3.5. Analytic Geometry
Let
U
V) by (1.5.4),
so {^{c} = X^{c} = 0U, V} is a unipotent cover on X.
Next we consider the general case. Suppose t: T --> X
is a map that is disjoint with U and V. We prove that T
is an initial map by contradiction. So we assume that T is non-initial.
Replacing T by its reduced model red(T) if necessary
we may assume that T is reduced. Since U is an intersection
of disjunctable strong subobjects, there is a disjunctable strong subobject
U containing _{1}U such that t does not factor
through U. Then _{1}t(^{-1}U)
is a proper strong subobject of _{1}T. As T is reduced, t(^{-1}U)
is not unipotent, thus there is a non-initial map _{1}s: W --> T
which is disjoint with t(^{-1}U). The
non-initial map _{1}ts is disjoint with U. Since
_{1}t is disjoint with V, so is ts. Applying the same
procedure to ts and V we can find a disjunctable strong subobject
V containing _{1}V and a non-initial map r:
R --> W such that tsr is disjoint with V. We obtain a
non-initial map tsr which is disjoint with U
and _{1}V, but this is absurd, as _{1}U
implies that {_{1}
V_{1} = U V = XU} is a unipotent cover
on _{1}, V_{1}X, by the proceeding case.
U)
and f(^{-1}V).
U), f(^{-1}V)}
is a unipotent cover on Y, and f(^{-1}U)
f(^{-1}V) is a unipotent strong subobject of Y.
Since Y is reduced, we have f(^{-1}U)
f(^{-1}V) = Y as desired.
S(X) --> S(Y) sending each reduced strong subobject
U in X to the reduced model of f(^{-1}U).
It has a left adjoint which coincides with the restriction of fon ^{+1
}S(Y), denoted by red(f)
.
^{+1}
S(X) is a locale isomorphic
to the locale (X)
of analytic sieves.
(b) The functor sending each object RedX to S(X)
and each map ^{op}f: Y --> X to red(f)
is equivalent to the analytic topology ^{-1}
on _{A}.
A
X).
(b) Consider any map f: Y --> X. If u: U -->
X is a reduced strong mono we have f(^{-1}u)
= f(^{-1}u)
= r(f(^{-1}u))
= (r(f)(^{-1}u)).
This shows that the isomorphism in (a) is functorial, thus the functor
is equivalent to the framed topology Red.
_{A}
U, where each _{i}
--> UU is an analytic subobject of
_{i}X. Suppose t: T --> X is an epic map. Let s:
S --> U be the pullback of t along u, and let s:_{i}
S be the pullback of _{i} --> U_{i}t along u.
Since each _{i}u is analytic and _{i}t is epic, each
s is epic. Suppose _{i}s factors through a strong
subobject V of U. Then each s factors
through the strong subobject _{i}V
U of _{i}U. But _{i}s is
epic implies that _{i}V U. Thus _{i}
= U_{i}V contains the analytic cover U.
Since by assumption _{i}A is strict, U is
the colimit of analytic subobjects {U}, so {_{i}
U_{j}U} is not contained in any proper
subobject of _{i}U. Thus we have U = V. Therefore t is
epic by (1.1.3.e). This shows that u
is precoflat.
U(_{i} = V)_{i}.
According to (3.1.10) we have {^{c}U}_{i}
= V, so {U}
together with _{i}V form a unipotent cover on X. Since U
V = X and each U is coflat, we have
_{i}U(_{i} = U_{i}
X = U_{i} U
V) = (U)_{i}
U (U)_{i}
V = U _{i}
U.U contains each
U. Since _{i}U is disjoint with V, this implies
that {U} is a unipotent cover on _{i}U. Now consider
any map t: T -->. X which is disjoint with V. Then
t {U}.
Thus {_{i}t(^{-1}U)} is an analytic cover
on _{i}T contained in t(^{-1}U). Since the analytic
geometry is strict, there is no proper subobject of T containing
each t(^{-1}U). This implies that _{i}T
= t(^{-1}U). Thus t factors through U.
It follows that U = V. Similarly ^{c}V = U.
We have proved that ^{c}U and V are analytic subobject. Thus
(U, V) is an analytic cover on X. As the analytic geometry
is strict, X is the smallest subobject containing U and V.
But by (1.3.9) U + V --> X is a
mono which is the smallest subobject containing U and V.
This shows that U + V --> X is an isomorphism.
X in a
strict analytic geometry,
such that U are disjoint for all _{i}, U_{j}i
j, then the induced map
U
(where _{i} --> U_{i}U
is the sum of {_{i}U}) is an isomorphism.
_{i}
Let Q
be ideals of _{n}R such that Q + _{i}Q
= _{j}R for all i
j. Then R/()
is isomrophic to _{i}Q_{i}/_{i}RQ.
_{i} |