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