<<

. 35
( 60 .)



>>

Hom(C — A, B) Hom(C — A ’ A, B — A ’ A)
’ ’
= Hom(A— C, B — A ’ A) Hom(C, A— (B — A ’ A))—¦
’ ’

For the second construction, consider the canonical presentation

∈(PPB)
’ ’ ’ ’ ’ PPB ’∈B B
’’’’
PPPPB ’ ’ ’ ’ ’ ’’
’’ (1)
’’’’
PP(∈B)

of the object B in E op (Section 3.3). Writing f , g and h for the three maps and
B = PB, B = PPPB this becomes an equalizer
g
f ’ ’ PB

B ’ ’ PB ’ ’
’ (2)

h
in E .
5.4 Toposes are Cartesian Closed 185
For any objects B and C, let φ(B, C): Hom(C, PB) ’ Sub(B — C) be the

natural isomorphism. Because it is an isomorphism, the middle and lower hori-
zontal arrows in the following diagram are uniquely de¬ned:

Hom(A — C, g)
E
Hom(A — C, PB ) Hom(A — C, PB )
T T
’1
φ(B , A — C)’1
φ(B , A — C)

(3)
E Sub(B — A — C)
Sub(B — A — C)
T T
φ(B — A, C) φ(B — A, C)

E Hom(C, P(B — A))
Hom(C, P(B — A))

Let
g A : P(B — A) ’ P(B — A)

be the map induced by the bottom arrow of (3) via the Yoneda lemma. In the
same way, de¬ne hA : P(B — A) ’ P(B — A), and let the exponential B A be

de¬ned by requiring that

gA
B A ’ P(B — A) ’ ’ P(B — A)
’’
’ (4)
’’
’’
hA
be an equalizer.
It remains to prove that Hom(C, B A ) is naturally isomorphic, as a functor of
C, to Hom(A — C, B). This follows from applying Hom(A — C, ’) to diagram (2)
and Hom(C, ’) to diagram (4). Both give equalizer diagrams since Hom(C, ’)
preserves equalizers. But the parallel pairs of arrows being equalized in the
diagrams thus obtained are naturally isomorphic by (3) (for g) and the analog of
(3) for h. Thus the left sides must be naturally isomorphic, as required.
Corollary 2. Logical functors preserve exponentials.
The most common de¬nition of topos in the literature is that it is a cartesian
closed category with ¬nite limits and a subobject classi¬er. Exercises 3 and 4 of
this section show that our de¬nition is equivalent to the usual one.

Exercises 5.4

1. Show that for any object X of any topos, PX = „¦X .
186 5 Properties of Toposes
2. Identify the counit AB — B ’ A in the category of sets. (Hint: this counit

in any cartesian closed category is called ev.)

3. Let f : B ’ C be an arrow in a topos. Show that Pf : PC ’ PB corresponds
’ ’
by the adjunction de¬ning „¦B to the arrow

ev —¦ (1 — f ): PC — B ’ „¦


where ev is the arrow of Exercise 2.

4. Use Exercise 3 to prove that a cartesian closed category with a subobject
classi¬er and all ¬nite left limits is a topos.

5. Construct the following isomorphisms in any Cartesian closed category in such
a way that they are natural in both variables.
(a) (B — C)A ∼ B A — C A .
=
(b) (C A )B ∼ C (A—B) .
=

6. Show that in a cartesian closed category, if f, g ∈T B A and y ∈T A, then
(f, g)(y) = (f (y), g(y)). (This notation assumes that the isomorphism in Exer-
cise 5(a) is the identity map.)

7. Show that Cartesian closed categories are models of an LE-theory.


5.5 Exactness Properties of Toposes
In this section we deduce a number of facts about maps in a topos, most of which
have to do in some way with colimits.

Epi-mono factorizations

Lemma 1. If f : A ’ B is an epimorphism in a topos, and i: B ’ C is any
’ ’
map, then the induced map A —C A ’ B —C B is epi.

Proof. The functor A —C ’ commutes with colimits, hence preserves epimor-
phisms. Thus the induced map f — 1: A —C A ’ B —C A is epi. Similarly, the

induced map 1 — f : B —C A ’ B —C B is epi, so their composite is too.

