in E / § f — A. But then E /A has a faithful (because § f — A ’ A has global

’

support), logical functor into a topos in which f has a section and hence by

Proposition 1, f is a powerful epi.

248 7 Representation Theorems

AC and Booleanness

Our goal is to show that a topos satisfying AC, and ultimately a topos satis-

fying IAC, is Boolean. We require a lemma:

Lemma 5. If

EB

A

c c

ED

C

is a pullback and a pushout with all arrows monic, then its image under any near

exact functor ¦ from E to a topos F is a pushout as well as a pullback.

Proof. Since (2) is a pushout,

A’ B+C ’ D

’ ’

’’

is a coequalizer. The kernel pair of the arrow from B + C to D is (B + C) —D

(B + C), which is isomorphic to

B —D B + B —D C + C —D B + C —D C

B ’ D monic implies that B —D B = B and similarly C —D C = C. The fact

’

that (2) is a pullback implies that the other two summands are each A. The result

is that the kernel pair is the re¬‚exivized, symmetrized image of A in B + C. The

same considerations will apply to ¦ applied to (2). Thus

¦B + ¦A + ¦A + ¦C ’ ¦B + ¦C ’ ¦D

’ ’

’’

is a kernel pair and since the right arrow is epi, is also a coequalizer. It follows

that

¦A¦B + ¦C ’ ¦D ’

is a coequalizer and hence that ¦ of (2) is a pushout.

A topos E which satis¬es the Axiom of Choice is Boolean.

Theorem 6.

Proof. We must show that every subobject A of an object B has a complement.

Since a slice of a topos satisfying AC is a topos satisfying AC and subobjects of

B in E are the same as subobjects of 1 in E /B, it is su¬cient to consider the

case B = 1. So let A be a subobject of 1. Form the coequalizer

A ’ 1 + 1 ’ 1 +A 1

’ ’

’’

7.3 Morphisms of Sites 249

and ¬nd a right inverse f for the right arrow. Take the pullbacks C1 and C2 of f

along the two inclusions of 1 in 1 + 1; then we claim that the complement of A is

C = C1 © C2 as subobjects of 1. We know that A and C are subobjects of 1, so we

have a map A + C ’ 1, which we want to prove is an isomorphism. By Freyd™s

’

Theorem 7 and Exercise 5 of Section 7.1, there is a family of near-exact functors

¦: E ’ Set which collectively re¬‚ect isomorphisms. By Lemma 5, everything in

’

the construction of A + C is preserved by near-exact functors, so it is su¬cient

to prove that A + C ’ 1 is an isomorphism in Set. In Set, either A = … in

’

which case C1 = C2 = C = 1 which is the complement of A, or A = 1 in which

case one of C1 or C2 is empty, so the intersection is empty and is therefore the

complement of A. In each case the map A + C ’ 1 is an isomorphism.

’

Corollary 7. A topos satisfying IAC is Boolean.

Proof. Suppose A is a subobject of B in a topos satisfying IAC. According to

Proposition 4, the topos has a logical, faithful embedding into a topos satisfying

the Axiom of Choice. Since it is logical, it preserves the construction of ¬A. Thus

the equation A + ¬A = B is true in the original topos if and only if it is true

after the embedding. This follows from Theorem 6.

Exercises 7.2

1. Show that if f : A ’ B is a map for which § f has global support, then f B is

’

epi. (Hint: Slice by § f .)

2. Show that if G is a nontrivial group, SetG satis¬es IAC but not AC.

3. If A is a subobject of B, then A has a complement in B if and only if the

epimorphism B + B ’ B +A B is split.

’

4. If A is a subobject of B and C is any object whose support includes the

support of B, then A is complemented in B if and only if A — C is complemented

in B — C. (Hint: Adapt the argument used in Corollary 7.)

7.3 Morphisms of Sites

In this section, we state a theorem about extensions of left exact cover-preserving

functors to the sheaf category which will play the same role for theories with

cocones (treated in Chapter 8) that Theorem 4 of Section 4.3 plays for theories

with only cones. This theorem is also used in the proof of Deligne™s Theorem in

the next section.

We need some preliminary results.

250 7 Representation Theorems

Proposition 1. Let B and D be complete toposes, A a left exact generating

subcategory of B, and f : B ’ D a colimit-preserving functor whose restriction

’

to A is left exact. Then f is left exact.

Proof. Since A is left exact, it contains the terminal object 1, so f (1) = 1. Thus

we need “only” show that f preserves pullbacks.

In the proof below we systematically use A and B with or without subscripts

to refer to objects in A or B respectively. The proof requires several steps.

