
1
B
i
:
TB
i
→
B
i
, so that for each 1
≤
j
≤
m
,(b)
f
σ
−
1
(j)
=
B
σ(j)
◦
σ
−
1
1
g
jσ(j)
◦
is
A
j
:
A
j
→
B
σ(j)
(it is equal to (a) for
i
=
(j)
, that is,
j
=
σ
1
(i)
).
is
−
B
◦
is
A
, that is,
f
is a
composition of three isomorphisms, and hence it is an isomorphism
f
Consequently, from point 3 of Definition
23
,
f
=
g
◦
:
A
B
as
well.
Let us show that this is valid also when
A
and
B
are not strictly complex. Let
A
and
B
be the strictly complex objects obtained from
A
and
B
by elimination of
all
0
. Then, from point 4 of Lemma
9
and commutativity (point 2 of Lemma
9
),
⊥
A
is an isomorphism with
TA
TA
and
f
2
:
B
f
1
:
A
B
is an isomorphism
TB
,so
that
f
:
A
→
B
for
f
TA
TB
is an isomorphism and
f
=
f
2
◦
f
◦
f
1
is a
composition of the isomorphisms and hence an isomorphism as well.
TB
. Thus, we obtain
TA
TB
, i.e.,
TA
with
TB
TA
TB
Notice that in this proposition we provided the
sufficient
conditions for monic,
epic and isomorphic arrows in
DB
and not the necessary conditions: we have seen
other cases of monic and epic arrows in the case of complex arrows, for example,
[
id
A
, id
A
]:
→
A
, which
are not
isomorphic. We can have the cases when
A
and
B
are isomorphic objects also when
TA
=
TB
(as in the case when
C
is iso-
morphic to the separation-composed object
C
A
A
0
⊥
in Corollary
9
. In what follows,
