# Equivalent definition of exactness of functor?

I'll use the following definition:

(Def) A functor $F$ is exact if and only if it maps short exact sequences to short exact sequences.

Now I'd like to prove the following (not entirely sure it's true but someone mentioned something like this to me some time ago):

Claim: $F$ is exact if and only if it maps exact sequences $M \to N \to P$ to exact sequences $F(M) \to F(N) \to F(P)$

Proof:

$\Longleftarrow$: Let $0 \to M \to N \to P \to 0$ be exact. Then $0 \to M \to N$, $M \to N \to P$ and $N \to P \to 0$ are exact and hence $0 \to F(M) \to F(N)$, $F(M) \to F(N) \to F(P)$ and $F(N) \to F(P) \to 0$ are exact. Hence $0 \to F(M) \to F(N) \to F(P) \to 0$ is exact.

$\implies$: This is direction I'm stuck with. I am trying to do something like this: Given $M \to N \to P$ exact, we have that $0 \to ker(f) \to M \to im(f) \to 0$ is exact. Hence $0 \to F(ker(f)) \to F(M) \to F(im(f)) \to 0$ is exact. Then I want to do this again for the other side of the sequence and stick it back together after applying $F$ to get the desired short exact sequence.

How does this work? Perhaps I need additional assumptions on $F$? Thanks for your help.

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This equivalence helps to explain why we have only the term exact rather than the two terms short exact, long exact. On the other hand it is noteworthy that right-exact is a distinct term for a reason, the well known example $\otimes$ shows that it is weaker for a functor to preserve right-exact sequences than it is for it to preserve exact sequences. – Karl Kronenfeld May 20 '13 at 6:35

Any exact sequence can be broken down into short exact sequences (the $C_i$ are kernels/images):
So, since your functor $F$ preserves short exact sequences, you can apply $F$ and the diagonal sequences will remain exact. It's now a general fact that in any such diagram, if the diagonals are exact, then the middle terms are exact as well (by diagram chasing).
EDIT: If $f_i\colon A_i\to A_{i+1}$, then $C_i=\ker(f_i)$ which by exactness is isomorphic to $\operatorname{im}(f_{i-1})$.
Would you mind being more explicit about what the $C_i$'s are? – azarel Jun 13 '12 at 14:37
I think $C_i$ is the kernel of $A_{i} \to A_{i+1}$ (which is also the image of $A_{i-1} \to A_i$ for $i > 1$) – Joel Cohen Jun 13 '12 at 14:43
$C_7$ is a cokernel, by the way. You probably don't need this anywhere else. – Dylan Moreland Jun 13 '12 at 15:04