(i). Let B = Ai . Then for any A, f preserves the pullback

E Ai

A —B Ai

c c

EB

A

where the right arrow is coordinate inclusion.

Proof. We have

A —B Ai ∼ A

=

and

f (A —B Ai ) ∼ f A ∼ f (A) —f B f (Ai )

= =

because f commutes with sums and so do pullbacks in a topos. Both these

isomorphisms are induced by the coordinate inclusions A —B Ai ’ A. Thus the

’

sum of all the maps f (A —B Ai ) ’ f A —f B f Ai is an isomorphism, so all the

’

individual ones are.

A is the full subcategory of B consisting of sums of objects of A ,

(ii). If

A has ¬nite limits and the restriction of f to A is left exact.

then

Proof. Suppose we are given

Ak

c

E

Aj Ai

For each j, de¬ne Aji so that

E Ai

Aji

c c

E

Aj Ai

7.3 Morphisms of Sites 251

is a pullback, and similarly de¬ne Aki for each k. Because pullbacks distribute

over sums,

E

i,j,k Aji —Ai Akl Ak

c c

E

Aj Ai

is a pullback. This proves that pullbacks and hence all limits exist in A . Since

sums and all the pullbacks used in the preceding construction are preserved by

A.

f , f preserves these pullbacks, and therefore all ¬nite limits in

We will henceforth assume that A is closed under sums.

(iii). f preserves pullbacks of diagrams of this form:

E B2

B1 —A B2

c c

EA

B1

In particular, f preserves monos whose target is in A .

Proof. By Exercise 8, Section 6.8, B1 is the colimit of objects Ai , i running over

some index set, and B2 is the colimit of objects Aj , j running over a di¬erent

index set. For each i and j, form the pullback Ai —A Aj . Then B1 —A B2 is the

colimit of these pullbacks for i and j ranging over their respective index sets.

Then we calculate

f (B1 —A B2 ) ∼ f (colim(Ai —A Aj ) ∼ colim f (Ai —A Aj )

= =

∼ colim f Ai —f A f (A ) ∼ f (B1 ) —f A f (B2 )—¦

= =

j

(iv). f preserves pullbacks

E A2

A1 —B A2

c c

EB

A1

in which the two maps to B are joint epi.

Proof. Let A = A1 + A2 . Under the hypothesis, A ’ B is an epi whose kernel

’

pair is the disjoint sum

E = A1 —B A1 + A1 —B A2 + A2 —B A1 + A2 —B A2

252 7 Representation Theorems

E is an equivalence relation whose transitivity is equivalent to the existence

of an arrow E —A E ’ E (the ¬ber product being the pullback of one projection

’

against the other) satisfying certain equations. This pullback is of the type shown

in (iii) to be preserved by f . It follows that f E is transitive on f A. The fact

that E is a subobject of A — A implies in a similar way that f E is a subobject

of f (A — A) ∼ f (A) — f (A). Symmetry and re¬‚exivity are preserved by any

=

functor, so f E is an equivalence relation on f A. Hence it is the kernel pair of its

coequalizer. Since f preserves colimits, that coequalizer is f A ’ f B. So letting

’

Aij = Ai —B Aj and Cij = f Ai —f B f Aj for i, j = 1, 2, both rows in

E E fB

f (A1,1 ) + f (A1,2 ) + f (A2,1 ) + f (A2,2 ) E fA

= =

c Ec c

E fB

C1,1 + C1,2 + C2,1 + C2,2 E fA

are kernel pairs, so the left vertical arrow is an isomorphism, from which it follows

that each component is an isomorphism. The second component is the one we

were interested in.

(v). Given the pullback

E A2

A1 —B A2

c c

EB

A1

the induced map f (A1 —B A2 ) ’ f (A1 ) —f B f (A2 ) is monic.

’

Proof. This follows from the diagram

E f (A1 — A2 )

f (A1 —B A2 )

∼

=

c c

E f A1 — f A2

f A1 —f B f A2

where the horizontal arrows are induced by the mono A1 —B A2 ’ A1 — A2 (and

’

so are mono by (iii)) and the fact that if the composite of two arrows is monic

then so is the ¬rst one.

7.3 Morphisms of Sites 253

(vi). Given the pullback

E A2

A1 —B A2

c c

EB

A1

the induced map f (A1 —B A2 ) ’ f (A1 ) —f B f (A2 ) is epic.

’

Proof. Since A is closed under sums, every object B has a presentation A ’ B

’

(epi). Form the following diagram

E A2

A1 —A A2

d

d