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completeness of categories of triple algebras. In this section, we investigate co-
completeness, with results which are informative but less satisfactory. First we
need a lemma.
9.3 Colimits of Triple Algebras 305
Lemma 1. Let B be a category and I a small diagram. Then the diagonal
functor ∆: B ’ B I has a left adjoint L, if and only if every functor F : I

’ B has a colimit in B, and in that case LF is that colimit.

Proof. The isomorphism HomB (LF, B) ∼ HomBI (F, KB ), where KB is the con-
=
stant functor, applied to the identity on LF yields a cocone from F to LF which
is universal by de¬nition.
The converse is true by pointwise construction of adjoints.
Dually, a right adjoint, if it exists, takes a functor to its limit.
Theorem 2. [Linton] Let U = B ’ C be of descent type and I a small

category. In order that every D: I ’ B have a colimit, it su¬ces that every

U D: I ’ C have colimits and that B have coequalizers.

Note that there is no claim that any colimits are created by U , only that they
exist.
Proof. We give a slick proof under the additional assumption that C is cocom-
plete. See Linton [1969c] for a proof without this hypothesis. Apply Theorem 3(b)
of Section 3.7 and Lemma 1 to the diagram
∆E
BI
B


c ∆E c
CI
C'
colim
where the colimit functor on the bottom exists by Exercise ?? of Section 1.7.
Corollary 3. Let B ’ C be of descent type and C have ¬nite colimits. Then

B has ¬nite colimits if and only if it has coequalizers.
A similar statement can be made for countable colimits, all colimits, etc.
[Linton] Let T be a triple on Set. Then SetT is cocomplete.
Proposition 4.
Proof. Given
d0
(A, a) ’ ’ (B, b)
’’
’’
’’
d1
we must construct its coequalizer. Since the underlying functor from SetT creates
limits, B — B underlies a unique T-algebra. Let E ⊆ B — B be the intersection
of all subsets of B — B which
(a) are T-subalgebras,
306 9 More on Triples
(b) are equivalence relations, and
(c) contain the image of R = (d0 , d1 ): A ’ B.


E is clearly a subset satisfying all three conditions. Thus in Set,

E’ C

’’

is a contractible pair because any equivalence relation in Set is. Thus U , being
tripleable, must lift the coequalizer of this contractible pair to a coequalizer in
SetT which is also the coequalizer of d0 and d1 . Hence SetT has all colimits by
Corollary 3.

When do triple algebras have coequalizers?

Given a triple T in a category C , to force C T to have coequalizers seems to
require some sort of preservation properties for T . For example, if T preserves
coequalizers and C has them, then C T will have them too (Exercise 1). However,
that happens only rarely.
Theorem 7 below is an example of what can be proved along these lines. We
need another result ¬rst; it says that the quotient of an algebra map is an algebra
map, if by quotient we mean the image of a regular epimorphism.
Lemma 5. If C is regular, then the classes E of regular epis and M of all
monos form a factorization system.
Proof. This has the same proof as Theorem 2 of Section 5.5. (The assumption
of regularity makes r in diagram (1) of Section 5.5 an epi).
Given an arrow F : C ’ B in such a category, its image via the factorization

into a regular epi followed by a mono is called its regular image.
One de¬nes equivalence classes of epis the same way as for monos”if C ’ A

and C ’ B are epi, they are equivalent if there are (necessarily unique) maps

for which
C
 d
  d
 
© d

E
A' B
commutes both ways.
A category C is regularly co-well-powered if for each object C there is
a set R consisting of regular epimorphisms of C with the property that every
regular epimorphism of C is equivalent to one in R.
9.3 Colimits of Triple Algebras 307
Proposition 6. Let C be a regular, regularly co-well-powered category with
coequalizers, and T a triple which preserves regular epis. Then the regular image
of an algebra morphism is a subalgebra of its codomain.
Notice that the condition that T preserve regular epis is automatically satis¬ed
in categories such as Set and categories of vector spaces over ¬elds in which every
epi is split.
Proof. Let an algebra map (C, c) ’ (B, b) factor as in the bottom line of the

following diagram. The vertical arrows make coequalizers as in diagram (13),
Section 3.3. The arrow a, and hence T a, exists because of the diagonal ¬ll-in
property of the factorization system.
E T 2A
E E T 2B
T 2C

µC µA µB
Tc Ta Tb
c c c c cc
E TA
E E TB
TC

c a b
c c c
EA
EE EB
C
The associative law for algebra follows because it works when preceded by the
epimorphism T 2 C ’ T 2 A. A similar diagram with ·™s on top gives the unitary

law.
Proposition 7. Let C and T satisfy the hypotheses of Proposition 6, and suppose
in addition that C is complete. Then C T has ¬nite colimits.
Proof. We construct coequalizers in B = C T and use Corollary 3.
Given a parallel pair
1
d0
(B, b) ’ ’ (C, c)
’’
’’
’’
d
let C ’ Ci run over all regular quotients of C which are algebras and which

coequalize d0 and d1 . Then form the image d: C ’ C0 which is a subalgebra of

Ci . Clearly d coequalizes d0 and d1 . If f : (C, c) ’ (C , c ) coequalizes d0 and

d1 , the image of f is among the Ci , say Cj . Then
pj inclusion
C0 ’ ’ Ci ’ ’ Cj ’ ’ ’ ’ ’ c
’’ ’’’’
is the required arrow. It is unique because if there were two arrows, their equalizer
would be a smaller subobject of C0 through which C ’ ’ Ci factors.
The following theorem provides another approach to the problem.
308 9 More on Triples
Theorem 8. Suppose C has ¬nite colimits and equalizers of arbitrary sets of
maps (with the same source and target). Let T be a triple in C which preserves
colimits along countable chains. Then B = C T has coequalizers.
Proof. Again we use Corollary 3. Let (2) be given, and let e: C ’ C0 be the

