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.
Proof. (a) Suppose X = SiXi
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 (SiUi
® X, SiVi
® X) is a sum. Let r be
the map SiVi ®
V and s the map SiUi
® U. Then v°r
= Sivi and
u°s = Siui.
Thus the pullback of Sivi
along v is r and the pullback of Siui
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.
Proposition 3.8.3. Suppose A
is a complete and cocomplete, well-powered and co-well-powered analytic
category. Then
Proof. (a) Suppose {Vi} is a family of von
Neumann regular subobjects of an object X. Let W = SiVi
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).
Proposition 3.8.4. Suppose A
is a locally disjunctable analytic category.
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.
Proposition 3.8.5. In a locally disjunctable
analytic category the followings are equivalent for a von Neumann regular
object X .
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 ¹ 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 Ú 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). ––n |