then U3 —¦ U2 —¦ U1 is tripleable.

Proof. Easy.

The following proposition is a di¬erent sort, imposing hypotheses on the cat-

egories involved which imply the existence of an adjoint. A pointed category is

a category with an object 0 which is both initial and terminal. This implies that

for any two objects A and B there is a (necessarily unique) arrow from A to B

that factors through 0. The category of groups, for example, is pointed.

Proposition 2. Let A be a complete cocomplete pointed category, let U : B

’ A be tripleable, and let C be a small category. De¬ne V : Func(C , B) ’ A

’ ’

by taking a functor Ψ to the product of all objects U Ψ(C) of A over all objects

C of C. Then V is tripleable.

Outline of proof. We will refer repeatedly to the following diagram, in which S

is the set of objects of C and i is the inclusion. In the direction from Func(C , B)

3.5 Su¬cient Conditions for Tripleability 129

to A , both routes represent a factorization of V , and the square commutes.

Func(C , UE)

Func(C , B) ' Func(C , A )

Func(C , F ) T

Func(i, B) Func(i, A )

Kan

(VTT) E

c c

E Func(S, A ) '

Func(S, B) A

Func(S, U ) diagonal

F is the left adjoint of U , so Func(C , F ) os the left adjoint of Func(C , U ). The

left adjoint of Func(i, A ) exists by Kan extensions because A is cocomplete

takes a functor F : S ’ A to the product

(see Section 1.9). The functor ’

of all its values; it is easy to see that the diagonal map (takes an object of A

to the constant functor with that object as value) is the left adjoint. Thus by

composition, V has a left adjoint.

The left vertical arrow preserves colimits because colimits (like limits) are

computed pointwise. Thus it satis¬es the coequalizer condition of CTT. Note that

it might not have a left adjoint, although it will have one (use Kan extensions) if

B is also cocomplete. Func(S, U ) satis¬es the coequalizer condition of PTT by

a similar argument. The pointedness of A implies that satis¬es VTT. This

works as follows: If T : S ’ A and C is an object of C (element of S), then

’

’ T of C (which is the product of all the

there is a canonical embedding T C ’

objects T C ) induced by the identity map on T C and the point maps from T C

to all the objects T C . This makes the composite

proj

TC ’’ T ’ ’ ’ TC

’’

the identity. This enables one to transfer the maps involved in a U -split coequal-

izer in A up to Func(S, A ), verifying VTT. Thus V factors into a composite of

functors satisfying the coequalizer conditions of CTT, PTT and VTT in that

order, so it satis¬es the PTT.

The observation using pointedness above, applied to V instead of to , yields

the proof that V re¬‚ects isomorphisms. It follows that V satis¬es the CTT, hence

is tripleable.

Exercises 3.5

1. Show that if M is any monoid, the underlying functor to Set from the category

of actions by M on sets and equivariant maps is tripleable.

130 3 Triples

2. Same as preceding exercise for the category of R-modules for any ¬xed ring

R.

3. Show that the functor L of Exercise 14 of Section 1.9 is tripleable.

4. (a) Show that any map 1 ’ 0 in a category is an isomorphism.

’

(b) Show that if a category has 1 and equalizers and if Hom(1, ’) is never

empty, then 1 is initial.

3.6 Morphisms of Triples

In this section, we de¬ne a notion of morphism of triples on a given category in

such a way that functors between Eilenberg-Moore categories that commute with

the underlying functors correspond bijectively with morphisms of triples. In the

process of proving this, we will describe (Proposition 1) a method of constructing

morphisms of triples which will be used in Section 3.7.

Let T = (T, ·, µ) and T = (T · , µ ) be triples on a category C . A morphism

of triples ±: T ’ T is a natural transformation ±: T ’ T making diagrams

’ ’

(1) and (2) below commute. In (2), the notation ±2 denotes ±± in the sense of

Exercise 6 of Section 1.3. Thus ±2 is the morphism T ± —¦ ±T which, because ± is

a natural transformation, is the same as ±T —¦ T ±.

Id

d

· d· (1)

d

© d

‚

ET

T ±

±2 E

TT TT

(2)

µ µ

c c

ET

T ±

The following proposition gives one method of constructing morphisms of triples.

