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d
dc
В‚
T
commutes.
The concept of free triple on R is clearly analogous to the concept of free
monoid on a set, with composition of functors playing the role of Cartesian prod-
uct of sets. In one special situation, we can construct the free triple on R in
very much the same way as for the free monoid. If C has countable sums and R
preserves them, let T = id + R + R в—¦ R + R в—¦ R в—¦ R + В· В· В·. Then T в—¦ T в€ј T , and
=
with the identity map and the obvious map onto the п¬Ѓrst summand serving as Вµ
and О· respectively, one obtains a triple which is easily seen to be the free triple
generated by R. (See Exercise 1).
To get a more general construction, we form the category which will be C T .
Let (R: C ) denote the category whose objects are pairs (C, f ) where C is an
object of C and f : RC в€’ C. A morphism f : (A, a) в€’ (A , a ) in (R: C ) is an
в†’ в†’
arrow f : A в€’ A for which
в†’
RfE
RA RA

Uf
c c
EA
A
commutes. This is a (non-full) subcategory of the comma category (R, C ). If R
should happen to be the functor part of a triple, then the category of algebras
for the triple is a full subcategory of (R: C ).
The canonical underlying functor U : (R: C ) в€’ C takes (A, a) to A and
в†’
f to f .
Proposition 1. The underlying functor U:(R: C ) в€’ C satisп¬Ѓes the condition
в†’
of the PTT except possibly for the existence of a left adjoint. Thus if it has a left
adjoint, it is tripleable.
Proof. It is clear that U reп¬‚ects isomorphisms. If
d0
(A, a) в€’ в€’ (B, b)
в€’в†’
в€’в€’
в€’в†’
d1
9.4 Free Triples 311
lies over a contractible coequalizer

в€’в†’ в€’d
A в€’в†’ B в†ђв†’ C
в€’в€’
в†ђв€’ в€’
s t
then c = d в—¦ b в—¦ Rt is a structure on C for which d is a morphism in (R: C ) which
is the required coequalizer.
Proposition 2. Let R: C в€’ C be a functor and T = (T, О·, Вµ) a triple on
в†’
C . Then there is a one to one correspondence between natural transformations
from R to T and functors О¦: C T в€’ (R: C ) which commute with the underlying
в†’
functors.
Proof. Given a natural transformation О±: R в€’ T , deп¬Ѓne О¦ to take an algebra
в†’
(A, a: T A в€’ A) to (A, a в—¦ О±A), and a morphism to itself. It is easy to see that
в†’
this gives a functor.
Going the other way, let О¦ be given. Since (T A, ВµA) is a T-algebra, О¦(T A, ВµA) =
(T A, П†A) for some arrow П†A: RT A в€’ T A. Now given f : A в€’ B, T f : (T A, ВµA)
в†’ в†’
в€’ (T B, ВµB) is a morphism between (free) algebras, so О¦T f = T f : (T A, П†A)
в†’
в€’ (T B, П†B) must be a morphism in (R: C ). It is immediate from the deп¬Ѓnition
в†’
that this is the same as saying that П†: RT в€’ T is a natural transformation,
в†’
whence so is О± = П† в—¦ RО·: R в€’ T .
в†’
We must show that this construction is inverse to the one in the preceding
paragraph. For this we need
Lemma 3. Any functor О¦: C T в€’ (R: C ) which commutes with the underlying
в†’
functors preserves the coequalizers of U T -contractible coequalizer pairs.
Proof. Such a functor clearly takes a U T -contractible coequalizer pair to a U -
contractible coequalizer pair, where U : (R: C ) в€’ C is the canonical underlying
в†’
functor. Then by Proposition 1 and the fact that U T is tripleable, О¦ must preserve
the coequalizer.
Now suppose that О¦ is given, О± is constructed as above, and О¦ is constructed
from О±. We must show that О¦ and О¦ agree on T-algebras; to do this, we use the
standard technique of showing they agree on free algebras, so that by Lemma 3
they must agree on coequalizers of U -contractible diagrams of free algebras; but
those are all the algebras by Proposition 4 of Section 3.3.
A free algebra has the form (T A, ВµA), and Вµ: (T 2 A, ВµT A) в€’ (T A, ВµA) is a
в†’
2
T-algebra morphism. Thus Вµ: (T A, П†T A) в€’ (T A, П†A) is a morphism in (R: C ),
в†’
whence П† в—¦ RВµ = Вµ в—¦ П†T . Then
О¦ (T A, ВµA) = (T A, ВµA в—¦ О±T A) = (T A, ВµA в—¦ П†T A в—¦ RО·T A)
= (T A, П†A в—¦ RВµA в—¦ RО·T A) = (T A, П†A) = О¦(T A, ВµA)
312 9 More on Triples
Thus О¦ and О¦ agree on free algebras, and so since both U T and U preserve
coequalizers, О¦ and О¦ must agree on all algebras. (Since both U T and U create
coequalizers of U -contractible coequalizer pairs, О¦ and О¦ do not merely take an
algebra to isomorphic objects of (R: C ), but are actually the same functor.)
Conversely, suppose we start with a natural transformation О±, construct a
functor О¦, and then from О¦ construct a natural transformation О± . As before, let
О¦(T A, ВµA) = (T A, П†A). Since by deп¬Ѓnition О¦(T A, ВµA) = (T A, ВµA в—¦ О±T A), we
have П† = Вµ в—¦ О±T ). Then

