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: