Let C be a coherent analytic geometry. 

Proposition 5.6.1. Any prime of an object is contained in a maximal prime. 

Proof.  Suppose P is a prime of an object X. Let S be the set of prime subobject of X containing P. It is non-empty. Let C be an ascending chain in S. For each prime Q in C let X_Q be the localization of X at Q. Then X_Q is a local pbject with the a minimal prime above Q. We obtain a decending equence of non-initial fractions of X. The intersection U of these fractions is non-initial as the initial object is finitely presented. Let O be any prime of U. Then the strong image of O in each X_Q is a prime containg its minimal prime. Thus the strong image of O in X contains each Q. By Zorn's lemma, S has a maximal element, which is a maximal prime object containing P. 
 
Propisition 5.6.2. Suppose f: Y --> X is an epi. Then any maximal prime of X is in the image Spec(f)(Spec(Y)) of Spec(f). 
 
Proof. Suppose P is a maximal prime of X. Let i: X_P --> X be the localization of X at P. Then X_P is an object with a unique prime object (i.e. quasi-local simple object) above P. Let i': T --> Y and f': T --> Xp be the pullback of f and i. Since f is epic and i is a fraction (thus coflat), the pullback f' is epic, thus non-initial. It follows that i' is non-initial. Let Q be any prime of T. Then f⁺¹(i⁺¹(Q)) = (fi')⁺¹(Q) = (if')⁺¹(Q) = P. Thus P is in the image of Spec(f). 

A subset U of Spec(X) is called stable under specialization (resp. generalization) if any prime contained in (resp. containing) a prime in U is in U. 

Proposition 5.6.3. Let f: Y --> X be a map. Then Spec(f)(Spec(Y)) is closed iff it is stable under specialization. 

Proof. One direction is clear. Suppose Spec(f)(Spec(Y)) is stable under specialization. Replacing X by the strong image f⁺(Y) we may assume that f is an epi. We have to show that any prime of X is in Spec(f)(Spec(Y)). By (5.6.1) P is contained in a maximal prime Q, which is in the image by (5.6.2), thus P is in the image as f is stable under specialization. 

Proposition 5.6.4. Suppose X is a von Neumann regular object. Then a subset U of Spec(X) is closed iff U = Spec(f)(Spec(Y)) for a map f: Y --> X. 

Proof. If U is a closed then by the definition of the topology on Spec(X) it is the image of Spec(f) for a regular mono f to X. Conversely, for any map f, the image Spec(f)(Spec(Y)) is stable under specialization as any prime in a von Neumann regular object is simple, thus maximal. So Spec(f)(Spec(Y)) is closed by (5.6.3). 

Example 5.6.4.1. If A is simply atomic Stone geometry the a map is unipotent iff it induces a surjective function on the spectrums. Thus a map is epiciff it is unipotent by (5.6.4) and the fact that A is reduced. It follows that any epi is universal and any map is coflat.

Definition 5.6.5. (a) Suppose f: Y --> X is a map. A map g: Z --> X is called a pre-complement of f if g is disjoint with f, and any non-initial map to X that is disjoint with f is not disjoint with g. 
(b) A map f: Y --> X is called pre-complementary if it has a pre-complement.

Note that a map f is a pre-complement of a map g iff g is a precomplement of f.

Proposition 5.6.6. If f: Y --> X is a coflat quasi-complemenetary map then Spec(f)(Spec(Y)) is an open subset. 

Proof. Suppose g: Z --> X is a map which is disjoint with f and {f, g} is a unipotent cover. Then the assumption implies that Spec(f)(Spec(Y)) = Spec(X) - Spec(g)(Y). Consider a prime P of Spec(f)(Spec(Y)). Let XP be the locaization of X at P. Applying (5.2.13) (Going Up Theorem) the image of Spec(XP) in Spec(X) is in Spec(f)(Y). Thus XP -- X is disjoint with  g. The fraction XP is an intersection of a collection {Ui} of analytic subobjects of X containing the residue of P. Let Vi be the pullback of Ui along g. Then the intersection of Vi is the pullback of Xp along g, which is initial. Since the initial object is finite presentable, Vi is initial for some i. Thus Ui is dijoint with g, which implies that Spec(Ui) is contained in Spec(f)(Spec(Y)). Since Spec(Ui) contains P, this shows that Spec(f)(Spec(Y)) is an open subset of Spec(X). 

Proposition 5.6.7. Suppose any finitely presentable map in A is quasi-complementary. If f: Y --> X is a coflat finitely presentable map then Spec(f) is an open map. 

Proof. Consider an open subset U of Spec(Y). Any point p of U is contained in an analyitc subset V of U. The analyitc mono v: V --> Y is coflat and finitely presentable, so the composite  fv is coflat finitely presentable, thus is also quasi-complementary by assumption. By (5.6.6)  fv(V) is an open subset of X, thus is an open neighborhood of f(p) contained in f(U). This shows that f(U) is an open subset, so f is an open map.