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of commutative diagram where we have ¬xed XG2 = (XT 2 , XT µ) in the ¬rst variable ”
—¦
G (XG2 , E) in AT
(XG2 , Y ) — (XG2 , E)
y y

±
±—±


G (XT, E)
(XT, Y ) — (XT, E) in A
—¦



The following elements match up under the vertical isomorphisms:

G
(a, y 0 s) 1y s


±
±—±


G sT · ψ
(a, ξs)

ξs and sT · ψ correspond to the compositions of s with 0 and 1 under adjointness (cf.
description of the nonhomogeneous coboundary operator in §2). Thus, nonhomogeneously,
G Y in A is (E, ψ)˜, then a is determined by the relation
if a: XT

G E.
a —¦ ξs = sT · ψ
(5) in the set of maps XT

Of course, (5) could be used to de¬ne a directly, since the two maps

ξ llllS X ‚‚‚‚ s
‚‚‚
l
llll ‚‚‚
A
llS E
XT ‚‚‚
lll
‚‚‚ l
‚ lll ψ
sT ‚A
ET
28 JONATHAN MOCK BECK

G X. The argument we sketched shows that
are equal after composition with p: E
this de¬nition of a is consistent with the earlier, homogeneous, de¬nition, in the not
necessarily tripleable case. Since a is a 1-cocycle, we have the coboundary zero relation
GY.
(aT · θ) —¦(ξT · a) = Xµ · a in the group of maps XT 2
G Z 1 (X, Y ) on maps is as follows (with Z 1 treated
The e¬ect of ˜: PO(X, Y )
G (E , ψ ), then the corresponding coboundary b =
nonhomogeneously). If f : (E, ψ)
G a is given by the formula
(f )˜: a
G E in A,
(6) b —¦ s = sf as maps X
where s and s are the sections of E and E .
Now we shall de¬ne the inverse functor

Z 1 (X, Y )
PO(X, Y ) o (relative to U T ).
˜’1

G Y is any 1-cocycle, let a˜’1 be the principal T-algebra given by X — Y
If a: XT
GX
as an object in A, with left Y -operation (in A) on the second factor, πX : X — Y
G X — Y as section, and the composition
as projection, (X, 1): X
ξ—θa
G XT — Y T GX —Y
π
(X — Y )T
as T-structure. Here π is induced by the projections, π = (πX T, πY T ), and ξ — θa is an
abbreviation which we shall use consistently for the map (t0 , t1 )(ξ—θa) = (t0 ξ, t1 θ —¦ t0 a) for
G XT , t1 : A G Y T in A. Diagrammatically, ξ —θa is the composition
any maps t0 : A
(ξ,a)—θ X—(π2 ,π1 )
GX —Y —Y GX —Y —Y GX —Y
X—mult.
XT — Y T
G (X, ξ) is indeed
One can verify easily that the projection πX : (X — Y, π(ξ — θa))
a T-algebra map, although the section (X, 1) is only a map in A (unless a = 1, or a
coboundary, of course, as follows from the theorem; if a = 1, then π(ξ — θa) is the T-
structure on the product algebra (X, ξ) — (Y, θ) ). The rest of the veri¬cation that a˜’1
is a (Y, θ)-principal T-algebra will be left till the end.
G a is a map of 1-cocycles, we set b˜’1 = X — Y —¦ b
If b: a
—¦b
a˜’1 = (X — Y, π(ξ — θa)) G (X — Y, π(ξ — θa )) = a ˜’1
X—Y


G X,
where X — Y —¦ b is given by the formula (t0 , t1 )(X — Y —¦ b) = (t0 , t1 —¦ t0 b) for t0 : A
G Y in A. We will sketch later a proof that this is a map of principal algebras.
t1 : A
We can now show that
G ˜˜’1 , ˜’1 ˜ = Z 1 (X, Y ) .
PO(X, Y )
p
G (X, ξ) be a (Y, θ)-principal algebra with section s: X G E. Then we
Let (E, ψ)
G X — Y in A,
know that E
(p,σ)
GX —Y GE

E
29
TRIPLES, ALGEBRAS AND COHOMOLOGY

˜˜’1 , all we have to do is show
where σ —¦ ps = E, (x, y)„ = y —¦ xs. To prove that PO
that the A-isomorphism
(p,σ)
G (X — Y, π(ξ — θa))
(E, ψ)
is a T-algebra map. Consider the diagram
(p,σ)T
G (X — Y )T
ET
ET π
 