О± = П† в—¦ RО· = Вµ в—¦ О±T в—¦ RО· = Вµ в—¦ T О· в—¦ О± = О±

Theorem 4. If U : (R: C ) в€’ C has a left adjoint F , then the resulting triple
в†’
is the free triple generated by R.
Proof. The comparison functor (R: C ) в€’ C T is an equivalence, so its inverse
в†’
corresponds via Proposition 2 to a morphism О·: R в€’ T . If О»: R в€’ T is a natural
в†’ в†’
transformation, the composite of the corresponding functor with the comparison
functor yields a functor from C T to C T which by Theorem 3 of Section 3.6
corresponds to natural transformation О±: T в€’ T . Since О± в—¦ О· corresponds to the
в†’
same functor from C T to (R: C ) as О», they must be equal.
The converse of Theorem 4 is true when C is complete:
Proposition 5. Let C be a complete category and R an endofunctor on C
which generates a free triple T. Then U : (R: C ) в€’ C has a left adjoint and the
в†’
Eilenberg-Moore category C T is equivalent to (R: C ).
Proof. To construct the left adjoint we need a lemma.
If C is complete, then (R: C ) is complete.
Lemma 6.
Proof. The proof is the same as that of Theorem 1 of Section 3.4, which does not
require the existence of an adjoint.
Now let B be any set of objects of (R: C ). Let B # be the full subcategory
of (R: C ) consisting of all subobjects of products of objects in B. Then the
composite B # в€’ (R: C ) в€’ C has an adjoint F by the Special Adjoint Functor
в†’ в†’
Theorem, the objects of B being the solution set.
ВЇ
Now form B from B # by adding any object (B, b) for which there is a mor-
phism (B0 , b0 ) в€’ (B, b) for which B0 в€’ B is a split epi. If a map C в€’ B in
в†’ в†’ в†’
9.4 Free Triples 313
C is given, it lifts via the splitting to a unique map C в€’ B0 for which
в†’
EB
C
d В
d В
d В
В‚
d В
B0
commutes. This lifts to the unique map F C в€’ (B0 , b0 ) given by the deп¬Ѓnition
в†’
of left adjoint. Composition with B0 в€’ B gives a map F C в€’ (B, b), and this se-
в†’ в†’
quence of constructions gives an injection from HomC (C, B) to HomB (F C, (B, b)).
ВЇ
The fact that B0 в€’ B is split makes it surjective, so that we have shown that
в†’
ВЇВЇв†’
the underlying functor U : B в€’ C has a left adjoint.
If
(B , b ) в€’ (B, b)
в†’
в€’в†’
ВЇ
is a U -contractible coequalizer diagram with both algebras in B, then it has
ВЇ
a coequalizer (B , b ) in (R: C ) which by deп¬Ѓnition belongs to B. Thus since
U : (R: C ) в€’ C satisп¬Ѓes the requirements of PTT except for having a left adjoint,
в†’
ВЇ
and U is a restriction of U to a subcategory which has the requisite coequalizers,
ВЇ
U must be tripleable. Let T = (T , О· , Вµ ) be the resulting triple.
The inclusion C T в€’ (R: C ) corresponds to a natural transformation R в€’
в†’ в†’
T which by the deп¬Ѓnition of free triple gives a morphism T в€’ T of triples which
в†’
makes
ET
R
d В
d В
d В
dВ
T
commute, the top transformation being the one given by the deп¬Ѓnition of free
triple. By the correspondence between morphisms of triples and morphisms of
triple algebras (Theorem 3 of Section 3.6), this yields the commutative diagram
W V
E E (R: C )
T
CT
B
s
dd В
T
d
d В
d
d В
d
d В  (в€—)
F T d dU T F T UT В U
d
d В
d
d В
d
d В
dd
В‚ c
В
C
314 9 More on Triples
in which the top row is the inclusion.
ВЇ
Thus every object and every map of B is in the image of C T . Since we began
with any set of objects, it follows that V is surjective on objects and maps.
ВЇ
Now take an object B of (R: C ) and a B contained in (R: C ) as constructed
above which contains B. We have, in the notation of diagram (в€—),
Hom(V F T C, B) в€ј Hom(V F T C, V W B) в€ј Hom(F T C, W B)
= =
в€ј Hom(C, U T B) в€ј Hom(C, U B),
= =
which proves that U has a left adjoint.
The last statement in the proposition then follows from Proposition 1 and
Theorem 4.
Proposition 7. Let C be complete and have п¬Ѓnite colimits and colimits of
countable chains. Let R be an endofunctor of C which commutes with colimits of
countable chains. Then R generates a free triple.
Proof. By Theorem 4, we need only construct a left adjoint to U : (R: C ) в€’ C .
в†’
Let C be an object of C . Form the sequence
T e0 E T e1 E E В·В·В·
RC0 RC1 RC2

c1 c2 e3
c c c
E C1 E C2 E C2 E В·В·В·
C = C0 e0 e1 e3
in which each square is a pushout and C1 = C0 + RC0 . Let C = colim Cn . Then
RC = colim RCn and there are induced maps e: C в€’ C and c : RC в€’ C .
в†’ в†’
Lemma 8. For any diagram
B
g
c
EC
A
f
there is an arrow f : C в€’ B for which
в†’
Rf E
RC RB

b
c
c c
f EB
C
s
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