0 1
coequalizer in C of d and d .
For i > 0, de¬ne each Ci in the diagram below to be the colimit of everything
that maps to it.

T 2 e0 E 2 T 2 e1 E
T 2e E T C
2
TC T C1 ···
20
dd d d d d
dd d d dd
T c µC d T eµC T cd T e0 µC T cd T e1 µC
d d d
0 1
d d d d d d
T (ec)
dd d d dd
dd dd dd
cc ‚‚ ‚ ‚ ‚‚
E T C0 E T C1 E T C2 · · ·
TC
Te T e0 T e1
T d T d T
d d
d c0 ·C1 d c1
c ·C0 ·C2
d d
d d
d d
c ‚ ‚
E C0 E C1 E C2 · · ·
C e e0 e1
It follows that ci+1 —¦ T ci = ci+1 —¦ T ei µCi .
If f : (C, c) ’ (B, b) coequalizes d0 and d1 , then there is a unique g: C0 ’ B
’ ’
for which g —¦ e = f . Then
g —¦ e —¦ c = f —¦ c = b —¦ Tf = b —¦ Tg —¦ Te
Also,
b —¦ T gT (e —¦ c) = b —¦ T f —¦ T c = f —¦ c —¦ T c
= f —¦ c —¦ µC = b —¦ T f —¦ µC = b —¦ T g —¦ T e —¦ µC,
and by a similar calculation,
b —¦ T g —¦ ·C0 —¦ e —¦ c = b —¦ T g —¦ T e
It follows that there is a unique g1 : C1 ’ B for which g1 —¦ e0 = g and g1 —¦ c0 = b —¦ T g.

Now assume gi : Ci ’ B has the property that gi —¦ ei’1 = gi’1 and b —¦ T gi’1 =

gi —¦ ci’1 . Then,
b —¦ T gi —¦ T ci’1 = b —¦ T (b —¦ T gi’1 ) = b —¦ T b —¦ T 2 gi’1
= b —¦ µB —¦ T 2 gi’1 = b —¦ T gi’1 —¦ µCi’1
= b —¦ T (gi —¦ ei’1 ) —¦ µCi’1 = b —¦ T gi —¦ T ei’1 —¦ µCi’1 ,
9.4 Free Triples 309
and similar but easier calculations show that

b —¦ T gi —¦ T ei’1 = gi —¦ ei’1

and
b —¦ T gi —¦ ·Ci —¦ ci’1 = b —¦ T gi —¦ T ei’1
This means that there is a unique gi+1 : Ci+1 ’ B for which gi+1 —¦ ei = gi and

b —¦ T gi = gi+1 —¦ ci .
Now we go to the colimit of the chain. Let C = colim Ci , so T C =
colim T Ci’1 and T 2 C = colim T 2 Ci’2 . We get g : C ’ B whose “restric-

tion” to Ci is gi . The maps ci induce a map c : T C ’ C . This is an algebra

structure map making g an algebra morphism:
(a) g —¦ c = b —¦ T g because gi —¦ ci = b —¦ T gi’1 .
(b) c —¦ T c = c —¦ µC because ci —¦ T ci’1 = ci —¦ T ei’1 —¦ µCi’1 and the ei commute
with the transition maps in the diagram.

However, we are not done. This map g need not be unique. To make it
unique, we pull a trick similar to the construction in the proof of the Adjoint
Functor Theorem. We have a map e : (C, c) ’ (C , c ) induced by the ei which

is an algebra morphism. The equalizer of all the endomorphisms m of (C , c )
for which me = e is the coequalizer. This equalizer exists because it exists in
C and tripleable functors create limits. By copying the argument of the Adjoint
Functor Theorem, one gets a map e : C ’ E which is the required coequalizer.



Exercises 9.3

1. Show that if T is a triple in C , C has coequalizers and T preserves them, then
C T has coequalizers.

2. Show that Theorem 8 may be generalized to show that if C has ¬nite colimits
and equalizers of arbitrary sets of maps and if T is a triple in C which preserves
colimits along chains indexed by some cardinal ±, then B = C T has coequalizers.


9.4 Free Triples
Given an endofunctor R: C ’ C on a category C , the free triple generated

by R is a triple T = (T, ·, µ) together with a natural transformation ±: R ’ T

310 9 More on Triples
with the property that if T = (T , · , µ ) is a triple and β: R ’ T is a natural

transformation, then there is a triple morphism T ’ T for which

±E
R T
d
βd

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