(pT,σT )
G XT — Y T
ET
ψ ξ—θa
 
GX —Y
E (p,σ)

Since ψp = pT · ξ ( p is a T-algebra map) and
(pT, σT )(ξ — θa) = (pT · ξ, (σT · θ) —¦(pT · a))
G Y . We let them both
we have to prove that ψσ = (σT · θ) —¦(pT · a) as maps ET
GE :
operate on ψps: ET
(σT · θ) —¦(pT · a) —¦ ψps = (σT · θ) —¦(pT · a) —¦(pT · ξs)
= (σT · θ) —¦ pT · (a —¦ ξs)
= (σT · θ) —¦(pT · sT · ψ)
= (σT · θ) —¦((ps)T · ψ)
= (σ —¦ ps)T · ψ
=ψ ,
using Lemma 4, and
ψσ —¦ ψps = ψ(σ —¦ ps)

Veri¬cation of f ˜˜’1 = f : E G E will be left out; it also uses Lemma 4.
To prove ˜’1 ˜ = Z 1 (X, Y ), let a be a 1-cocycle. Then a˜’1 ˜ is determined by the
equation
(X, 1)T · π(ξ — θa) = a˜’1 ˜ —¦ ξ(X, 1) ,
(X, 1) being the section in a˜’1 . Computing, we have
(X, 1)T · π(ξ — θa) = (XT, 1T )(ξ — θa)
= (ξ, (1T · θ) —¦ a)
= (ξ, a)
30 JONATHAN MOCK BECK

G Y , the operation ( )T · θ preserving the group
because 1T · θ is the neutral map X
law (adaptation of Lemma 2 to the non-abelian case), whereas


ξ(X, 1) = (ξ, 1XT ),


by naturality of neutral maps. Manifestly, a˜’1 ˜ = a.
It remains, ¬nally, to prove that the values of ˜’1 are principal T-algebras and maps
thereof. We begin by showing that the proposed T-structure π(ξ — θa) on a˜’1 is asso-
ciative. This is done by computing both sides of the diagram


µ
G (X — Y )T
(X — Y )T T
π
πT
 
(XT — Y T )T XT — Y T
(ξ—θa)T ξ—θa
 
GX —Y
(X — Y )T π(ξ—θa)




We present the result for the left side in the table which follows. Each entry on the
right is the accumulated composition of the maps down to that stage.


(X — Y )T T
πT

(XT — Y T )T (πX T, πY T )T
(ξ—θa)T

(X — Y )T (πX T · ξ, (πY T · θ) —¦(πX T · a))T
π

XT — Y T (πX T T · ξT, ((πY T · θ) —¦(πX T · a))T )
ξ—θa

X —Y (πX T T · ξT · ξ, ((πY T · θ) —¦(πX T · a))T · θ —¦(πX T T · ξT · a))
31
TRIPLES, ALGEBRAS AND COHOMOLOGY

which is (πX T T ·ξT ·ξ, (πY T T ·θT ·θ) —¦(πX T T ·aT ·θ) —¦(πX T T ·ξT ·a)) by the multiplicative
property of ( )T · θ. The right side of the diagram is, similarly3 ,

(X — Y )T T
µ

(X — Y )T
π

XT — Y T (πX T T · Xµ, πY T T · Y µ)
ξ—θa

X —Y (πX T T · Xµ · ξ, (πY T T · Y µ · θ) —¦(πX T T · Xµ · a))

Since ξ and θ are associative, and a satis¬es the 1-cocycle identity, both sides of the
diagram give the same map. Thus the structure π(ξ — θa) is associative.
To prove the structure is unitary, consider the diagram

(X—Y )·
G (X — Y )T
X — Yu
uu
uu
uu
uu
uu
X·—Y · uuu
π
uu
u7 
XT — Y T

ξ—θa


X —Y

The composition ·π(ξ — θa) = (X· · ξ, (Y · · θ) —¦(X· · a)). Since ξ and θ are unitary,
G Y is the trivial map 1. (In a sense, the cocycle
it su¬ces to prove that X· · a: X
a is already “normalized”.) This follows from the following peculiar computation. Let
G XT T denote either of X· · XT ·, X· · X·T . Then
X··: X

X·· · Xµ = X· ,
X·· · aT · θ = X· · XT · · aT · θ
= X· · a · Y · · θ
= X· · a ,
X·· · ξT = X· · X·T · ξT
= X·.

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