I feel like this question has surely already been asked, but I wasn't able to find a formulation which made my search fruitful, so here goes. Let $C, D$ be two abelian categories, and F a (left/right exact) functor between them, admitting a sequence of (right or left) derived functors $F_n$.
Initially I was almost sure $F(A)=0$ implied that $F_n(A)=0, \forall n$. However trying to write down an explicit proof yielded some difficulties, and now I am not even sure it is true.
My idea is as follows, WLOG suppose we are deriving on the left, the right case being dual. Let $0\to A \to B\to C\to 0$ be any short exact sequence (and let $A$ be as above an object such that $F(A)=0$). Now assuredly, applying F yields a short exact sequence $0\to 0\to F(B)\to F(C)\to 0$ (because F is right exact, and any map from the 0 object is injective). Because of the intution that deriving functors measure failure of exactness, in this case exactness being preserved, the long exact sequence should be pieces of the form $0\to F_n(B)\to F_n(C)\to 0$.
I am however not managing to translate this sentiment rigorously (I am not even quite sure what I would want to write). Any help is appreciated, thanks a lot.
This is simply not true. $\mathrm{Hom}_\mathrm{Ab}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})=0$, while $\mathrm{Ext}^1_\mathrm{Ab}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})=\mathbb{Z}/n\mathbb{Z}.$ In your idea, the issue is that there is absolutely no reason to expect that if at first the long exact sequence looks like $0\to F^0(B)\to F^0(C),$ then it will look like $0\to F^i(B)\to F^i(C)\to 0$ later on.