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conditions for tripleability they are also useful in showing that certain functors
are not tripleable, as we will illustrate.

Completeness of categories of algebras

If T = (T, ·, µ) is a triple in a category C , then the category C T of algebras
is “as complete as C is”, in the following sense:
Theorem 1. Let T be a triple in C . Then U T : C T ’ C creates limits. Hence

any tripleable functor re¬‚ects limits.
Proof. In the following, we write U for U T for simplicity. Let D: I ’ C T be

a diagram, and let C be the limit of U —¦ D in C . We must ¬nd an algebra
structure map c: T C ’ C making (C, c) the limit of D, and that structure must

be unique. Let n be an object of I . Then Dn is a structure (U D(n), ±n). Let
βn: C ’ U (D(n)) be the element of the limit cone corresponding to n. Then

the elements (±n) —¦ (T βn) form a cone from T C to U D and so induce an arrow
c: T C ’ C which is the required algebra structure. The fact that the resulting

structure is a T-algebra follows from the jointly monic nature of the elements βn.
3.4 Properties of Tripleable Functors 123
For example, associativity follows from the diagram
E T 2 U Dn
T 2C

c c cc
E T U Dn

c ±n
c c
E U Dn

To see that (C, c) is the limit, suppose we have a cone with elements

γn: (C , c ) ’ Dn = (U Dn, ±n)

Applying U , this gives a cone from C to U Dn, which induces an arrow C ’ C ’
by the fact that C is a limit. The left square in the following diagram must then
commute because the βn are a jointly monic family. That means that the map
C ’ C is an algebra map as required.


e ±n
e (1)
c c c
If U : B ’ C is tripleable and C is complete, then so is B.
Corollary 2. ’
Note that U in the preceding corollary need not create limits, since B might
be equivalent to but not isomorphic to C T .
In Section 9.3 we will describe su¬cient conditions for proving B cocomplete.

Banach spaces

The fact that a subobject is a subalgebra if and only if it is closed under the
operations (Exercise 4 of 3.2) is a useful necessary condition for tripleability. We
apply this criterion to Banach spaces.
Let Ban denote the category whose objects are real Banach spaces and whose
morphisms are linear maps which to not increase the norm. Let U : Ban ’ Set ’
be the functor that takes a space B to its unit ball {b ∈ B | |B| ¤ 1} and maps
124 3 Triples
to their restriction. We will show that U re¬‚ects isomorphisms and has a left
adjoint, but is nevertheless not tripleable.
U re¬‚ects isos: Suppose f : B ’ C is such that U f is an isomorphism. The

proof repeatedly uses the fact that for any b ∈ B, b/ |b| is in U B. Thus f is
injective (if f (b) = 0 then U f must take b/ |b| to 0) and surjective (if c ∈ C then
c/ |c| is the image of some b ∈ U B, so c is the image of b |c|). It is also necessary
to show that f ’1 preserves the norm, which will follow if we show that f preserves
the norm. Let b ∈ B and n = |f (b/ |b|)|. Then

f (b/(n |b|) = (1/n)f (b)/ |b|)

must be in U C because its norm is 1. Since f is injective and U f is surjective
onto U C, b/(n |b|) is in U B, whence n ≥ 1. Since f does not increase norms,
n ¤ 1, so n = 1. Thus f preserves norms, as required.
The left adjoint to U assigns to a set X the set F X of all functions f : X ’ R

for which Σ(x∈X) |f (x)| ¤ ∞. The norm |f (x)| = Σ |f (x)| makes F X a Banach
space. It is in fact l1 (X), where X is regarded as a measure space with atomic
The identities for an algebra for the triple induced by F and U imply by a
somewhat long but straightforward agrument that for a given Banach space C,
the induced algebra structure c = U C: U F U C ’ U C is de¬ned for f ∈ F U C

c(f ) = Σ(x∈X) x |f (x)|
Then if C is the closed interval [’1, 1], the only f ∈ F U C for which c(f ) = 1 is
the function with value 1 at 1 and 0 elsewhere, and the only f with c(f ) = ’1
are the functions with value ’1 at 1 and 0 elsewhere, and with value 1 at ’1 and
0 elsewhere. Letting B = (’1, 1), this implies that c(F U B) ⊆ B, which means
by Exercise 4 of 3.2 that there is an algebra structure on the open interval that
agrees with c. But the open interval (’1, 1) with the usual addition and scalar
multiplication is not the closed ball of any Banach space, so U is not tripleable.
Even worse is the three-point algebra [0, 1]/(0, 1), which is the coequalizer of
the inclusion of (0, 1) in [0, 1] and the zero map. It is instructive to work out the
algebra structure on this algebra.

Tripleability over Set

A useful necessary condition for tripleability over Set is the following propo-
3.4 Properties of Tripleable Functors 125
Proposition 3. If U : B ’ Set is a tripleable functor, then the pullback of a

regular epi in B is a regular epi.
Proof. If c is a regular epi in B, then it is the coequalizer of its kernel pair
(Corollary 2 and Exercise 6 of Section 1.8). This kernel pair induces an equiva-
lence relation in Set which, like any such, is split. Hence by Proposition 2(c) of
Section 3.3, c is a U -contractible coequalizer. Hence U c is a regular epi in Set.
Let b be the pullback of c along an arrow f . Then U b is a regular epi in Set since
U preserves pullbacks and the pullback of a regular epi is a regular epi in Set.
U b is then the class map of an equivalence relation and so b is a U -contractible
coequalizer in B. Since U is tripleable, the PTT implies that b is regular.
Corollary 4. The category Cat of small categories and functors is not tripleable
over Set.
(Note that this means there is no functor from Cat to Set which is tripleable”
in particular, neither of the functors which take a category to its set of objects
or its set of arrows.)
Proof. Let 1 denote the category with one object and one arrow, and 2 the
category with two objects and exactly one non-identity morphism f , going from
one object to the other. There are two functors from 1 to 2, and to form their
coequalizer is to identify the domain and codomain of f . In this coequalizer, we
must have the arrows f 2 , f 3 , and so on, and there is no reason for any equalities
among them, so the coequalizer is the monoid (N, +) regarded as a category with
one object. Thus the arrow 2 ’ N is a regular epi. See Exercise 12.

