2.1. Normal Sieves   In this chapter we assume that A is an arbitrary category with a strict initial object 0. By an initial map we mean a map with the initial object 0 as domain. Recall that a set S of maps to an object X is called a sieve on X if S contains any map to X that factors through a map in S.  Suppose S and T are two sets of maps to an object X. We say that S is disjoint with T if any map in S is disjoint with any map in T. Denote by ØS the set of maps to X which is disjoint with S.  Denote by J(S) the sieve on X generated by S. (i.e. the smallest sieve on X containing S). Let 1X (resp. 0X ) be the sieve on X generated by the identity map (resp. the initial map) on X.  Proposition 2.1.1. Suppose S and T are two set of maps to X.  (a) ØS is a non-empty sieve which contains the initial map to X; S Ç ØS consists of only initial map; Ø(S È ØS) = 0X,; ØØ(S È ØS) = 1X; S Í ØØS.  (b) S Í T implies that ØT Í ØS and ØØS Í ØØT.  (c) ØS = ØØØS; T Í ØØS iff ØS Í ØT; T Í ØØS iff ØØT Í ØØS.  (d) A map t is in  ØØS iff any non-initial map to X which factors through t is not disjoint with S; if T is a sieve then T  Í ØØS iff any non-initial map in T is not disjoint with S.  (e) ØS = ØJ(S) and ØØS = ØØJ(S)  (f) (ØØS) Ç (ØØT) = ØØ (J(S) Ç J(T)).  (g) If S and T are sieves then (ØØS) Ç (ØØT) = ØØ (S Ç T).  Proof. (a), (b) (d) and (e) are obvious; (c) follows from (a) and (b).  (f) We have J(S) Ç J(T) Í J(S), so ØØ (J(S) Ç J(T)) Í ØØJ(S) = ØØS by (e). Similarly ØØ (J(S) Ç J(T)) Í ØØJ(T) = ØØT. Thus ØØ (J(S) Ç J(T)) Í (ØØS) Ç (ØØT). We prove the other direction using (d). Assume  t: Y ® X is a non-initial  map in (ØØS) Ç (ØØT). To see that t is in ØØ (J(S) Ç J(T)), it suffices to prove that for any non-initial map s: Z ® Y, s°t is not disjoint with J(S) Ç J(T). Since t Î ØØS, we have s°t Î ØØS, so by (d) s°t is not disjoint with S. Thus there is a non-initial map u: U ® Z such that u°s°t factors through a map in S. Since t Î ØØT, we have u°s°t Î ØØT, so by (d) u°s°t is not disjoint with T. Thus there is a non-initial map v: V ® U such that v°u°s°t factors through a map in T. It follows that v°u°s°t Î J(S) Ç J(T), which implies that s°t is not disjoint with J(S) Ç J(T) as desired.  (g) follows from (f) as J(S) = S and J(T) = T if S and T are sieves. n  A sieve S of maps to X is a normal sieve if S = ØØS. Clearly 1X and 0X are normal sieves.  Proposition 2.1.2.  (a) ØS is a normal sieve for any set S of maps to X .  (b) Any intersection of normal sieves is a normal sieve.  (c) If {Si} is a collection of sets of maps to X then ØØ(Èi Si) is the smallest normal sieve containing each Si.  Proof. (a) follows from (2.1.1.c).  (b) Suppose S is an intersection of a set of normal sieve {Si} on X. Then Ç Si Í Si implies ØØ(Ç Si) Í ØØSi = Si, thus ØØ (Ç Si) Í Ç Si, which implies the equality ØØ (Ç Si) = Ç Si because the other direction is trivial.  (c) If S is a normal sieve which contains each Si, then Èi Si Í S implies ØØ(Èi Si) Í ØØS = S. n  Consider a map f: Y ® X. If S is a set of maps to X we denote by f*(S) the inverse image of S under f, which consists of all the maps z: Z ® Y such that f°z is in S. If S is a sieve on X then f*(S) is a sieve on Y.  Denote by Â(X) the set of normal sieves on an object X.  Proposition 2.1.3. (a) f*(ØS) = Øf*(S) for any sieve S on X.  (b) If S is normal then f*(S) is normal.  (c) The function  f*: Â(X) ® Â(Y) preserves intersections.  Proof. (a) Suppose z: Z ® Y is a map in Øf*(S). If w: W ® Z is a map such that f°z°w Î S, then z°w Î Øf*(S) Ç f*(S), so z°w is an initial map by (2.1.1.a); thus w in an initial map, which implies that f°z Î ØS, i.e. z Î f*(ØS). This shows that Øf*(S) Í f*(ØS). The other direction is trivial.  (b) If S is normal then S = ØØS, by applying (a) twice we obtain f*(S) = f*(ØØS)= ØØf*(S), i.e. f*(S) is normal.  (c) follows from (2.1.2.b) as f* preserves intersection of sieves. n  Proposition 2.1.4. (a) Â(X) is a complete boolean algebra with Ù = Ç.  (b) If f: Y ® X is a map then f*: Â(X) ® Â(Y) is a morphism of complete boolean algebras.  Proof. (a) We already know by (2.1.2.b) that Â(X) is a complete lattice with Ù = Ç. Suppose S is a normal sieve and {Tk} is a set of normal sieves on X. We prove the infinite distributive law   S Ç (Úk Tk) = Úk (S Ç Tk). We only need to verify the relation Í because the other direction is always true. Since Úk(S Ç Tk) = ØØ(Èk S Ç Tk) by (2.1.2.c) and S Ç (Úk Tk) is a sieve, we only need to prove that any map in u: U ® X in S Ç (Úk Tk) is not disjoint with Èk S Ç Tk by (2.1.1.d). As u Î Úk Tk = ØØ(Èk Tk) by (2.1.2.c), u is not disjoint with a Tk for some k by (2.1.1.d), and u Î S implies that u is not disjoint with S Ç Tk as required.  The normal sieve ØS is a complement of any normal sieve S in Â(X) by (2.1.1.a). Thus Â(X) is a complete boolean algebra.  (b) f* preserves arbitrary intersection (2.1.3.c) and complement by (a), thus also preserves arbitrary join. n  It follows from (2.1.4) that Â is a functor from A to the (meta)category of boolean locales, called the boolean functor on A.     [Next Section][Content][References][Notations][Home]