2. NON ADDITIVE EXACT FUNCTORS 11
Then there is a unique morphism 6 M F\{M) F2{M) completing this diagram
into a commutative one. This morphism does not depend on the choice of the
resolution 2.5: indeed, if
v v P'^M
is another resolution of M by objects of P, then as ^/ is an epimorphism and as
P is projective, there is a morphism a : P P' such that ip' o a = tp. Thus if
6'M : F\(M) » F2(M) is built using the second resolution, I have
«M?i W = ^ F i ( ^ ) ^ i ( a ) = F2W)0PF1(a) = ...
= W ) * 2 ( a ) 0
P
= F
2
0 / 0 # P = ^ M ^ i W
So #M = #M s m c e Fi(ip) is an epimorphism. Moreover if M = P is already in P, I
can choose the resolution P—P-^M—-0ofM, and then
#M = 0MFi{l) = F2(l)^p = #p
Now it is clear that the maps 6M giye a well defined natural transformation ex-
tending 6, and that this extension is unique.
To complete the proof of the theorem, I have to prove the existence of an extension
F' of F. For any object of A, I choose an exact sequence (2.5). Since F' must be
right exact, and coincide with F on V, the sequence
F(Q 0 P)
A J
M F(P)
F
^ \ F\M)
0
must be exact. So I can define F'(M) as the cokernel of AF(p). Of course, I must
make this definition functorial with respect to M.
I will show that if / : M M' is a morphism in *4, then there is a well defined
morphism Fi : F'(M) F'(Mf), which is moreover functorial with respect to / :
this follows from standard arguments on projective resolutions.
IfQ
/ v,
P'^M'
0 is any exact sequence with P' and Q' in P, then there
is a map a : P P ' such that ^ ' o a = / o ^ , because P is projective and i// is an
epimorphism. Now ip' o a o / ? = 0, so as Ken// = Imy/, and as Q is projective, there
is a map b : Q Q' such that (p' ob = ao ip.
Now I have the following diagram with exact rows
P(Q © P)
AF(^)
F(P)
F(a)
F{Q' © P '
A*V)
F(P')
J^'WO
F'ty/)
F'(M)
F'(M')
0
This diagram is commutative since
b 0
AP((^/)i?
0
( F ( ^ , 1 ) - F ( 0 , 1 ) ) F
F(p'ob,a)-F(0,a)
... = P(ao^a)-F(0,a ) = P(a) (V(, 1)-F(0,1) ) = F ( a ) o A % )
It follows that there is a morphism F} : F'{M) -+ F'(M') such that F'f o F'(V0 =
P'(?//)oF(a).
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