Definition 3.7.1. An analytic category is coflat if any map is coflat (or equivalently, any epi is stable). Proposition 3.7.2. Suppose A
is a coflat analytic category. Then
Proof. (a) is true because any stable epi is unipotent.
In the following we assume A is a coflat disjunctable analytic category. Proposition 3.7.3. (a) Any normal mono
is analytic.
Proof. (a) Suppose u: U --> X is a normal mono.
It is strong by (3.7.2.c). Since any strong mono is
disjunctable, the complement uc of u exist, which
is normal, thus also strong. Since u is normal, u is the
complement of uc (i.e. u = (uc)c
), which implies that u is analytic.
Recall that the class of analytic (resp. normal) monos is a subnormal divisor A (resp. N) on A (see (2.6.6)). Recall that is the boolean functor from A to the (meta)category of complete boolean algebras, sending each object X to the set (X) of normal sieves on object X (see (2.1.4)). Proposition 3.7.4. N = A and is equivalent to the subnormal framed topology determined by N. Proof. We already know that N = A by (3.7.3). To prove the second assertion, consider a normal sieve U on an object X. Consider any map t: T --> X in U with the strong image m: e(T) --> X. Since any map is coflat, by (1.5.2) we have t = m. Thus t = m. Since U is a normal sieve, it contains t, thus it also contains the normal mono m. This indicates that U is generated by the normal mono. Thus U is a N-sieve for the normal divisor N. Since any N-sieve is normal, we see that coincides with the set of N-sieves. Corollary 3.7.5. The boolean functor is a framed topology which coincides with its generic and analytic topologies. 末 Corollary 3.7.6. Suppose A
is reduced. The following
notions are the same:
In the following we assume X is an object such that R(X) is complete. Proposition 3.7.7. Any normal sieve on X is generated by a normal mono. Proof. Consider a normal sieve U. Let u: U --> X be the intersection of all the normal monos v to X such that the sieve sie(v) generated by v contains U. Then u is normal. and sie(u) contains U. We prove that U = sie(u). Suppose t: T --> X is a map in sie(u) which is disjoint with U. Its strong image m: S --> X satisfies the similar properties (as u is a strong mono). Thus m U which implies that U = U m = sie(mc). It follows that u factors through mc. Since m factors through u, we see that m factors through mc. Thus m is a initial map, and t is also initial. This shows that U dominates sie(u). But U is normal, thus U = sie(u). 末 If D is a divisor on A we denote by D(X) the set of D-subobjects of X. Corollary 3.7.8. (X) = N(X) = A(X) = A(X). 末 |