We are indebted to Felipe Gago-Couso for ¬nding the gap in the statement and

proof in the ¬rst edition and for ¬nding the correct statement.

3.6 Morphisms of Triples 131

Proposition 1. In the notation of the preceding paragraphs, let σ: T T ’ T

’

be a natural transformation for which

·T E

T TT

d

(3)

idd σ

d

dc

‚

T

and

µT Tµ

E TT E

TTT TT T TT

σ σ

Tσ σT

(4)

c c c c

ET ET

TT TT

σ µ

(a) (b)

commute. Let ± = σ —¦ T · : T ’ T . Then ± is a morphism of triples.

’

Proof. That (1) commutes follows from the commutativity of

· ET

Id

· T·

c ·T c

E TT

T (5)

d

d

σ

id d

d

‚

dc

T

In this diagram, the square commutes because · is a natural transformation and

the triangle commutes by (3).

132 3 Triples

The following diagram shows that (2) commutes.

µ ET

TT

d d

d d

dT ·

TT· d 1

d d

d d

‚ ‚

µT E

E TT

T· T 2 TTT

Tσ

d

d

T· TT T· T

3 6 d id (6)

d

d

c TT T· c c ‚

Tµ E

E TT TT TT E TT T

σ

TT T TT

σ

4 5 7

σT σT T σT

c c

c c

E T TT ET T ET

TT

Tσ

T T· µ

In this diagram, square 1 commutes because µ is a natural transformation,

squares 2 and 3 because T µ is and squares 4 and 5 because σ is. The commu-

tativity of square 6 is a triple identity and square 7 is diagram 4(b). Finally,

diagram 4(a) above says that σ —¦ µT = σ —¦ T σ which means that although the

two arrows between T T T and T T are not the same, they are when followed by

σ, which makes the whole diagram commute.

Squares 1 through 5 of diagram (6) are all examples of part (a) of Exercise 6,

Section 1.3. For example, to see how square 1 ¬ts, take B, C and D of the

exercise to be C , F = id, G = T , H = T 2 and K = T , and κ = · , µ = µ. Then

square 1 is · µ.

Corollary 2. With T and T as in Proposition 1, suppose σ: T T ’ T is such

’

that σ —¦ σT = σ —¦ T µ, σ —¦ µT = µ —¦ T σ, and σ —¦ T · = id. Then ± = σ —¦ ·T : T

’ T is a morphism of triples.

’

Proof. This is Proposition 1 stated in Catop (which means: reverse the functors

but not the natural transformations).

3.6 Morphisms of Triples 133

Theorem 3. There is a bijection between morphisms ±: T ’ T and functors

’

V : C T ’ C T for which

’

FE T

CT C

d

(7)

UTd UT

d

dc

‚

C

commutes. The bijection preserves composition.

Proof. Suppose ±: T ’ T is a morphism of triples. De¬ne U ± : C T ’ C T by

’ ’

U ± (A, a: T A ’ A) = (A, a —¦ ±A)

’

and for an algebra map f : A ’ B, U ± f = f . U ± (A, a) is a T-algebra: The

’

unitary law follows from (1) and the other law from the commutativity of this

diagram:

T ·AE Ta E

TTA TT A TA

d

d 1

2d 2 ±A

±T A

±A

d

dc

d

‚ c

ET A (8)

µA TTA

Ta

a

µA

3 4

c c c

ET A EA

TA a

±A

Here triangle 1 commutes by de¬nition of ±2 , square 2 because ± is a natural

transformation, square 3 is diagram (2), and square 4 because ± is a structure

map (diagram (1), Section 3.2).

Seeing that U ± is a functor is left as an exercise, as is the functoriality of the

operation which associates U ± to ±.

Conversely, suppose that V : C T ’ C T is a functor making (7) commute.

’

If we apply V to the free algebra (T A, µ A), the result must be a T-algebra

(T A, σA) with the same underlying object T A. The fact that σA is a T-algebra

structure map means immediately that σ is a natural transformation and it sat-

is¬es the hypotheses of Proposition 1. Hence ± = σ —¦ T · : T ’ T is a morphism

’

of triples, as required.

134 3 Triples