1.5. Beck-Chevalley Condition For any object 0 --> X as zero and 1:_{X}
X --> X as one. An object Z has exactly one strong subobject
(i.e. 0) iff it is initial.
_{Z} = 1_{Z}If u)
the pullback of u along f. Then f:^{-1}
R(X) --> R(Y) is a mapping preserves meets such
that f(^{-1}0)_{X} = 0
and _{Y}f(^{-1}1)_{X} = 1
(i.e. _{Y}f is ^{-1}bounded ). Also f
has a left adjoint ^{-1}f:^{+1} R(Y) --> R(X)
sending each strong subobject v: V --> Y to the strong image
of the composite f.v: V --> X . If V = Y then
f(^{+1}Y) is simply the strong image of f.
f: Y --> X are two maps and f
is coflat. Let y: Y be the pullback of
_{1} --> Yx along f. Then f(^{-1}x(^{+1}X))_{1}
= y(^{+1}Y)_{1}.
X)
and _{1}m: x(^{+1}X)_{1} --> X
be the epi-strong-mono factorization of x. Form the pullback
y = m and _{1}e_{1}T = f(^{-1}x(^{+1}X)).
The map _{1}m as the pullback of the strong mono _{1}m
is a strong mono. The map t as the pullback of the coflat map f
is coflat. Thus the pullback e of the epi _{1}e along
t is also epi. Therefore (e) is
the epi-strong-mono factorization of _{1}, m_{1}t. Thus
y(^{+1}Y)_{1} = T = f(^{-1}x(^{+1}X))_{1}.
f: Y --> X
, then the strong image of x is disjoint with f.
x and f is 0
. The map e:_{1} 0 = Y is an isomorphism
as it is an epic strong mono. It follows that _{1} --> Tf and m are
disjoint.
Proposition 1.5.3. If f:
Y --> X is a coflat map, then
f:^{-1} R(X) --> R(Y) is a morphism
of bounded lattice.
u:
U --> X and v: V --> X are two strong subobjects. Let
x: U + V --> X be the map induced by u and v.
Then x(^{+1}U + V) is the join of U and V
. Since sums are stable, the pullback Y of _{1}U + V
along f in the following pullback diagram is f(^{-1}U)
+ f(^{-1}V).
e:_{1} Y is epic, _{1} --> TT =
f(^{-1}U)
f(^{-1}V). Applying (1.5.1) we obtain T =
f(^{-1}x(^{+1}U + V)) = f(^{-1}U
V). This shows that f(^{-1}U
V) = f(^{-1}U)
f(^{-1}V).
Proposition 1.5.4. If f:
Y --> X is a coflat mono, then f
is the identity ^{-1}f^{+1}R(Y) --> R(Y).
U = u(^{+}U) = f((^{-1}fu)(^{+1}U))
= f(^{-1}f(^{+1}U)). Proposition 1.5.4 is a special case of the following
V = n(^{-1}U) p(^{+1}V) = (pv)(^{+1}V)g(^{+1}U) = (gu)(^{+1}U).x = gu:
U --> X one has
pv)(^{+1}V) = f((^{-1}gu)(^{+1}U)).p(^{+1}n(^{-1}U)) = p(^{+1}V)
= (pv)(^{+1}V) = f((^{-1}gu)(^{+1}U))
= f(^{-1}g(^{+1}U)). |