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{i | Mi (A + B) = Mi A + Mi B = Mi A} ∈ f

whence N (A + B) = N A = N A + N B.
This theorem demonstrates, for instance, that the category of ¬elds is not the
models of any regular theory and similar such negative results. As for positive
results, there are few. Beck™s tripleableness theorem and its variants give a posi-
tive result in terms of a chosen underlying set functor, or equivalently, in terms
of chosen representing sketch. Exercise 6 shows that the category of models of
an LE sketch might be regular without being the category of models of an FP

Exercises 8.4

1. Let f be an ultra¬lter. Show that if J ∪ K ∈ f then either J ∈ f or K ∈ f.

2. Prove Lemma 2. (Compare Lemma 6 of Section 4.4 for a variation on this

3. Prove that ¬ltered colimits of regular functors are regular.

4. Show that the de¬nition of regular epimorphic family given in this section is
the same as the one given in Section 6.7 in the presence of pullbacks. Show that
the class of C used in the de¬nition in this section may be restricted to being in
a generating family.

5. (a) Show that the category of posets and order-preserving maps is a subcat-
egory of the category of graphs which is closed under products and subobjects.
(b) Show that the map which takes the graph below to the graph obtained
by identifying A with A and B with B

(but not identifying the edges) has a kernel pair which is (i) an equivalence relation
and (ii) is de¬ned on a poset.
290 8 Cocone Theories
(c) Use (a) and (b) to show that equivalence relations are not e¬ective in the
category of posets.

6. An orthodox semigroup is a semigroup in which the product of any two
idempotents is idempotent. Let O be the category of orthodox semigroups and
semigroup homomorphisms.
(a) Show that O is the category of models of an LE-sketch.
(b) Show that O is regular. (Hint: Show that O is closed in the category
of semigroups under subobjects and products and that such a subcategory of a
regular category is regular.)
(c) Show that O does not have e¬ective equivalence relations. (Hint: Any
free semigroup is orthodox.)

7. In this exercise and the next, the categories of groups considered are under-
stood to have all group homomorphisms as arrows. Show that the category of
torsion groups is the category of models of a geometric theory. Can it be the
category of models of a coherent theory? (Hint: For the ¬rst part, consider the
theory of groups augmented with types

Gn = {x ∈ G | xn = 1}

which is an equalizer and hence in the left exact theory. Require that {Gn ’ G}

be a cover.)

8. (a) Show that the category of cyclic groups is not the category of models of
any coherent theory. (Hint: Consider a non-principal ultrapower of Z.)
(b) Show that the category of ¬nite cyclic groups is not the category of models
of any geometric theory.
More on Triples
This chapter consists of four independent sections providing additional results
about triples. Everything may be read immediately after Chapter 3 except for
Lemma 5 of Section 9.3, which depends on a fairly easy result from Chapter 5.

9.1 Duskin™s Tripleability Theorem
In this section we state and prove a theorem of Duskin (Theorem 1) giving nec-
essary and su¬cient conditions for a functor U : B ’ C to be tripleable, under

certain assumptions on the two categories involved.
If B is an equationally de¬ned class of algebras (i.e. models of a single sorted
FP theory) and C is Set then Birkhoªs theorem on varieties says that B is
“closed under” products, subobjects and quotients by equivalence relations in B.
The ¬rst two closure properties mean little more than that, in our language, U
creates limits. The third condition means that U creates coequalizers of parallel
pairs which become equivalence relations in C = Set.
Duskin™s Theorem is motivated by the idea that a functor U which satisfy
a categorical version of Birkhoªs theorem ought to be tripleable. We begin by
studying equivalence relations in a category. Throughout this section we study a
functor U : B ’ C with left adjoint F .

Equivalence pairs and separators

An equivalence pair on an object C is a parallel pair f, g: B ’ C for which

(a) f and g are jointly monic, which means that for any elements x and y of
B de¬ned on the same object, if f (x) = f (y) and g(x) = g(y) then x = y; and
(b) the induced pair of arrows from Hom(X, B) ’ Hom(X, C) is an equivalence

relation in Set for any object X. When B has products, f and g are jointly monic
if and only if the arrow (f, g) is monic, and are an equivalence pair if and only if
(f, g): B ’ C — C is an equivalence relation (see Exercise 18 of Section 1.7).

Maps f, g: B ’ C form a U -contractible equivalence pair if U f and U g

are an equivalence pair which is part of a contractible coequalizer diagram in C .

292 9 More on Triples
A separator of a parallel pair f, g: B ’ C is the limit

[(b, b ) | f (b) = f (b ) and g(b) = g(b )]

Thus it is the intersection of the kernel pairs of f and g. In particular, a
category which has separators of all parallel pairs has kernel pairs of all maps.

