4.3. Extensive Topologies 

Suppose A is a coherent analytic category. 

Recall that (1.7.4.c) a mono is locally direct if it is an intersection of direct monos. 

Proposition 4.3.1. Any composite of locally direct mono is a locally direct mono. 

Proof. Suppose f: Y --> X is an intersection of direct monos {fi: Yi --> X |iI}, and we may assume that fi is a cofiltered system. Assume Y = U + V where U and V are direct subobjects of Y. Then the unique maps U --> 1 and V --> 1 induce a unique map t: Y --> 1 + 1. Since 1 + 1 is finitely copresentable, t factors through Y --> Yr for some r in I in a map s: Yr --> 1 + 1. The pullbacks of the injections 1 --> 1 + 1 along s induces a direct sum Yr = Ur + Vr, and U is the pullback of Ur along Y --> Yr. Since Yr is a direct factor of X and Ur is a direct factor of Yr, Ur is also a direct factor of X. This shows that any direct factor of Y is induced from a direct factor of X. Consequently any locally direct factor of Y is also a locally direct factor of X.  

Definition 4.3.2. (a) A map f: Y --> X is called indirect if it does not factors through any proper direct mono. 
(b) A non-initial object is called indecomposable if it has exactly two direct subobjects. 
(c) An indecomposable component of an object X is a locally direct indecomposable subobject of X

Proposition 4.3.3. (a) Any map can be factored uniquely as an indirect map followed by a locally direct mono. 
(b) The class of indirect monos is closed under composition. 
(c} If P --> X is an indirect map and P is indecomposable, then X is indecomposable. 
(d} Any non-initial object has an indecomposable component. 
(e} An indecomposable subobject is a indecomposable component iff it is a maximal indecomposable subobject. 

Proof. (a) Consider a map f: Y --> X. Let u: U --> X be the intersection of all the direct monos to X such that f factors through. Then u is a locally direct mono, and the induced map g: Y --> U is indirect by (4.3.1). The uniqueness is obvious. 
(b) The proof is similar to that of (3.5.4.a). 
(c) Suppose U + V = X is a direct sum and U is proper. Then f-1(U) is a proper direct of P, thus f-1(U) = 0, i.e. f is disjoint with U, thus f factors through V. Since f is indirect, we must have V = X, so U = 0 as desired. 
(d) Clearly any simple object is indecomposable. Any non-initial object has a simple subobject by (4.2.2.d), whose indirect image in X is indecomposable and locally direct by (c). 
(e) is an easy consequence of (b).  

Proposition 4.3.4. The extensive topology on A is spatial and strict

Proof. A non-initial object is indecomposable iff it has exactly two direct subobjects. Thus a non-initial object is indecomposable iff it determines a one-point-space in the extensive topology. Clearly any simple object is indecomposable. Thus the direct topology is spatial by (2.6.5.a) and (4.2.2.d). By (4.2.2.e) any direct cover of an object X has a finite subcover. To see that the extensive topology is strict it suffices to consider a finite direct cover {Ui}: i = 1, ..., n of X. Suppose Vi is the complement of each Ui. Let W1 = U1, W2 = V1  U2 , ..., Wi = V1  V2 ...  Vi-1  Ui. Then {Wi} is a direct cover of X, with Wi  Wj = 0 for i < j, and X =  Wi. Let z: Z --> X be the sum of ui: Ui --> X, and let s: X =  Wi --> Z be the map induced by the inclusion Wi --> Ui Then z°s is the identity of X. Thus z is a retraction, hence a regular epi. This shows that {Ui} is a strict direct cover. Thus the direct topology is strict.  

If X is any object we denote by B(X) the poset of direct subobjects of X. The following Proposition 4.3.5 holds for any extensive category: 
Proposition 4.3.5. B(X) is a Boolean algebra.  

Proof. (a) First we show that B(X) is a lattice. Suppose U and V are two direct subobjects of an object X. Since finite sums are stable, we have 

X = (U + Uc)  (V + Vc) = ( V) + (Uc  V) + ( Vc) + (Uc  Vc). 
( V) + (Uc  V) + ( Vc)
are direct objects and 
( V) + (Uc  V) + ( Vc) = (Uc  Vc)c.
( V) + (Uc  V) + ( Vc
is contained in V. Since ( V) + ( Vc) = U  and  ( V) + (Uc  V) = V, it also contains U and V. Thus 
( V) + (Uc  V) + ( Vc) = V.

