to 1 + X + X 2 + X 3 + · · · by the map which takes each summand on the left to

the same thing on the right.

Lifting

If T1 and T2 are triples on a category C , a lifting of T2 to C T1 is a triple

T— = (T2 , ·2 , µ— ) on C T1 for which

— —

2 2

9.2 Distributive Laws 301

(L1) U T1 —¦ T2 = T2 —¦ U T1 ,

—

(L2) U T1 ·2 = ·2 U T1 : U T1 ’ U T1 —¦ T2 and

— —

’

(L3) U T1 µ— = µ2 U T1 : U T1 —¦ (T2 )2 ’ U T1 —¦ T2 —¦

— —

’

2

In understanding L2 and L3, observe that by L1, U T1 —¦ T2 = T2 —¦ U T1 and

—

U T1 —¦ (T2 )2 = T2 —¦ U T1 —¦ T2 = T2 —¦ U T1 .

— — 2

—

Note that given a functor T2 satisfying (L1), rules (L2) and (L3) determine

what ·2 and µ— must be, and the resulting T— = (T2 , ·2 , µ— ) is trivially a triple.

— — —

2 2 2

A lifting of T2 is equivalent to having a natural way of viewing T2 A as a T1

algebra whenever A is a T1 algebra, in such a way that ·2 and µ2 are algebra

morphisms.

—

Given a distributive law »: T1 —¦ T2 ’ T2 —¦ T1 , de¬ne T2 (A, a) = (T2 A, T2 a —¦ »A),

’

—

and de¬ne T2 to be the same as T2 on algebra morphisms.

T2 is the functor part of a lifting of T2 to C T1 .

—

Proposition 1.

Proof. (L1) is clear, so (L2) and (L3) determine ·2 and µ— .

—

2

—

These diagrams show that T2 (A, a) is a T1 algebra.

· 1 T2 A E

T2 A T1 T2 A

rr

rr

=

r2 ·1 A

T »A

rr

c j

rc

'

T2 A T2 T1 A

T2 a

T1 »A T1 T2 a

E E

2

T1 T2 A

T1 T2 A T1 T2 T1 A

»T1 A »A

c c

E T2 T1 A

2

µ 1 T2 A T2 T1 A

T2 T1 a

T2 µ 1 A T2 a

c c c

E T2 T1 A E T2 A

T1 T2 A a

»A

Furthermore,

—

·2 A —¦ a = T2 a —¦ »A —¦ T1 ·2 A

—

by (D1) and naturality, so ·2 is an algebra morphism, and

2 2

µ2 A —¦ T2 a —¦ T2 »A —¦ »T2 A = T2 a —¦ »A —¦ T1 µ2 A: T T2 A ’ T2 A

’

by D3 and naturality, so µ— is an algebra morphism.

2

302 9 More on Triples

Compatibility

The following de¬nition captures the idea that a triple with functor T2 —¦ T1 is

in a natural way the composite of triples with functors T2 and T1 .

The triple T = (T2 —¦ T1 , ·, µ) is compatible with triples T1 = (T1 , ·1 , µ1 ) and

T2 = (T2 , ·2 , µ2 ) if

(C1) · = T ·2 —¦ ·1 = ·1 T —¦ ·2 : id ’ T1 —¦ T2 —¦

’

2

(C2) µ —¦ T2 T1 ·2 T1 = T2 µ1 : T2 —¦ T1 ’ T2 —¦ T1 —¦

’

2

(C3) µ —¦ T2 ·1 T2 T1 = µ2 T1 : T2 —¦ T1 ’ T2 —¦ T1 —¦

’

2

(C4) µ2 T1 —¦ T2 µ = µ —¦ µ2 T1 T2 T1 : T2 —¦ T1 —¦ T2 —¦ T1 ’ T2 —¦ T1 , and

’

2

(C5) T2 µ1 —¦ µT1 = µ —¦ T2 T1 T2 µ1 : T2 —¦ T1 —¦ T2 —¦ T1 ’ T2 —¦ T1 —¦

’

Given triples T1 and T2 on a category C and a lifting T— on C T1 , we get

2

—

F2 F

’’ ’ C T1 ’’1’ C

’’

’’ ←’

’’’

—

(C T1 )T2 ←

—

U2 U1

— —

whence F2 —¦ F1 is left adjoint to U1 —¦ U2 , so

— — —

T2 —¦ T1 = T2 —¦ U1 —¦ F1 = U1 —¦ T2 —¦ F1 = U1 —¦ U2 —¦ F2 —¦ F1

produces a triple T = (T2 —¦ T1 , ·, µ) on C .

