2. NON ADDITIVE EXACT FUNCTORS
So 5 o A = 1, and in particular 5 is an epimorphism. So if # : G o F(N) — P is
any map in B, then
So D and D' have the same image, and the sequence (2.4) is exact. This completes
the proof of the proposition. •
2.3.2. Products and sums. If A and A! are abelian categories, then their prod-
uct A x A' is also abelian. If / : M — T V and / ' : M' — • TV' are maps in „4 and
*4', I will denote by
the associated couples in A x A', and
[/, /'] : [M, M'] - • [TV,
the associated morphism. The image of [M, M'} under
a functor F will be denoted by F[M,
instead of F([M, M']).
PROPOSITION 2.10. Let F : A -* B and F' : A! — B' be right exact functors
between abelian categories. Then
FxF' :AxA! -BxB'
is right exact.
This is obvious, since a product of exact sequences is exact. •
2.11. Let F,F' : A — B be right exact functors between abelian
categories. Then F 0 F is right exact.
PROOF. The functor F e F factors as
where A is the diagonal functor, mapping the object M to [M, M] and the morphism
/ to [/,/], and E is the direct sum functor mapping the object [P, Q] to P © Q,
and the morphism [/, g] to / 0 g. Those two functors are obviously additive and
exact, so the corollary follows from proposition 2.9. •
2.3.3. Pairings. If A, A', and B are abelian categories, a pairing F : Ax A' — B
is just a biadditive functor: for any object M of A, the functor F[M, — ] : A! — B
is additive, and for any object M' of A', the functor F[—, M'] : A — B is additive.
Note that F itself is not additive in general.
2.12. Let A, A!, andB be abelian categories, and F : Ax A! —• B
be a pairing. The following are equivalent:
1. The functor F is right exact.
2. For any objects M of A and M' of A!', the (additive) functors F[M, —] and
F[—,M'\ are right exact.
3. For any exact sequences
F[M, N'] © F[N, M'} 4 F[N, N'} -J11Z4 p[L, L'} - 0