An epi-mono factorization of an arrow f : A ’ B in a category is a repre-

sentation f = m —¦ e where m is mono and e is epi. It is easy to see that in a topos,
the representation is essentially unique (Exercises 1 and 3). The codomain of e
is called the image of f (see Exercise 5).
5.5 Exactness Properties of Toposes 187
Theorem 2. If f : A ’ B is any arrow in a topos, then f has an epi-mono

factorization.
Proof. Given f : A ’ B, construct its kernel pair (h, k) and the coequalizer q: A

’ C of its kernel pair. Then because f coequalizes its kernel pair, there is an

arrow i as in Figure 1 below, and i has a kernel pair (u, v). Since q coequalizes
the kernel pair of f , there is an arrow r as in the ¬gure.
hE fE
EA
A—A B
k
q (1)
=
r
c uE c c
EB
EC
C —C
i
v
By Lemma 1, r is an epimorphism. Since q coequalizes h and k, u —¦ r = v —¦ r,
so u = v. Thus, because the two arrows in its kernel pair are the same, i is a
monomorphism. The required factorization is then f = i —¦ q.

Exactness properties

Proposition 3. Every epimorphism in a topos is regular.
Proof. Assume f in Figure 1 is epi. Then the map i must also be epi. It is mono,
as we proved, so is regular mono (Corollary 4 of Section 2.3). But a map which is
both regular mono and epi is an isomorphism. Hence f is a coequalizer, namely
of its kernel pair, as required.
Proposition 3 and preceding results imply that a complete topos is a regular
category (see Section 1.8.
The set of subobjects of an object forms a partially ordered set, and so one
may ask when a pair of subobjects has an intersection (greatest lower bound) or
union (least upper bound). In a category with ¬nite limits, intersections always
exist: the intersection of subobjects B and C of D is the pullback
EB
A

(2)
c c
ED
C
in which all the arrows are monomorphisms because the pullback of a mono is a
mono.
Unions, however, are harder. In general categories, they are not colimits in a
simple way (see Exercise 11). In a topos, however, the situation is quite simple.
188 5 Properties of Toposes
Proposition 4. In a topos, the union of any two subobjects B and C of an
object D is the image of the arrow from B + C to D induced by the inclusions of
B and C in D:
ED
B+C
d  

 
d (3)
 
d


B∪C
Proof. Trivial.
Proposition 5. Suppose that (2) is a pullback diagram in which all arrows are
mono. Then the induced arrow B +A C ’ D is monic.

Proof. The idea of the proof is to show that B +A C is isomorphic to B ∪ C, so
that the induced arrow is the inclusion. Construct the kernel pair of the arrow
from B + C to D in diagram (3):

(B + C) — (B + C) ’ B + C ’ B ∪ C
’ ’ (4)
’’
The resulting diagram is a coequalizer, by Theorem 2 and Proposition 4. Since
the pullback commutes with colimits,

(B + C) —D (B + C) B —D B + B —D C + C —D B + C —D C
= B + B © C + C © B + C—¦

However, a map coequalizes the two arrows B + B © C + C © B + C ’ B + C

if and only if it coequalizes the two arrows B © C ’ B + C (because the two

arrows agree on the ¬rst and last components and interchange the middle two).
So
B©C ’ B+C ’ B∪C
’ ’
’’
is a coequalizer. This means that
EB
B©C

(5)
c c
E B∪C
C
is a pushout (as well as a pullback) diagram, hence B ∪ C = B +A C, as required.

As an application of Proposition 5, we have
5.5 Exactness Properties of Toposes 189
Theorem 6. A functor between toposes which preserves monomorphisms, ¬nite
products and cokernel pairs is left exact.
Proof. Let F : E ’ E satisfy the conditions of the theorem. It is su¬cient to

show that F preserves equalizers. So let
A’ B’ C

’’ ’
be an equalizer. An easy argument using elements shows that this is equivalent
to the following diagram being an equalizer.
A’ B ’ B—C

’’ ’
Then the two maps to B — C are monic (split by the projection onto B). It
is easily seen that we have a pullback
E EB
A
c c


c c
E E B—C
B
It follows from Proposition 5 that
B +A B ’ B — C

is monic. Applying F , we have
E
E E FB E FB — FC
FA
T T T


E
E E FB T
E F B +F A F B
FA

<<

. 35
( 60 .)



>>