Now let M )’ N denote the submonoid of even integers. Let 2 (as opposed

to 2) denote 1 + 1, that is the category with two objects and no nonidentity
arrows. It is easy to see that
c c
c c
is a pullback and that the top arrow is not a regular epi.
(It is a notable phenomenon that a functor may merge objects and therefore
make arrows compose with each other that never dreamed of composing before
the functor was applied. This makes colimits in Cat hard to understand and
is probably the main reason why the oft-expressed notion that a category is a
“monoid with many objects” has only limited fruitfulness”in contrast to the
very suggestive idea that a category is a generalized poset.)
126 3 Triples
On the other hand, Cat is tripleable over the category Grph of graphs. The
left adjoint L to the underlying functor U : Cat ’ Grph is de¬ned by making LG

(for a graph G) the category whose nonidentity arrows are all composable paths
of arrows of G. (Exercise 11 of Section 1.9.) Composition is concatenation of
paths. Identity arrows are empty paths.
That U re¬‚ects isomorphisms is the familiar fact that the inverse of a bijective
functor is a functor.
Finally, suppose
A’ B’ (3)
’ ’
is a pair of functors for which

UA ’ UB ’ ’ G
’ ’ (4)

is a contractible coequalizer. In particular, it is an absolute coequalizer, so ap-
plying U L gives a coequalizer

U Ld
U LU A ’ U LU B ’ ’ ’ U LG
’ ’ ’’ (5)

If b: LU B ’ B is the map corresponding to the given category structure on B

(taking a composable path to its composite), then by the coequalizer property
there is a map g making the following diagram commute.
U Ld

g (6)
c c
and it is easy to see from the universal property of coequalizers that both hori-
zontal arrows are surjective. A straightforward check using that surjectivity then
shows that de¬ning composition of arrows a and b of U LG by a —¦ b = g(a, b) and
identity arrows as images of identity arrows in U B makes G a category. Then d
is a functor which is the required coequalizer.
Grph is also tripleable over sets; this follows from Exercise 10 below and
Exercise 1 in Section 3.5. Thus the composite of two tripleable functors need not
be tripleable. For another example, see Exercise 11 below. In the next section we
will state theorems giving circumstances under which the composite of tripleable
functors is tripleable.
3.4 Properties of Tripleable Functors 127
Exercises 3.4

10. Show that Grph is isomorphic to the category of set-valued actions of the
monoid M with multiplication table


(Hint: If M acts on X, X is the set of arrows of a graph and sx and tx are its
source and target.)

11. (a) Show that the inclusion of the category of torsion-free Abelian groups
into the category of Abelian groups is tripleable. (Hint: Try Exercise 7 of Sec-
tion 3.3.)
(b) Show that the category of Abelian groups is tripleable over Set.
(c) Show that the underlying set functor from the category of torsion-free
Abelian groups to Set has a left adjoint (hint: restrict the free-Abelian-group
(d) Show that the functor in (c) re¬‚ects isomorphisms.
(e) Show that the functor in (c) is not tripleable. (Hint: The two maps
(m, n) ’ m + 2n and (m, n) ’ m from Z • Z ’ Z determine a U -split equiva-

lence relation but are not the kernel pair of anything).

12. Using the notation of Corollary 4, we have the commutative diagram

2 N
Hd G

We have already seen that F is a regular epi. Show that G is a regular epi and
H is not, so that the composite of regular epis need not be one. This destroys a
very reasonable sounding conjecture.
128 3 Triples
3.5 Su¬cient Conditions for Tripleability
We de¬ne two useful su¬cient doncitions on a functor which make it tripleable,
which in addition allow us to give circumstances under which the composite of
tripleable functors is tripleable.
Such a composite can fail to be tripleable if the ¬rst functor applied fails to
lift contractible coequalizers to contractible coequalizers. However, the composite
might still be tripleable if the second functor lifts all coequalizers. This motivates
the following two de¬nitions.
A functor U : B ’ C satis¬es the hypotheses of the “VTT” (Vulgar Triplea-

bility Theorem) if it has a left adjoint, re¬‚ects isomorphisms, and if any re¬‚exive
U -contractible coequalizer pair is already a contractible coequalizer in B. Since
contractible coequlizers are preserved by any functor, it follows that if U satis¬es
the VTT it must be tripleable.
U satis¬es the hypotheses of the “CTT” (Crude Tripleability Theorem) if U
has a left adjoint and re¬‚ects isomorphisms, B has coequalizers of those re¬‚exive
pairs (f, g) for which (U f, U g) is a coequalizer and U preserves those coequalizers.
Such a functor is clearly tripleable.
Proposition 1. Suppose U1 : A ’ B and U2 : B ’ C .
’ ’
(a) If U1 and U2 both satisfy CTT (respectively VTT) then so does U2 —¦ U1 .
(b) If U1 satis¬es CTT, U2 satis¬es PTT and U3 : C ’ D satis¬es VTT,


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