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).