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E UB
UA C
s
st s
in C .
112 3 Triples
Let U : B ’ C be tripleable and
Proposition 3. ’

d0
B ’’ B
’’ (8)
’’
’’
d1
be a U -contractible coequalizer pair. Then (8) has a coequalizer d: B ’ B in

B and
U d0 Ud
U B ’ ’ ’ U B ’’
’’ ’’ U B (9)
’’’
’’
1
Ud
is a coequalizer in C .
Before beginning the proof, we need some terminology for commutative dia-
grams. A diagram like
eE
A EB
f

k l
gEc
c
EB
A
h
is said to commute serially if g —¦ k = l —¦ e and h —¦ k = l —¦ f . (The arrows are
matched up in accordance with the order they occur in the diagram). In more
complicated diagrams the analogous convention will be understood. For example,
f0 E
A EB
f1
k0 k1 l0 l1
g0
c c Ec c
EB
A
g1
will be said to commute serially if gi —¦ kj = lj —¦ fi for all four possible choices of
i, j = 0, 1.
Proof. Proof of Proposition 3. (This proof was suggested by J. Beck). As in
Proposition 1, let B = C T . Thus we are given a pair

d0
(C , c ) ’ ’ (C, c)
’’ (10)
’’
’’
d1
3.3 Tripleability 113
and we suppose
d0 d
1‚ (11)

dE
C C C
ss
st
is a contractible coequalizer in C . Then by Proposition 2(b), all three rows of
the following diagram (in which we have not yet de¬ned c ) are contractible
coequalizers.
T T d0 E TTd E
E TTC
TTC TTC
1
TTd
µC
µC µC
Tc
Tc Tc
cc Ec c cc
T d0 Td E TC (12)
E TC
TC
T d1
c
c c
c c c
d0 EE EC
EC
C
d
d1
The lower left square commutes serially because d0 and d1 are algebra maps
by assumption. This implies that d —¦ c coequalizes T (d0 ) and (T (d1 ). Using the
fact that the middle row is a coequalizer, we de¬ne c to be the unique arrow
making the bottom right square commute. By the proof of Proposition 2(a) (“To
get the induced map, compose with the contraction”) we have
c = d —¦ c —¦ Ts
a fact we will need in the proof of Proposition 4.
We ¬rst prove that c is a structure map for an algebra. The upper right square
commutes serially, the square with the µ™s because µ is a natural transformation,
and the one with T c and T c because it is T of the bottom right square. Using
this and the fact that c is a structure map, we have
T 2d = c T 2d
c µc Tc
—¦ —¦ —¦ —¦


so, since T 2 d is an epimorphism (it is split),
c Tc = c µc
—¦ —¦


The unitary law for algebras is obtained by replacing the top row of (12)
by the bottom row and the vertical arrows by ·c , ·c, and ·c , and using a
114 3 Triples
similar argument. We know d is an algebra map because the bottom right square
commutes.
Finally, we must show that (C , c ) is a coequalizer, so suppose f : (C, c) ’’
(E, e) coequalizes d0 and d1 . Then because C is their coequalizer in C , there is
a unique arrow u: C ’ E for which f = u —¦ d. We need only to prove that the

right square in
T uE
T dE
TC TC TE

c e
c
c c c
EC EE
C u
d
commutes. This follows from the fact that the left square and the outer rectangle
commute and T d is epi.
Note that we have actually proved that U T creates coequalizers of U T -contracti-
ble pairs.

Algebras are coequalizers

A parallel pair is said to be re¬‚exive if the two arrows have a common right
inverse. A re¬‚exive coequalizer diagram is a coequalizer of a re¬‚exive parallel
pair.
Let T be a triple on C . Then for any (A, a) in C T ,
Proposition 4.
µA
’ ’ ’ (T A, µA)
’’
(T T A, µT A) ’ ’ ’ (13)
’’
Ta
is a re¬‚exive U -contractible coequalizer pair whose coequalizer in C is (A, a).
Proof. The associative law for µ implies that µA is an algebra map, and the
naturality of µ implies that T a is an algebra map. It follows from the identities
for triples and algebras that
µA a
Ta ‚ (14)

E TA
TTA A
·A
·T A s
s
is a contractible coequalizer. Thus by Proposition 3, there is an algebra structure
on A which coequalizes µA and T a in C T . As observed in the proof of that
proposition, the structure map is
a —¦ µA —¦ T ·A = a
3.3 Tripleability 115
as claimed.
The common right inverse for T a and µA is T ·A:

T a —¦ T ·A = T (a —¦ ·A) = T (id) = id

by the unitary law for algebras and

µA —¦ T (·A) = U ( F A —¦ F ·A) = U (id) = id

by Exercise 15, Section 1.9. The naturality of µ implies that T · is an algebra
map.
Corollary 5. If, given
U
B ←’
’’ C
’’ (15)
’’
F
F is left adjoint to U , then for any object B of B,

FUB
FUFUB ’ ’ ’ ’ FUB
’’’ (16)
’’’’
’’’
FU B
is a re¬‚exive U -contractible coequalizer pair.
Proof. U F is an algebra structure map by Exercise 1 of Section 3.2 and the
de¬nition of C T . Thus by Proposition 4, it is the coequalizer of the diagram
underlying (16). The common right inverse is F ·U .

Beck™s Theorem

We will now prove a number of lemmas culminating in two theorems due to
Beck. Theorem 9 characterizes functors U for which the comparison functor is
full and faithful, and Theorem 10 characterizes tripleable functors.
In the lemmas, we speak of an adjoint pair (15) with associated triple T in C
and cotriple G in B.
Lemma 6. The diagram

·T C
·C ’ ’ ’ T 2C
’ ’’
C ’’’’ T C ’ ’ ’ (17)
’ ’’
·T C

is an equalizer for every object C of C if and only if ·C is a regular mono for all
objects C of C .
116 3 Triples
Proof. If (17) is an equalizer then obviously ·C is regular mono. Suppose ·C is
regular mono; then we have some equalizer diagram of the form

d0
·C
’’ T C ’ ’ C
’’
C ’’ (18)
’’
’’
d1
It is su¬cient to show that for any element w of T C, d0 and d1 agree on w if
and only if ·T C and T ·C agree on w.
(i) If d0 —¦ w = d1 —¦ w, then w = ·C —¦ g for some g, so

·T C —¦ w = ·T C —¦ ·C —¦ g = T ·C —¦ ·C —¦ g = T ·C —¦ w

(ii) If ·T C —¦ w = T ·C —¦ w, then T ·T C —¦ T w = T 2 ·C —¦ T w. But

T ·T C
T ·C ’ ’ ’ ’ T 3C
’’’
2
TC ’’ ’ T C ’ ’ ’ ’
’ ’’ (19)
’’’
T 2 ·C

is a contractible equalizer, with contractions µC and µT C. Thus T w = T ·C —¦ h
for some h, so that T d0 —¦ T w = T d1 —¦ T w. Hence in the diagram below, which
commutes because · is a natural transformation, the two composites are equal.

d0 E
wE
EC
C TC
1
d
·T C (20)
·C ·C
c c T d0E c
E TTC E TC
TC
Tw 1
Td
It follows from the fact that ·C is (regular) mono that d0 —¦ w = d1 —¦ w.
Dually, we have
B is a regular epi for every object B of B if and only if

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