12 SERGE BOUC
This morphism does not depend on the choice of a and 6, since if a' : P P'
is another map such that ^' o ar f o ^, then -0' o (a a') = 0, so there is a map
c : P - Q' such that a - a' = p' o c. Now
F(ac,)
is a map from F(P) to F(Q' © P'),
and
A F ^ ' ) ^ , ) = (F(V',l)-F(0,l))F(^=F(p'oc+a')-F(a') = F(a)-F(a')
This proves first that Fj- is well defined. So if F 1, as I can take a = 1 and 6 = 1 , 1
have F[ = 1. A similar argument shows that F ^ FjoF'g,so the correspondence
M H+ F'(M) and / h- Fj- is a functor from .4 to B.
Moreover if P is in P, then I can choose the following exact sequence for P
P-^ P - i P
The associated sequence is
F ( P © P)
A F
^
Q
^ F(P)
F
^ F'(P)
As AP(0) = 0, I have Ff{P) ~ F{P). Moreover, if / : P
P, then as the diagram
0
T-,
1
0
P' is a morphism in
P
/I
P '
P
/ i
P'
P
/ i
P'
is commutative, the following commutative diagram
AF(0) _ i
F(P®P) F(P) F(P)
f 0
0 /
F(f)
F'(f)
F{P'®P')
AF(-°\
F(P')
1
^ 0
0
shows that F'{f) = F(f), and this induces an isomorphism between F and the
restriction of F'toV. Finally, the diagram
0 , , 1
F(P)
F(Q ® P)
F(P)
1
F(P)
F'
^m
F{P]
zm
rw
0
0
shows that F^ = F ' O ) .
It remains to check that the functor F' is right exact: denote by
V M —• P M
M - + 0
the chosen resolution by objects of V for the object M of A. Suppose moreover as
before that if M P is in P, then this sequence is
P ^p 1+p
0
so that F\P) can be identified with F(P).
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