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
(
k Tk)
= k (S
Tk).
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.