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 S the set of maps to X which is disjoint with S.  Denote by  sie(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) S is a non-empty sieve which contains the initial map to X; S  S consists of only initial map; (S  S) = 0X,; (S  ØS) = 1X; S  S.  (b) S  T implies that T  S and S  T.  (c) S = S; T  S iff S  T; T  S iff T  S.  (d) A map t is in  S iff any non-initial map to X which factors through t is not disjoint with S; if T is a sieve then T  S iff any non-initial map in T is not disjoint with S.  (e) S = sie(S) and S = sie(S)  (f) (S)  (T) =  (sie(S)  sie(T)).  (g) If S and T are sieves then (S)  (T) =  (S  T).  Proof. (a), (b) (d) and (e) are obvious; (c) follows from (a) and (b).  (f) We have sie(S)  sie(T)  sie(S), so  (sie(S) Ç sie(T))  sie(S) = S by (e). Similarly  (sie(S)  sie(T))  sie(T) = T. Thus  (sie(S) Ç sie(T))  (S)  (T). We prove the other direction using (d). Assume  t: Y --> X is a non-initial  map in (S)  (T). To see that t is in  (sie(S)  sie(T)), it suffices to prove that for any non-initial map s: Z --> Y, st is not disjoint with sie(S)  sie(T). Since t  S, we have st  S, so by (d) st is not disjoint with S. Thus there is a non-initial map u: U --> Z such that ust factors through a map in S. Since t  T, we have ust  T, so by (d) ust is not disjoint with T. Thus there is a non-initial map v: V --> U such that vust factors through a map in T. It follows that vust  sie(S)  sie(T), which implies that st is not disjoint with sie(S)  sie(T) as desired.  (g) follows from (f) as sie(S) = S and sie(T) = T if S and T are sieves.  A sieve S of maps to X is a normal sieve if S = S. Clearly 1X and 0X are normal sieves.  Proposition 2.1.2.  (a) S 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 (i 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  Si  Si implies ( Si)  Si = Si, thus  ( Si)   Si, which implies the equality  ( Si) =  Si because the other direction is trivial.  (c) If S is a normal sieve which contains each Si, then i Si  S implies (i Si)  S = S.  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 fz is in S. If S is a sieve on X then f*(S) is a sieve on Y.  Denote by (X) the set of normal sieves on an object X.  Proposition 2.1.3. (a) f*(S) = f*(S) for any sieve S on X.  (b) If S is normal then f*(S) is normal.  (c) The function  f*: (X) --> (Y) preserves intersections.  Proof. (a) Suppose z: Z --> Y is a map in f*(S). If w: W --> Z is a map such that fzw  S, then zw  f*(S)  f*(S), so zw is an initial map by (2.1.1.a); thus w in an initial map, which implies that fz  S, i.e. z  f*(S). This shows that f*(S)  f*(S). The other direction is trivial.  (b) If S is normal then S = S, by applying (a) twice we obtain f*(S) = f*(S)= f*(S), i.e. f*(S) is normal.  (c) follows from (2.1.2.b) as f* preserves intersection of sieves.  Proposition 2.1.4. (a) (X) is a complete boolean algebra with  = .  (b) If f: Y --> X is a map then f*: (X) --> (Y) is a morphism of complete boolean algebras.  Proof. (a) We already know by (2.1.2.b) that (X) is a complete lattice with  = . Suppose S is a normal sieve and {Tk} is a set of normal sieves on X. We prove the infinite distributive law   S  (k Tk) = k (S  Tk). We only need to verify the relation  because the other direction is always true. Since k(S  Tk) = (k S  Tk) by (2.1.2.c) and S  (k Tk) is a sieve, we only need to prove that any map in u: U --> X in S  (k Tk) is not disjoint with k S  Tk by (2.1.1.d). As u  k Tk = (k Tk) by (2.1.2.c), u is not disjoint with a Tk for some k by (2.1.1.d), and u  S implies that u is not disjoint with S  Tk as required.  The normal sieve S is a complement of any normal sieve S in (X) by (2.1.1.a). Thus (X) is a complete boolean algebra.  (b) f* preserves arbitrary intersection (2.1.3.c) and complement by (a), thus also preserves arbitrary join.  It follows from (2.1.4) that  is a functor from A to the (meta)category of boolean locales, called the boolean functor on A.     [Next Section][Content][References][Notations][Home]