Proposition 2. With the notation of the preceding paragraph, the triple T is

compatible with T1 and T2 .

Proof. We will check one of the hardest compatibility conditions, namely C5, and

leave the rest to you. In the diagram, we have carried out the replacement of

——

U2 F2 U1 by U1 U2 F2 and replaced µ by its de¬nition so that the diagram that has

to be shown to commute is

—— ——

U1 U2 F2 F1 U1 U2 F2 1 F1

E U1 U2 F2 F1 U1 U2 F2 F1

—— —— —— ——

U1 U2 F2 F1 U1 U2 F2 F1 U1 F1

—— —— —— ——

U1 U2 F2 1 U2 F2 F1 U1 F1 U1 U2 F2 1 U2 F2 F1

c c

E U1 U2 F2 U2 F2 F1

———— ————

U1 U2 F2 U2 F2 F1 U1 F1 ————

U1 U2 F2 U2 F2 1 F1

U1 U2 — F2 F1 U1 F1

— —

U1 U2 — F2 F1

— —

2 2

c c

E U1 U2 F2 F1

———— ——

U1 U2 F2 U2 F2 F1 ————

U1 U2 F2 U2 F2 1 F1

9.2 Distributive Laws 303

Now suppose that the triple T is compatible with T1 and T2 . De¬ne »: T1 —¦ T2

’ T2 —¦ T1 as the composite

’

·TT· µ

T1 T2 ’ ’ ’ 2 ’ T2 T1 T2 T1 ’ ’ T2 T1

’2’1 ’ 1

’’ ’

Proposition 3. » is a distributive law.

Proof. D1 and D2 follow from this diagram (note that T · = T2 T1 ·2 ·1 ).

· 2 T1 E T2 T1

T1

d

d=

T1 · 2 T· d

d

d

‚

c c

E T2 T1 T2 T1 E T2 T1

T1 T2 µ

·2 T1 T2 ·1

To get D4, consider this diagram.

T1 ·2 T1 T2 ·1 E T1 µ E

2

T1 T2 T1

T1 T2 T1 T2 T1 T2 T1

2

·2 T1 T2 T1 T2 T1 ·2 T1 T2 T1

T1 T2 ·1

c c c

E T2 T1 T2 T1 T2 T1 E T2 T1 T2 T1

2

T1 T2 T1

·2 T1 ·2 T1 T2 T1 T2 T1 µ

µ

µ1 T2 T1 µT2 T1

c c c

E T2 T1 T2 T1 E T2 T1

T1 T2 T1 µ

·2 T1 T2 T1

where the top left square commutes by de¬nition, the top right square by natu-

rality, and the bottom right square by a triple identity. The bottom left square

is the following square applied to T2 —¦ T1 :

2

·2 T1 T2 T1 · 2 T1 E

E T2 T 2

2

T1 T2 T1 T2 T1

1

d d

µ1d µ

T2 µ1d

d d

‚

d ‚©

d

E T2 T1

T1

304 9 More on Triples

Now the left and bottom route around square (1) is » —¦ µ1 T2 by de¬nition of

» and the fact that

2

µ1 T2 T1 —¦ T1 T2 ·1 = T1 T2 ·1 —¦ µ1 T2

by naturality, and the top and right route is T2 µ1 —¦ »T1 —¦ T1 » because

·1 T1 T2 ·1 TE µT1 T1 T2 T1 µ1

E

1 2 2

T1 T2 T1 T2 T1 T2 T1 T2 T1

d

d

d

d

·2 T1 T2 T1 d T1 T2 T1 µ1 T2 µ1

d

d

d

dc

‚ c

E T2 T1

T2 T1 T2 T1 µ

The left triangle is a triple identity and the square is C5.

We leave the rest to you.

Exercise 9.2

1. Prove that for the constructive processes described in the text, all composites

of length three in the following triangle are the identity.

distributive Proposition 1 E liftings

laws

d

s

d

Proposition 3d Proposition 2

d

d ©

compatible

triple structures

9.3 Colimits of Triple Algebras

Theorem 1 of Section 3.4 gives what is perhaps the best possible result on the