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