Duskin™s theorem

Theorem 1. [Duskin] Suppose B has separators and C has kernel pairs of split
epimorphisms. Then the following two statements are equivalent for a functor
U: B ’ C :

(i) U is tripleable.
(ii) U has an adjoint and re¬‚ects isomorphisms, and every U -contractible
equivalence pair has a coequalizer that is preserved by U .
Before we prove this theorem, notice what it says in the case of groups. An
equivalence pair d0 , d1 : H ’ G in Grp forces H to be simultaneously a subgroup

of G — G and a U -contractible equivalence relation on G (every equivalence pair
in Set is part of a contractible coequalizer diagram). It is easy to see that the
corresponding quotient set is the set of cosets of a normal subgroup of G, namely
the set of elements of G equivalent to 1. The canonical group structure on the
quotient set makes the quotient the coequalizer of d0 and d1 . Note that you
can show that the quotient is the coequalizer by showing that d0 and d1 are the
kernel pair of the quotient map; since the quotient map is a regular epi (it is the
coequalizer of the trivial homomorphism and the injection of the kernel of the
quotient map into its domain), it follows from Exercise 6 of Section 1.8 that the
quotient map is the coequalizer of d0 and d1 . That may not be the method of
proof that would have occurred to you, but it is the strategy of the proof that
The argument is even more direct in the case of compact Hausdor¬ spaces.
If R is an equivalence relation in that category on a space X then R is a closed
(because compact) subspace of X — X, so corresponds to a compact (because
image of compact) Hausdor¬ (because R is closed) quotient space X/R.
Proof. Proof of Theorem 1. Because of Beck™s Theorem and Proposition 3 of
Section 3.3, the proof that (i) implies (ii) is immediate.
To prove that (ii) implies (i), we ¬rst prove three lemmas, all of which assume
the hypotheses of Theorem 1 (ii).
9.1 Duskin™s Tripleability Theorem 293
Lemma 2. If U f : U A ’ U B is a split epimorphism, then f : A ’ B is a
’ ’
regular epimorphism.
Proof. Let h, k: S ’ A be the kernel pair of f . Then because U preserves limits,

(U h, U k) is the kernel pair of U f ; hence by Exercise 2 (h, k) is a U -contractible
equivalence pair and so by assumption has a coequalizer A ’ C in B. Thus

there is an induced map from C to B which necessarily becomes an isomorphism
under U . Thus because U re¬‚ects isomorphisms, f must be the coequalizer of h
and k.
Lemma 3. If U d, U e: U E ’ U B is an equivalence pair, then so is d, e.

Proof. For any object B of B, U B —¦ ·U B = idU B (Exercise 15 of Section 1.9),
so Lemma 2 implies that B is a regular epi. Then by Corollary 7 of Section 3.3,

FUFUB ’ ’ ’ ’ FUB ’ ’ B
’’’ ’’
is a coequalizer.
For any object C of C , the induced maps
Hom(C, U E) ’ Hom(C, U B)

are an equivalence relation in Set, and by adjointness so is
Hom(F C, E) ’ Hom(F C, B)

Putting the facts in the preceding two paragraphs together, we have that the
rows in
E Hom(F U B , E)
Hom(B , E) E Hom(F U F U B , E)

c c c c c c
E Hom(F U B , B)
Hom(B , B) Hom(F U F U B , B)

are equalizers and the right hand and middle columns are both equivalence re-
lations. It follows from an easy diagram chase that the left hand column is an
equivalence relation, too.
Lemma 4. In the notation of Lemma 3, if f : B ’ Y is a map for which U d

and U e form the kernel pair of U f , then d and e form the kernel pair of f .
Proof. The proof follows the same outline as that of Lemma 3. One has that for
any object C, the parallel pair in
Hom(F C, E) ’ Hom(F C, B) ’ Hom(F C, B )
’ ’
294 9 More on Triples
is the kernel pair of the right arrow, so that in the diagram below
E Hom(F U B , E)
Hom(B , E) E Hom(F U F U B , E)

c c c c c c
E Hom(F U B , B)
Hom(B , B) E Hom(F U F U B , B)

c c c
E Hom(F U B , B )
Hom(B , B ) Hom(F U F U B , B )
the middle and right hand vertical parallel pairs are kernel pairs of the corre-
sponding arrows and the horizontal sides are all equalizers. Then a diagram
chase shows that the left hand column is a kernel pair diagram too.
Now, to prove that (ii) of Theorem 1 implies (i). By Beck™s Theorem, we must
prove that if
A ’’ B
is a re¬‚exive U -split coequalizer pair, then it has a coequalizer which is preserved
by U .
In B, we construct the following diagram, in which (p0 , p1 ) is the kernel pair
of d1 and S is the separator of d0 —¦ p0 and d0 —¦ p1 . The object E and the arrows
into and out of it will be constructed later.
e0 E yE
p0 p1 u0 u1
c c d0 E c c
In C , we have diagram (5), in which c is the coequalizer of U d1 and U d2 with
contracting maps s and t and (q 0 , q 1 ) is the kernel pair of of the split epi c.
U e0 E w E UE
U p0 U p1 q0 q1
U d0
c c ‚c c cE
UA ' UB ' C
U d1 s

9.1 Duskin™s Tripleability Theorem 295
This is how the proof will proceed: (a) We will construct w and (b) show txhat
it is a split epi with (c) kernel pair (U e0 , U e1 ). It then follows from Lemma 3 that
(e0 , e1 ) is a U -contractible equivalence relation which must by hypothesis have a
coequalizer v: R ’ E from which we conclude that up to isomorphism, U E can

be taken to be C . (d) We then construct u0 , u1 for which u0 —¦ v = d0 —¦ p0 and
u1 —¦ v = d0 —¦ p1 and also U (u0 ) = q 0 and U (u1 ) = q 1 . Now c is the coequalizer
of (q 0 , q 1 ) (Exercise 6 of Section 1.8: if a regular epi has a kernel pair then it is
the coequalizer of its kernel pair), so (u0 , u1 ) is a U -split equivalence pair, hence


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