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responds to a terminal object of such a category. A category is complete if every
diagram in it has a limit. Dually, it is co-complete when every diagram has a col-
imit. A finite diagram (a functor in DB J in our case) is one that has a finite number
of objects, and a finite number of arrows between them. A category is finitely com-
plete (Cartesian) if it has a limit for every finite diagram. It is well known that any
category that has a terminal object and the pullbacks for every pair of morphisms,
than it has all finite limits. Let us show that DB is not finitely complete if we reduce
only to simple arrows and objects. For example, for the pullbacks. Let us suppose
that the following diagram, where A,B and C are simple databases and LimP is a
limit object of the diagram A
f
g
C
B , is a pullback (PB):
=
In order to have for every pair of arrows (m,h) , the commutativity f
m
g
h ,
i.e., f
h it must be true that
\ f)) and h
m
=
g
m
T ((T A
g)
(T A
f)
p A
p B
T ((T B
(T B
\
g)) , and from the commutativity m
=
k and h
=
k ,
p A and h
p B , it must be true that p A
\ f)) and
m
=
T ((T A
g)
(T A
p B
f)
g)) . In fact, from f
=
T ((T B
(T B
\
TA we have that
(T A
\ f) f
= f
g)
(T A
g
and hence for closed object f
T(f
\ f))
f)
g
=
g)
=
T (((T A
g)
(T A
=
= p A
f . Similarly p B
g and hence p A
\ f))
f
= f
T ((T A
g)
(T A
g
= p B
f
= h then there is no simple
p A
p B . However, if
g , i.e., f
=
g
m
p A ) and
LimP because we need k
= h (from h
k and h
p A
arrow k
:
D
=
similarly k = m and hence
m = h, which is a contradiction (the case when
m = h
with A
B is the case when it is a pullback with all simple objects, corresponding
to equalizers in Lemma 16 ). Consequently, LimP
=
\ f))
=
((T A
g)
(T A
×
((T B
f)
(T B
\
g)) and all arrows to and from it have to be complex , i.e., k
=
m,h
and
p B
and p A substituted by
1 ,p B
p A ,
1
[⊥
]
and
[
]
, respectively. Consequently, we
obtain:
Proposition 48
DB category is finitely (co)complete, i . e ., ( co ) Cartesian .
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