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.

1 Answer

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 $\operatorname{im}(f_{i-1})$.