8 SERGE BOUC
PROOF .
Suppose that F is right exact. Fix an object M' of A!. Now the
functor F[-,
Mf]
: M i- F[M, M'} factors as
F [ - M
,
] = F O ( K
/
X C A / / ) O A
where A is the diagonal functor A A x A as above, and CM is the constant
functor, equal to M' everywhere. So F[—, M'] is composed of three right exact
functors, hence it is right exact by proposition 2.9. A similar argument shows that
F[M, —] is right exact for any object M of A, so 1) implies 2) (note that this does
not depend on the fact that F is a pairing).
Now if 2) holds, and if
are exact sequences, I have the following commutative diagram
F[M,M'] - F[M,N'] - F[M,L'\ - 0
1 ih 1
F[N,M'\ -^
F[N,Nf]
X F[N,L'] - 0
1/ la id
F[L,M'] A F[L,N'] A F[L,L'} - 0
0 0 0
where a = F[^, 1],... h F[tp, 1]. The rows and columns of this diagram are exact.
To prove that 3) holds, I must show that the sequence
F[M,N'] 0 F[N,Mr] (h,9\ F[N,Nf] C°a ; F[L,L'] - 0
is exact. But c o a is the product of two epimorphisms, so it is an epimorphism.
And if 0 : F[N, N'} - P is any map in B such that 6 o (h,g) = 0, then 0oh = O
and 0 o g = 0. As the middle column is exact, the map 6 factors as 0 = 0' o a. Now
As / is an epimorphism, this gives 0' o e = 0, and as the bottom row is exact, the
map 0' factors as 0' = 0" oc. So 0 = 9" ocoa, and the image of (h,g) is the kernel
of c o a. So 2) implies 3).
Now suppose 3) holds. Let
be exact sequences in A and A!. The sequence
(F[p,i\,F[i,p'])
F W
^
U = F[M, TV'] © F[AT, M'] ^ 4 F[JV, W]
l
F[L, L'] - 0
is exact, and to prove 1), I must show that the sequence
V = F[M ®N,M'® N'} ^ F[N, N'] ^ \ F[L, L'} - 0
is exact. I will set D = (F[p, l],F[l,p']).
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