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f f (1)
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A PE-structure N = (N, 0, s) is a natural number object or NNO (or ob-
ject of natural numbers or natural numbers object) if for any PE-structure (A, a, t)
there is a unique morphism t(’) a: (N, 0, s) ’ (A, a, t). If we write (suggestively)

n (’)
t (a) for t a(n) when n ∈ N , then the de¬ning properties of a morphism of
PE-structure means that
(i) t0 (a) = a, and
(ii) tsn (a) = t —¦ tn (a).
It follows immediately that if we identify the natural number n with the global
element s —¦ s —¦ s · · · s —¦ 0 (s occurring in the expression n times) of N, then expressions
like tn (a) are not ambiguous. However, now tn (a) is de¬ned for all elements n of
N, not merely those global elements obtained by applying s one or more times to
In this section, we will derive some basic properties of natural number objects
and prove a theorem (Theorem 6 below) due to Freyd that characterizes them
by exactness properties. The proof is essentially the one Freyd gave; it makes
extensive use of his embedding theorems (Section 7.1).
We begin with Proposition 1 below, which says in e¬ect that any PE-structure
contains a substructure consisting of all elements tn (a). Note that this is a state-
ment about closure under a “countable” union in any topos, so it will not be a
surprise that the proof is a bit involved. Mikkelsen [1976] has shown that inter-
nal unions in PA, suitably de¬ned, always exist. (Of course, ¬nite unions always
exist). That result and a proof of Proposition 1 based on it may be found in
Johnstone [1977].
Proposition 1. If (A, a, t) is a PE-structure, then there is a substructure A of
A for which the restricted map a, t : 1 + A ’ A is epi.

The notation a, t is de¬ned in Section 1.8. A PE-structure (A , a , t ) for
which a , t is epi will be called a Peano system.
7.5 Natural Number Objects 259
Proof. We begin by de¬ning a natural transformation r: Sub(’ — A) ’ Sub(’ —

A) which takes U ⊆ A — B to

U © (Im(idB — a) ∪ (idB — t)(U ))

where (idB — t)(U ) means the image of

id — t
U )’ B — A ’’ B ’
’ ’ ’’ B — A

That r is natural in B follows easily from the fact that pullbacks commute
with coequalizers, hence with images. Note that if A ⊆ A then rA = A if and
only if A ⊆ Ima ∪ tA .
The function r induces an arrow also called r: PA ’ PA. Let E be the

equalizer of r and idPA . De¬ne C by the pullback

c c
E E PA — A
Here ∈ A is de¬ned in Exercise 2 of Section 2.1. In rest of the proof of
Proposition 1, we will repeatedly refer to the composites C ’ E and C ’ A
’ ’
of the inclusion with the projections.
In the following lemma, A ⊆ A corresponds to cA : 1 ’ PA.

Lemma 2. The following are equivalent for a subobject A )’ A.

(i) A ⊆ Ima ∪ tA —¦
(ii) rA = A
(iii) cA : 1 ’ PA factors through E by a map u: 1 ’ E.
’ ’
(iv) A can be de¬ned as a pullback


c c
for which the inclusion A )’ A = A ’ C ’ E — A ’ A, the last map being
’ ’ ’ ’
260 7 Representation Theorems
Proof. That (i) is equivalent to (ii) follows from the fact that in a lattice, A©B =
B if and only if B ⊆ A. That (ii) is equivalent to (iii) follows from the de¬nition
of E: take B = 1 in the de¬nition of r.
Assuming (iii), construct v: A ’ C by the following diagram, in which the

outer rectangle is a pullback, hence commutative, and the bottom trapezoid is
the pullback (2).
E E A=1—A
d d
dv d idA
u — idA
d d

d c d

E E E—A EA (4)

