Database Reference
In-Depth Information
Let us show that the pullback diagram (PB) commutes for this arrow
k
, i.e.,
p
B
◦
=
k
p
B
◦
k
h
(and
p
A
◦
k
=
m
). First of all, note that each ptp arrow
(s
xi
:
E
x
→
B
i
)
∈
k
is the union of the following set of compositions of the ptp arrows in
S
2
⊂
and
,
s
xi
=
{
p
B
p
jli
◦
k
xj i
:
k
xj i
∈
S
2
,p
jli
∈
p
B
in
E
x
→
B
i
|
and 1
≤
j
≤
M
}
.From
Lemma
7
, the information flux of this ptp arrow is
T
s
xi
p
jli
◦
k
xj i
|
p
jli
◦
k
xj i
∈
s
xi
=
T
s
xi
p
jli
∩
k
xj i
|
p
jli
◦
k
xj i
∈
=
T
s
xi
p
jli
∩
h
xi
|
p
jli
◦
k
xj i
∈
=
=
T
s
xi
h
xi
|
p
jli
◦
k
xj i
∈
T(h
xi
)
=
h
xi
,
=
h
xi
. Consequently,
p
B
◦
k
=
h
and hence
p
B
h
. Analogously,
p
A
that is,
s
x
i
=
◦
k
=
◦
k
=
m
, so that
k
is this unique arrow which satisfies the pullback's requirements.
Notice that in this definition of the pullbacks we considered only nonempty ar-
rows
f
and
g
. Notice that when
f
g
=
=∅
are empty morphisms, then the pullback
diagram corresponds to the standard product diagram. In this case,
D
ilj
=
TA
j
∩⊥
0
∪
TA
j
\⊥
0
=
TA
j
A
j
,
D
jli
=
TB
i
∩⊥
0
∪
TB
i
\⊥
0
=
TB
i
B
i
.
B
with the projections
p
A
1
, and
p
B
1
, id
B
]
Thus,
LimP
=
A
×
=[
id
A
,
⊥
]
=[⊥
.
