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it is a section, to see that it makes the triangle commute and is hence a section
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

c 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
¦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
where the right arrow is coordinate inclusion.
Proof. We have
A —B Ai ∼ A
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.
Proof. Suppose we are given

Aj Ai
For each j, de¬ne Aji so that
E Ai

c c
Aj Ai
7.3 Morphisms of Sites 251
is a pullback, and similarly de¬ne Aki for each k. Because pullbacks distribute
over sums,
i,j,k Aji —Ai Akl Ak

c c
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
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
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 )—¦
= =

(iv). f preserves pullbacks
E A2
A1 —B A2

c c
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
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
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


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