Thus B(X) is a lattice with 

V = U  V,
( V) + (Uc  V) + ( Vc) = V. 
(b) Next we show that B(X) is distributive. If W is another direct subobject of X, then 
X = W  Uc + W  U + Wc 
implies that 
(W  U)c = W  Uc + Wc
(W  U)c  W = [(W  Uc) + Wc W = W  Uc W + Wc  W = W  Uc.
(W  V)c  W = W  Vc.
We have 
    W  ( V
= W  [( V) + (Uc  V) + ( Vc)]  
= W  V + W  Uc  V + W  Vc  
= W  V + ( Uc)  V + W  ( Vc).  
= ( U)  ( V) + (( U)c  W)  V + U  ( ( V)c)  
= ( U)  ( V) + ( U)c  ( V) + ( U)  ( V)c  
= ( U)  ( V). 
This shows that B(X) is a distributive lattice. Clearly Uc is the complement U of U in the lattice B(X). Thus B(X) is a Boolean algebra.  

Remark 4.3.6. Consider the canonical functor J: FiniteSet --> A which preservs finite limits and sums. For each object X in A the pullback of the finite-limit-preserving functor homA (X, ~) along J is a finite-limit-preserving functor FiniteSet --> Set, thus determines a Boolean algebra, which is precsely B(X). This argument holds for any extensive category with a terminal object 1. 

Recall that (2.6.6) the extensive topology on A is the framed topology E generated by the divisor E of direct monos

Proposition 4.3.7. (a) A finite set {Ui} of direct subobjects of an object X is a unipotent cover iff the join of {Ui} is X
(b) E(X) is isomorphic to the frame of ideals of the Boolean algebra B(X). 
(c) There is a bijection between the set of indecomposable components and the set of prime ideals (or ultrafilters) of B(X). 

Proof. (a) Suppose {Ui} is a finite unipotent cover of X. Let V be the join of {Ui}. Then Vc is disjoint with each Ui, so Vc = 0 and V = X. Conversely, suppose the join of a finite set {Ui} of direct subobjects is X. Suppose t: T --> X is disjoint with each Ui. Then t factors through each (Ui)c. Thus t factors through  (Ui)c = ( Ui)c = Xc = 0. Thus t is initial, i.e. {Ui} is a unipotent cover over X
(b) It follows easily from (a) that if U is an E-sieve then the collection of direct monos in U is an ideal of B(X), and the resulting map E(X) --> B(X) is an isomorphism. 
(c) If P is an indecomposable component of X then the set V of direct subobjects containing P is a prime filter (ultrafilter) of B(X). For if U is a direct subobject not containing P, then Uc contains P as it is indecomposable. Conversely, if V is a ultrafilter of B(X) then the intersection P of direct subobjects in V is non-initial as 0 is finitely copresentable, and P is locally direct. If U is a proper direct subobject of P then U is also a locally direct subobject of X, thus there is a proper direct subobject W of X such that P is a proper direct subobject contains U. Since V is an ultrafilter, W is not in V implies that Wc is in V. It follows that P is contained in Wc. Thus W = 0, which implies that U = 0. This shows that P is an indecomposable component of X. Suppose V and V' are two different ultrafilters of B(X) and P and P' are the corresponding indecomposable components. If T is a direct subobject in V not in V' then Tc is in V'. Thus P is contained in T and P' is contained in Tc, which implies that P and P' are disjoint. This shows that E(X) is isomorphism to B(X).  

Corollary 4.3.8. (a) The space pt(E(X)) of E(X) is homeomorphic to the space of prime ideals of the Boolean algebra B(X), thus is a Stone space (see [Johnstone 1982, p.62-75]). 
(b) The extensive topology on a coherent analytic category is a strict metric site, which may be interpreted as the functor sending each object X to the Stone space of the indecomposable components of X.  

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