© c
E E PA — A
The last part of (iv) follows immediately from the preceding diagram.
Now in the following diagram, II, III, IV and the left rectangle are pullbacks
and III is a mono square. Hence I and therefore the top rectangle are pullbacks
by Exercise 12, Section 2.2.
p1 E
E E 1—A 1

u — idA II
c c
E E E—A EE (5)
C p1

c c c
Thus (iii) implies (iv).
Given (iv), let u: 1 ’ E be the bottom arrow in (3). Then in (5), III is a

pullback and so is the rectangle I+II, so I is a pullback because II is a mono
square. Hence the rectangle I+III is a pullback. Clearly rectangle II+IV is a
pullback, so the outer square is a pullback as required.
Let D be the image of C ’ E — A ’ A. We will show that D is the
’ ’
subobject required by Proposition 1.
7.5 Natural Number Objects 261
Lemma 3. The statement
(A) D ⊆ Ima ∪ tD
is true if and only if
(B) the map e: P1 ’ C in the following pullback is epi.

e EC
c c

Proof. We use Freyd™s near-exact embedding Theorem 7 of Section 7.1. By
Lemma 2, the statement (A) is the same as saying that two subobjects of A
are equal, a statement both preserved and re¬‚ected by a near-exact faithful func-
tor. (Note that the de¬nition of r involves almost all the constructions preserved
by a near-exact functor). Since statements (A) and (B) are equivalent in the
category of sets (easy), they are equivalent in a power of the category of sets
since all limits and colimits are constructed pointwise there. It thus follows from
the near-exact embedding theorem that (A) is equivalent to (B).
Now we will do another transference.
Lemma 4. Suppose that for every global element c: 1 ’ C the map f : P2 ’ 1
’ ’
in the pullback
f E1

c (7)

c c
1+C E1+A EA
is epic. Then (B) is true.
Proof. We ¬rst observe that by a simple diagram chase, if e: P ’ C is not epi,

then neither is the left vertical arrow in the following pullback diagram.
E P —C

e — idC (8)
c c
E C —C
(c, idC )
262 7 Representation Theorems
Now if e in (6) is not epi, its image in the slice E /C is not epi either. The
observation just made would then provide a global element of C in a diagram of
the form (7) in which the map P ’ 1 is not epic. The lemma then follows from

the fact that all constructions we have made are preserved by the logical functor
E ’ E /C.

Lemma 5. (B) is true.
Proof. We prove this by verifying the hypothesis of Lemma 4. Let c be a global
element of C. De¬ne C by requiring that

c c
1 C
be a pullback, and P3 by requiring that the top square in
E 1+C

c c
E 1+C
f 1+A

c c
be a pullback. The bottom square is (7) with c replaced by the global element
of C induced by c and the de¬nition of C . This square is easily seen to be a
pullback, so the big rectangle is a pullback.
Because g —¦ h epi implies g epi, it su¬ces to show that P3 ’ 1 is epic. We

can see that by factoring the big rectangle in (10) vertically:
E P4 E 1+C

c c c
Here P4 is de¬ned so that the right square is a pullback. The middle arrow is
epi by Lemma 2 and Lemma 3, so the left arrow is epi as required.
7.5 Natural Number Objects 263
By Lemma 5, D satis¬es property (i) of Lemma 2. By Lemma 2, any subobject
A which has that property factors through C, hence through its image D. Since
Ima ∪ tD also has property (i) of Lemma 2 (easy), it must be that D = Ima ∪ tD.
This proves Proposition 1.
Theorem 6. [Freyd] A PE-structure (A, a, t) for which
(i) a, t is an isomorphism and
(ii) the coequalizer of idA and t is 1
is a natural number object, and conversely.
The proof will make use of
Proposition 7. [The Peano Property] A PE structure (A, a, t) which satis¬es
requirements (i) and (ii) of Theorem 6 has no proper PE-substructures.
Proof. Let (A , a , t ) be a substructure. By going to a subobject if necessary we
may assume by Proposition 1 that the restricted a , t : 1 + A ’ A is epi. Since

this proposition concerns only constructions preserved by exact functors, we may
assume by Corollary 6 of Section 7.1 that the topos is well-pointed, hence by
Proposition 2 of the same section that it is Boolean.
Let A be the complement of A in A. If the topos were Set, it would follow
from the fact that a, t is an isomorphism on 1 + A and an epimorphism to A
on 1 + A that

(—) tA ⊆ A

Since sums, isomorphisms, epimorphisms and subobjects are preserved by
near-exact functors, using the near-exact embedding of Theorem 7 of Section 7.1,
(*) must be true here. Thus t = t + t where t and t are the restrictions of t to


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