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 Í u+1(U), clearly uc is a complement of the strong mono u+1(U) ® X, thus it is analytic. n  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 Ç t = 0 and g Ç s = 0. Since the composite f°g of coflat maps is coflat by (1.4.2), it suffices to prove that f°g = (t Ú f+1(s))c. We have  (f°g) Ç (t Ú f+1(s)) = (f°g)-1(t Ú f+1(s)) = (f°g)-1(t) Ú (f°g)-1(f+1(s))  = g-1(f-1(t)) Ú g-1(f-1(f+1(s))) = g-1(f-1(t)) Ú g-1(s) = (g Ç (f Ç t)) Ú (g Ç s) = 0 by (1.5.3) and (1.5.4). Next consider a map r: W ® X such that r × X (t Ú 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))) = (f°k)-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 = f°g°v. This shows that f°g = (t Ú f+1(s))c, i.e. f°g is analytic. n    Proposition 1.6.4. If u: U ® X and v: V ® X are two disjunctable strong subobjects of X. Then uc Ç vc = (u Ú v)c.  Proof. Let w = u Ú 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 Ç Vc ® Uc and t: Uc Ç Vc ® Vc be the pullback of uc and vc. Then s and t are analytic monos by (1.6.2), and r = uc°s = vc°t = uc Ç vc: Uc Ç Vc ® X is an analytic mono by (1.6.3). We prove that r = wc. We have  r Ç w = r-1(W) = r-1(u Ú v) = r-1(u) Ú r-1(v) = s-1((uc)-1(u)) Ú t-1((vc)-1(v)) = s-1(uc Ç u) Ú t-1(vc Ç v) = s-1(0) Ú 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. n    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) Ç (u + v) = uc Ç u + vc Ç 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. n    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. n    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]
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