Consider an analytic category A.
Definition 3.8.1. An object is called von Neumann regular if any disjunctable strong mono to it is direct.
Proposition 3.8.2. (a) Any sum of von
Neumann regular objects is von Neumann regular.
(b) Any extremal quotient
of a von Neumann regular object is von Neumann regular.
Proof. (a) Suppose X = 
iXi
is the sum of a family of von Neumann regular objects {Xi}
with the direct maps {si: Xi --> X}.
Suppose v: V --> X is a disjunctable strong subobject of
X with the analytic complement u: U --> X. Let vi:
Vi = (si)-1(V)
--> V and ui: Ui = (si)-1(U)
--> U be the pullbacks of si along U and V
respectively. Then Vi is a disjunctable strong subobject
of Xi with the analytic complement Ui.
Since each Xi is von Neumann regular, Vi
is direct. It follows that (iUi
--> X, iVi
--> X) is a sum. Let r be the map iVi
--> V and s the map iUi
--> U. Then vr = ivi
and us = iui.
Thus the pullback of ivi
along v is r and the pullback of iui
along v is 0. Since finite sum is universal, r + 0
is an isomorphism. Thus r is an isomorphism, and v is a direct
mono.
(b) Suppose Y is a von Neumann regular object and f:
Y --> X is a extremal epi. We prove that X is von Neumann regular.
Suppose v: V --> X is a disjunctable strong mono with the
analytic complement u: U --> X. Let (r: V' = f-1(V)
--> V, v': V' --> Y and (s: U' = f-1(U)
--> U, u': U' --> Y be the pullbacks of f along v
and u respectively. Then f(u' + v') = (u
+ v)(r + s), and V' is a disjunctable strong mono with
the analytic complement U'. Since Y is regular, V'
is direct. Thus u' + v': U' + V' --> Y is an isomorphism.
Thus f = (u + v)(r + s)(u' + v')-1
factors through u + v. Since f is extremal epic, and u
+ v is a mono by (1.3.8.a), it follows
that u + v is an isomorphism. Thus v is direct. ––
Proposition 3.8.3. Suppose A
is a complete and cocomplete, well-powered and co-well-powered analytic
category. Then
(a) The union of any family of von Neumann regular subobjects is von
Neumann regular.
(b)The full subcategory of von Neumann regular objects is a coreflective
subcategory.
Proof. (a) Suppose {Vi} is a family of von
Neumann regular subobjects of an object X. Let W = iVi
and let V the union of the subobjects Vi. Then
W is von Neumann regular by (3.8.2.a). Let
w: W --> V be the map induced by the monos Vi
--> V. Then w is an extremal epi. Therefore V is von
Neumann regular by (3.8.2.b).
(b) For any object X let vnR(X) be union
of all the von Neumann regular subobjects of X. Any map f:
Y --> X from a regular subobject Y to X factors through
its extremal image in X which is a von Neumann regular subobject
contained in vnR(X). Thus any such map factors through
vnR(X). ––
Proposition 3.8.4. Suppose A
is a locally disjunctable analytic category.
(a) Any strong mono to a von Neumann regular object is locally
direct and normal.
(b) Any von Neumann regular object is reduced.
Proof. (a) Any strong mono to an object X in a locally
disjunctable analytic category is the intersection of disjunctable strong
monos. If X is von Neumann regular then any disjunctable strong
mono is direct, thus normal. Since any intersection of normal monos is
normal, any strong mono to X is locally direct and normal.
(b) By (a) any unipotent strong mono to a von Neumann regular object
X is normal, thus is an isomorphism. This implies that X
is reduced. ––
Proposition 3.8.5. In a locally disjunctable
analytic category the followings are equivalent for a von Neumann regular
object X .
(a) X is irreducible.
(b) X is quasi-primary.
(c) X is primary.
(d) X is integral.
(e) X is pseudo-simple.
(f) X is simple.
Proof. Since X is reduced, (a) - (d) are equivalent by
(3.2.6), and (e) and (f) are equivalent
by (3.3.9). Clearly (f) implies (a). Thus
we only need to show that (a) implies (f). Assume X 
0 is irreducible. Since X is von Neumann regular any disjunctable
strong mono v: V --> X is direct and X = V + Vc,
which implies that X = V 
Vc. Since X is irreducible, either V or Vc
is X. Thus V = X or V = 0. Since any strong subobject
of X is an intersection of disjunctable strong subobjects, we see
that X and 0 are the only strong subobjects of X.
Thus X is simple. This shows that (a) implies (f). ––