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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

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

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

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