3.8. Regular Objects 

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 XiXi 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 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).  

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