2. NON ADDITIVE EXACT FUNCTORS
Let i : [M, iV'] - [M 0 JV, M' 0 AT'] be the map in A x A' defined by i
[Q.ffl]
J
=
and
Similarly, let j : [AT, M'] - [M ® N,M' ® N'} be the map defined by
;?), (J)]. Let A : t/ - V defined by ^ = ( F ( I ) , F ( J ) ) .
Define moreover k : [M ®N,M' 0 AT'] - [M, N'} by fc = [(1,0), (y/, 1)
Z : [M © TV, M' 0 AT'] - [N, M'] by Z = [(0,1), (1,0)1. Let B : V -* U defined by
£ = $ * ) . Now
D o B = ( % l ] , F [ l ,
v
' ] ) o Q = . . .
... = F([
( P
,1][(1,0),(^,1)])+F([1,^][(0,1),(1,0)])=.. .
. . . = F ( [ ( ^ , 0 ) , ( Y / , 1 ) ] ) + F ( [ ( 0 , 1 ) , ( ^ , 0 ) ] )
Since F is biadditive, this is also equal to
F([{p, 0),
{ip1,0)])
+ F([{p, 0), (0,1)]) + F ( [ ( 0 , 1 ) , (*', 0)])
On the other hand, the map D' = AF[(/?, y/] is equal to
AF[p, p'] = F [(^, 1), (^, 1)] - F[(0,1), (0,1)] = ...
... = F([{p, 0), (,', 0)]) + F([(^, 0), (0,1)]) + F([(0,1), fa', 0)])
Thus D o B = D'. Moreover
BoA_(F(k)\(F{i)F()\(F(koi) F(koj
)
But
k o i
So F(k o i) = 1. Similarly
(1.0), (vM)
M
fc°i=[(l0),(^l)][(°).Q]
So F(k o j) = 0, since F is biadditive. Moreover
( 0 , 1 ) , ( 1 , 0 ) ' "
A / 0
(0,1), (1,0)
°J
0/
V1
J).G
= [i.i]
= [0,¥']
= [0,0]
= [1,1]
Finally, I have
5 o i
1 0
0 1
1
In particular B is an epimorphism. As D o B D\ the images of D and D' are
the same. So 3) implies 1), and this completes the proof.
COROLLARY
2.13. Let [M,N] *-» M ® N be a (biadditive) tensor product from
B x B to an abelian category C, which is right exact with respect to M and N. Then
1. If F and F' are right exact functors from A to B, so is F S F' : A C.
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