2
SERGE BOUC
to a finite right-free biset U. The case U H, when G is a subgroup of H, is the
usual tensor induction from it^-modules to it!iJ-modules. There is no essentially
new construction here, since the other cases correspond to restriction and inflation
of modules.
It should be noted however that even in those well known cases, the formalism
of bisets gives a single natural framework involving restriction, inflation and tensor
induction. The classical properties of those construction, such as Mackey formula,
transitivity of tensor induction, or composition of inflation and tensor induction,
are various aspects of a single simple composition formula. Similarly, the formula
giving the tensor induction of a direct sum, which is generally incomplete (and
evaluated only "up to terms induced from proper subgroups"), is nothing but a
generalization of the binomial formula, and can be written explicitly.
2. Non additive exact functors
2.1. Notations. If M, T V and P are objects of an abelian category A, and
/ : M 0 N —» P is a morphism, I will denote / by (/ o iM, f o iN)1 where %M and %N
are the canonical injections from M and N to M 0 N. Similarly, if g : P M 0 N
is a morphism, I will denote it by
(*M°^),
where SM and s^ are the canonical
surjections from M 0 N to M and N. With those notations, the usual rules of
matrix multiplication apply.
The identity morphism of M will generally be denoted by 1, and by
ICIM
if
some precision is needed. The zero morphisms will be denoted by 0.
2.2. Definition. First I observe that the classical definition of an exact func-
tor actually implies additivity:
LEMMA
2.1. Let A and B be abelian categories, and F be a functor from A to
B, which is not supposed to be additive. Suppose that for any exact sequence in A
the associated sequence
M^N^L^O
F(M)
F^]
F(N)
F^]
F(L)
is exact in B. Then F is additive.
PROOF.
Note that I don't suppose that the second exact sequence is the image
by F of the first one. In other words, I don't suppose that F(0) = 0. But it is
a consequence of the exactness of the second sequence: indeed, as F(ip) o F(ip) =
F(ip o cf)) = F(0) has to be zero, and as the identity of the zero object factors
through any (zero) morphism, the identity of F(0) has to be zero. Hence F maps
the zero object to the zero object.
Let M and N be objects of A. Applying the hypothesis to the sequence
, 0 / (0,1)
shows that the sequence
F
F(M) ^ L F(M 0 N)
F(-0,
*\ F(N)
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