2. NON ADDITIVE EXACT FUNCTORS 3
is exact. But the morphism F(Q) is a split monomorphism, and F(0,1) is a split
epimorphism. This proves that there are inverse isomorphisms
iM,N = ( £ [ £ Jj) : F(M ®N)^ F(M) © F(N)
3M,N
= (F Q , F ( J ) ) : F(M) 8 F(JV) - F( M 0 iV)
Now if f,g : M N are morphisms in .4, their sum f+g is obtained by composition
i \ / / o
1/ V ° a 7 (i,i)
Taking images by F gives the commutative diagram
F(1\
F(*
°
J / 1 0 ^ / F ( l , l )
F(M) ^ F{M®M) —± U F(N®N) v ' F(JV)
iMM 3MM lN,N 3N,N
1
F(M)
A
F{M)®F{M) —¥!_ F(N)®F(N) —^—» F(7V)
where the bottom row is obtained through the previous isomorphisms. Thus
' 1 \ / F ( l , 0 ) W l \ (F{1)\ (V
A = iM MoF\l)
=
{F(0,l))F\lJ
\F(l)J VI
and similarly S = (1,1). Moreover
- » ( ^ M ; K
0)
It follows that
/F(1,0)\ / / A / 0 \ N _ / F(/ ) F(0) \ _ / F(f) 0
^ ~ v/(0,1),/ r \o) *\g))-{ F(0) F(g) ) " V 0 F(fl)
Now the composition £ o p o A is equal to F(f) + F(g). It is also equal to F(f + g),
so F is additive.
I will modify the definition of right exactness to extend it to non-additive functors.
First I need the following notation:
NOTATION
2.2. If ip : M T V is a morphism in ^4, I can build the morphisms
(/?, 1) and (0,1) from M 0 N to N. So I have morphisms F(ip, 1) and F(0,1) in B
from F(M 0 N) to F(N). I denote by AF(£) their difference
AF(tp) = F(ip, 1) - F(0,1) : F{M 0 i V ) ^ F(N)
If ^ N L is a morphism in *4 such that ^ o £ ? = 0, then of course
F(^) o AF(^) = F(V O (y,, 1)) - F ( V O (0,1)) = 0
since ij; o (/?, 1) = (^ o £,V) = (0,^0 = V ° (0,1). This leads to the following
definition:
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