1.6. Analytic Monos  A mono uc: Uc --> X is a complement of a mono u: U --> X if u and uc are disjoint, and any map v: T --> X such that u and v are disjoint factors through uc (uniquely). The complement uc of a mono  u, if exists, is uniquely determined up to isomorphism.  Definition 1.6.1. (a) A mono f: U --> X is singular if it is the complement of a strong mono to X.  (b) A coflat singular mono is an analytic mono.  (c) A subobject u: U --> X of an object X is analytic if u is an analytic mono.  (d) A strong mono is disjuctable if it has a coflat complement.  Proposition 1.6.2. (a) The pullback of any analytic mono is analytic.  (b) The pullback of any disjunctable strong mono is disjunctable.  (c) Any coflat complement of a mono is analytic.  Proof. Suppose u: U --> X is a disjunctable strong mono with the coflat complement uc: Uc --> X. Suppose f: Y --> X is any map. Then the pullback f-1(u) of u along f is a strong mono. It is easy to verify that f-1(uc) = (f-1(u))c. Since the pullback f-1(uc) of coflat map is coflat by (1.4.2), it is analytic; so f-1(u) is disjunctable. This proves (a) and (b).  To see (c), suppose u: U --> X is a mono with the coflat complement uc: Uc --> X. By (1.5.2) uc: Uc --> X is also disjoint with u+1(U). Since U u+1(U), clearly uc is a complement of the strong mono u+1(U) --> X, thus it is analytic.  Proposition 1.6.3. Composite of analytic monos is analytic.  Proof. Let f: Y --> X and g: Z --> Y be two analytic monos. Let t: T --> X and s: S --> Y be two strong monos such that f = tc and g = sc. Then f t = 0 and g s = 0. Since the composite fg of coflat maps is coflat by (1.4.2), it suffices to prove that fg = (t f+1(s))c. We have  (fg) (t f+1(s)) = (fg)-1(t f+1(s)) = (fg)-1(t) (fg)-1(f+1(s))  = g-1(f-1(t)) g-1(f-1(f+1(s))) = g-1(f-1(t)) g-1(s) = (g (f t)) (g s) = 0 by (1.5.3) and (1.5.4). Next consider a map r: W --> X such that r X (t f+1(s)) = 0. We have r X t = 0 and r X f+1(s) = 0. Thus r factors through f in a map k: W --> Y. We have  k Y s = k-1(s) = k-1(f-1(f+1(s))) = (fk)-1(f+1(s))) = r-1(f+1(s)) = r X (f+1(s)) = 0. Thus k factors through g in a map v: W --> Z. Hence r = fgv. This shows that fg = (t f+1(s))c, i.e. fg is analytic.    Proposition 1.6.4. If u: U --> X and v: V --> X are two disjunctable strong subobjects of X. Then uc vc = (u v)c.  Proof. Let w = u v: W --> X be the join of u and v. Suppose uc: Uc --> X and vc: Vc --> X are the coflat complements of u and v respectively. Let s: Uc Vc --> Uc and t: Uc Vc --> Vc be the pullback of uc and vc. Then s and t are analytic monos by (1.6.2), and r = ucs = vct = uc vc: Uc Vc --> X is an analytic mono by (1.6.3). We prove that r = wc. We have  r w = r-1(W) = r-1(u v) = r-1(u) r-1(v) = s-1((uc)-1(u)) t-1((vc)-1(v)) = s-1(uc u) t-1(vc v) = s-1(0) t-1(0) = 0. On the other hand, suppose z: Z --> X is a map such that z X w = 0. Then z X u = z X v = 0. Thus z factors through uc and vc, therefore also factors through r. This shows that r = wc.    Proposition 1.6.5. Suppose u: U --> X and v: V --> Y are two disjunctable strong monos with coflat complements uc: Uc --> X and v: Vc --> Y respectively. Then u + v: U + V --> X + Y is a disjunctable strong mono with coflat complement: uc + vc: Uc + Vc --> X + Y.  Proof.  We know that uc + vc is coflat by (1.4.2.d) because uc and vc are coflat. We have (uc + vc) (u + v) = uc u + vc v = 0 + 0 = 0 by (1.3.4.a). Let t: M --> X + Y be a map such that t X+Y (u + v) = 0. Then t = tX + tY with tX and tY being the pullback of t along the injections X --> X + Y and Y --> X + Y respectively. Thus (tX + tY) X+Y (u + v) = 0 implies that tX X u + tY Y v = 0. It follows that tX u = tY v = 0. Thus tX can be factored through uc and tY can be factored through vc. This implies that t can be factored through uc + vc. Hence uc + vc is the complement of u + v.    Proposition 1.6.6. (a) Isomorphisms are analytic monos.  (b) Finite intersections of analytic monos are analytic monos.  (c) Finite sums of analytic monos are analytic monos.  (d) Injections of a sum are analytic monos.  Proof. (a) is obvious and (b) follows from (1.6.1) and (1.6.2).  (c) follows from (1.6.5).  (d) The injections of a sum are coflat strong monos by (1.4.5), and are complements of each other because finite sums are stable disjoint. Therefore they are analytic monos.    Example 1.6.7. Consider the category of affine schemes. Suppose A is a ring. From algebraic geometry we know that a strong mono to Spec(A) is precisely a closed immersion Spec(A/I) --> Spec(A) determined by an ideal I of A. Since by definition a singular mono is a complement of a strong mono, any singular mono of affine schemes must be an open embedding. By a result of Diers (cf. [D, p.42]) any singular mono of affine schemes is always coflat, so analytic monos of affine schemes are precisely open embedding of affine schemes. For instance, if a is any element of A, then Spec(A/(a)) --> Spec(A) is a disjunctable strong mono with the open embedding Spec(Aa) --> Spec(A) as the coflat complement, which is induced by the localization A --> Aa with respect to a.     [Next Section][Content][References][Notations][Home] 