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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 .
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