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в†’
then U3 в—¦ U2 в—¦ U1 is tripleable.
Proof. Easy.
The following proposition is a diп¬Ђerent 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
в€’ A be tripleable, and let C be a small category. Deп¬Ѓne V : Func(C , B) в€’ A
в†’ в†’
by taking a functor ОЁ to the product of all objects U ОЁ(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 Suп¬ѓcient 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 в€’ A to the product
(see Section 1.9). The functor в†’
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 satisп¬Ѓes 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 ) satisп¬Ѓes the coequalizer condition of PTT by
a similar argument. The pointedness of A implies that satisп¬Ѓes VTT. This
works as follows: If T : S в€’ A and C is an object of C (element of S), then
в†’
в†’ T of C (which is the product of all the
there is a canonical embedding T C в€’
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 в€’в†’ T в€’ в€’ в†’ TC
в€’в€’
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 satisп¬Ѓes the PTT.
The observation using pointedness above, applied to V instead of to , yields
the proof that V reп¬‚ects isomorphisms. It follows that V satisп¬Ѓes 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 п¬Ѓxed ring
R.

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

4. (a) Show that any map 1 в€’ 0 in a category is an isomorphism.
в†’
(b) Show that if a category has 1 and equalizers and if Hom(1, в€’) is never
empty, then 1 is initial.

3.6 Morphisms of Triples
In this section, we deп¬Ѓne 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, О·, Вµ) and T = (T О· , Вµ ) be triples on a category C . A morphism
of triples О±: T в€’ T is a natural transformation О±: T в€’ T making diagrams
в†’ в†’
(1) and (2) below commute. In (2), the notation О±2 denotes О±О± in the sense of
Exercise 6 of Section 1.3. Thus О±2 is the morphism T О± в—¦ О±T which, because О± is
a natural transformation, is the same as О±T в—¦ T О±.
Id
В  d
О·В  dО· (1)
В  d
В
В‚
ET
T О±

О±2 E
TT TT

(2)
Вµ Вµ
c c
ET
T О±
The following proposition gives one method of constructing morphisms of triples.
We are indebted to Felipe Gago-Couso for п¬Ѓnding the gap in the statement and
proof in the п¬Ѓrst edition and for п¬Ѓnding the correct statement.
3.6 Morphisms of Triples 131
Proposition 1. In the notation of the preceding paragraphs, let Пѓ: T T в€’ T
в†’
be a natural transformation for which

О·T E
T TT
d
(3)
idd Пѓ
d
dc
В‚
T
and
ВµT TВµ
E TT E
TTT TT T TT

Пѓ Пѓ
TПѓ ПѓT
(4)
c c c c
ET ET
TT TT
Пѓ Вµ
(a) (b)
commute. Let О± = Пѓ в—¦ T О· : T в€’ T . Then О± is a morphism of triples.
в†’
Proof. That (1) commutes follows from the commutativity of
О· ET
Id

О· TО·
c О·T c
E TT
T (5)
d
d
Пѓ
id d
d
В‚
dc
T
In this diagram, the square commutes because О· is a natural transformation and
the triangle commutes by (3).
132 3 Triples
The following diagram shows that (2) commutes.
Вµ ET
TT
d d
d d
dT О·
TTО· d 1
d d
d d
В‚ В‚
ВµT E
E TT
TО· T 2 TTT
TПѓ
d
d
TО· TT TО· T
3 6 d id (6)
d
d
c TT TО· c c В‚
TВµ E
E TT TT TT E TT T
Пѓ
TT T TT

Пѓ
4 5 7
ПѓT ПѓT T ПѓT

c c
c c
E T TT ET T ET
TT
TПѓ
T TО· Вµ
In this diagram, square 1 commutes because Вµ is a natural transformation,
squares 2 and 3 because T Вµ is and squares 4 and 5 because Пѓ 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 Пѓ в—¦ ВµT = Пѓ в—¦ T Пѓ which means that although the
two arrows between T T T and T T are not the same, they are when followed by
Пѓ, 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 п¬Ѓts, take B, C and D of the
exercise to be C , F = id, G = T , H = T 2 and K = T , and Оє = О· , Вµ = Вµ. Then
square 1 is О· Вµ.
Corollary 2. With T and T as in Proposition 1, suppose Пѓ: T T в€’ T is such
в†’
that Пѓ в—¦ ПѓT = Пѓ в—¦ T Вµ, Пѓ в—¦ ВµT = Вµ в—¦ T Пѓ, and Пѓ в—¦ T О· = id. Then О± = Пѓ в—¦ О·T : T
в€’ T is a morphism of triples.
в†’
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 О±: T в€’ T and functors
в†’
V : C T в€’ C T for which
в†’
FE T
CT C
d
(7)
UTd UT
d
dc
В‚
C
commutes. The bijection preserves composition.
Proof. Suppose О±: T в€’ T is a morphism of triples. Deп¬Ѓne U О± : C T в€’ C T by
в†’ в†’

U О± (A, a: T A в€’ A) = (A, a в—¦ О±A)
в†’

and for an algebra map f : A в€’ B, U О± f = f . U О± (A, a) is a T-algebra: The
в†’
unitary law follows from (1) and the other law from the commutativity of this
diagram:
T О·AE Ta E
TTA TT A TA
d
d 1
2d 2 О±A
О±T A
О±A
d
dc
d
В‚ c
ET A (8)
ВµA TTA
Ta

a
ВµA
3 4

c c c
ET A EA
TA a
О±A
Here triangle 1 commutes by deп¬Ѓnition of О±2 , square 2 because О± is a natural
transformation, square 3 is diagram (2), and square 4 because О± is a structure
map (diagram (1), Section 3.2).
Seeing that U О± is a functor is left as an exercise, as is the functoriality of the
operation which associates U О± to О±.
Conversely, suppose that V : C T в€’ C T is a functor making (7) commute.
в†’
If we apply V to the free algebra (T A, Вµ A), the result must be a T-algebra
(T A, ПѓA) with the same underlying object T A. The fact that ПѓA is a T-algebra
structure map means immediately that Пѓ is a natural transformation and it sat-
isп¬Ѓes the hypotheses of Proposition 1. Hence О± = Пѓ в—¦ T О· : T в€’ T is a morphism
в†’
of triples, as required.
134 3 Triples
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