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=
1
≤
j
≤
M
A
j
,
B
Proof
Let us consider a pullback diagram (PB) above with
A
=
1
≤
i
≤
K
B
i
,
C
=
1
≤
l
≤
N
C
l
and
E
=
1
≤
x
≤
N
1
E
x
for
M,K,N,N
1
≥
1. Then a
limit cone of this pullback is composed by limit object (see the ptp arrows in
S
A
and
S
B
in the diagram bellow)
LimP
=
(S
E
∪
S
A
∪
S
B
)
where
S
E
={
D
ilj
=
f
g
D
jli
=
(f
jl
∩
(f
jl
∪
g
il
)
∪
(T A
j
\
g
il
)
|
f
jl
∈
,g
il
∈
,i
=
j
and
TB
i
=
TA
j
}
,
(T A
j
\
f
jl
)
f
g
S
A
={
D
ilj
=
(T A
j
∩
g
il
)
∪
|
f
jl
∈
,g
il
∈
,TB
i
=
TA
j
}
and
S
B
=
f
g
D
jli
=
(T B
i
∩
f
jl
)
{
∪
(T B
i
\
g
il
)
|
f
jl
∈
,g
il
∈
,TB
i
=
TA
j
}
, and with outgoing
arrows
p
A
=
p
ilj
:
S
A
and
D
ilj
→
A
j
|
D
ilj
∈
S
E
∪
p
B
=
p
jli
:
S
B
.
D
jli
D
jli
∈
→
B
i
|
S
E
∪
From the fact that
f
jl
◦
p
A
f
◦
f
g
p
ij l
:
=
D
ilj
→
C
l
|
f
jl
∈
,g
il
∈
and
g
il
◦
,
p
B
g
◦
f
g
p
jli
|
=
f
jl
∈
,g
il
∈
f
◦
p
A
→
g
◦
p
B
f
p
ij l
∈
we obtain the bijection
σ
:
, such that for each
f
ji
◦
,
σ(f
ji
◦
p
ij l
)
=
g
il
◦
p
jli
with
σ
f
ji
◦
p
ij l
=
g
il
∩
p
jli
=
g
il
∩
f
jl
=
p
ij l
∩
f
jl
=
f
ji
◦
p
ij l
p
A
=
p
B
f
◦
g
◦
and, consequently we have the same set of ptp arrows
, i.e., the com-
p
B
.
Suppose that for another two arrows
m
:
E
→
A
and
h
:
E
→
B
the diagram
(PB) commutes, i.e.,
f
p
A
mutativity of the square diagram (PB), that is,
f
◦
=
g
◦
h
.For
m
h
it is obvious, so we will consider
nontrivial cases when both arrows have a nonempty set of ptp arrows. Let us define
the sets (see the commutative diagram bellow composed of the ptp arrows of this
pullback diagram),
S
1
={
k
xj i
:
E
x
→
D
ilj
|
m
xj
∈
◦
m
=
g
◦
=
=∅
m
,p
ilj
∈
p
A
with
k
xj i
=
m
xj
⊆
p
ilj
}
p
B
with
k
xj i
=
h
xi
⊆
p
jli
}
k
xj i
:
D
jli
|
h
,p
jli
∈
, and
S
2
={
E
x
→
h
xi
∈
. Then
we define the arrow
k
:
E
→
LimP
as the canonical representation (from Lemma
8
)
composed by the set of ptp arrows
k
S
1
∪
S
2
.