I have read that for abelian categories P, Q, it is not true in general that every additive functor $F: P \rightarrow Q$ is exact. However, I also read that if P is taken to be the category of vector spaces over a base field k, then the statement becomes true.
- Could you give me a hint why this extra constraint makes a difference? edit B/c all short exact sequences are split.
As a first step, I tried to show that if $0 \rightarrow A \xrightarrow{\text{f}} B \xrightarrow{\text{g}} C \rightarrow 0$ is a short exact sequence in P, then $F(f): F(A) \rightarrow F(B)$ is a monomorphism. I guess there should be a way of exploiting the fact that $f: A\rightarrow B$ is a monomorphism itself, but I couldn't make any progress in this direction.
- Is this way of approaching the problem viable? Could you give me a hint on how to proceed?
Edit
- I could show that (i) F(f) is a monomorphism, (ii) F(g) is a epimorphism, but I'm stuck on proving that (iii) Im F(f) = Ker F(g). Could you please provide me with a hint for this part?
A short exact sequence $0\to A\to B \to C\to 0$ can be extended to a biproduct diagram by adding appropriate maps $C\to B\to A$ if and only if it is split. Furthermore, the restriction of a biproduct diagram $A\leftrightarrows A\oplus C \leftrightarrows C$ to the rightward-pointing arrows is always (split) short exact. So given a short exact sequence of vector spaces, choose an extension of it to a biproduct diagram. The property of being a biproduct diagram is determined by equations preserved by all additive functors, which is why they preserve biproducts; thus the image in $Q$ is still a biproduct diagram, which restricts to a split short exact sequence which is the image of your original sequence.