Right derived functor and $F$-acyclic objects I need help regarding a question that came up connected to an exercise (I.19 on page 75 in Sheaves on Manifolds by Kashiwara and Schapira) I am currently working on.
Given a left-exact functor $F: \mathcal{A}\rightarrow\mathcal{B}$ between abelian categories, suppose its derived functor $RF$ exists. Then the full subcategory $\mathcal{J}$ of $F$-acyclic objects (i.e. those with $R^k F(X)=0$ for $k\neq 0$) forms an $F$-injective subcategory, hence we can compute $RF$ on complexes in $\mathcal{J}$.
Suppose now $F$ has cohomological dimension $n$. I want to show that, given an exact sequence
$$0\rightarrow X_0\rightarrow \cdots \rightarrow X_n\rightarrow 0$$ in $\mathcal{A}$ with $X_i\in \mathcal{J}$ for $i<n$, then $X_n$ is in $\mathcal{J}$ as well.
I came up with a solution that does not seem right to me, but I cannot find out where I'm wrong:
Since $\mathcal{J}$ is $F$-injective, the sequence
$0\rightarrow X_0\rightarrow\cdots\rightarrow X_n\rightarrow 0$ defines an $F$-injective resolution for $X_n$, i.e. a quasi-isomorphism $X_\ast \rightarrow X_n$ in $K^+(\mathcal{A})$ where 
$$X_{\ast} =[ \cdots\rightarrow 0\rightarrow X_0\rightarrow\cdots\rightarrow X_{n-1}\rightarrow 0\rightarrow \cdots ]$$
(here $X_{n-1}$ is at position $0$).
Now $R^kF(X_n)=R^kF(X_\ast)=H^kF(X)=0$ for $k>0$, hence $X_n$ is $F$-acyclic as $R^kF(X_n)=0$ for $k<0$ is always true in this case.
My problem is that I did not even use the fact that $F$ has finite cohomological dimension, so I don't think my argument is right.
(Note that in the book, the above sequence is not assumed to be exact on the left, however in class the situation was presented like I wrote it down. The context is the proof of the Poincaré-Verdier duality.)
 A: I don't see any mistakes in your proof. In fact, here is another one which does not use hyperderived functors (so without the fact that $R^kF(X_*)=H^kF(X_*)$ for a complex of acyclic objects) :
Claim : If $X_0\subset X_1$ are $F$-acyclic, then so is $X_1/X_0$. This follows from the long exact sequence associated to the short exact sequence $0\rightarrow X_0\rightarrow X_1\rightarrow X_1/X_0\rightarrow 0$.
Now, in your situation : $0\rightarrow X_0\rightarrow X_1\rightarrow ...\rightarrow X_{n-1}\rightarrow X_n\rightarrow 0$, if every $X_i$ except maybe $X_n$ is acyclic, then so is $X_n$. Indeed, you can split your sequence in short ones :
$$ 0\rightarrow X_0\rightarrow X_1\rightarrow \operatorname{Im}d^1\rightarrow 0$$
$$ 0\rightarrow \operatorname{Im}d^1\rightarrow X_2\rightarrow \operatorname{Im}d^2\rightarrow 0$$
$$ 0\rightarrow \operatorname{Im}d^2\rightarrow X_3\rightarrow \operatorname{Im}d^3\rightarrow 0$$
$$...$$
$$ 0\rightarrow \operatorname{Im}d^{n-2}\rightarrow X_{n-1}\rightarrow X_n\rightarrow 0$$
Now by the claim, $\operatorname{Im}d^1$ is acyclic as a quotient of two acyclics, then $\operatorname{Im}d^2$ is acyclic as a quotient of two acyclics, and so on... thus $X_n$ is also acyclic.

However, as you have noted, the sequence is not assumed to be exact on the left. And this is the crucial point ! So let $X_0\rightarrow X_1\rightarrow ...\rightarrow X_{n-1}\rightarrow X_n\rightarrow 0$ be a long exact sequence such that each $X_i$ are acyclic except maybe $X_n$ and assume that $F$ has cohomological dimension $\leq n$. Write $K$ for the kernel of $X_0\rightarrow X_1$, you can again split the long exact sequence into short ones : 
$$0\rightarrow K\rightarrow X_0\rightarrow\operatorname{Im}d^0\rightarrow 0$$
$$0\rightarrow \operatorname{Im}d^0\rightarrow X_1\rightarrow\operatorname{Im}d^1\rightarrow 0$$
$$...$$
$$0\rightarrow \operatorname{Im}d^{n-2}\rightarrow X_{n-1}\rightarrow X_n\rightarrow 0$$
Let $k>0$, then from the long exact sequences you get :
$$0=R^kF(X_{n-1})\rightarrow R^k F(X_n)\rightarrow R^{k+1}F(\operatorname{Im} d^{n-2})\rightarrow R^{k+1}F(X_{n-1})=0$$
$$0=R^{k+1}F(X_{n-2})\rightarrow F^{k+1}F(\operatorname{Im}d^{n-2})\rightarrow R^{k+2}F(\operatorname{Im}d^{n-3})\rightarrow R^{k+2}F(X_{n-2})=0$$
$$...$$
$$0=R^{k+n-1}F(X_0)\rightarrow R^{k+n-1}F(\operatorname{Im}d^0)\rightarrow R^{k+n}F(K)\rightarrow R^{k+n}F(X_0)=0$$
Thus, you have isomorphisms $$R^kF(X_n)\overset{\sim}\rightarrow R^{k+1}F(\operatorname{Im}d^{n-2})\overset{\sim}\rightarrow...\overset{\sim}\rightarrow R^{k+n-1}F(\operatorname{Im}d^0)\overset{\sim}\rightarrow R^{k+n}F(K)$$
But the last object is $0$ since $k+n>n$ and $F$ has cohomological dimension $\leq n$. So $R^kF(X_n)=0$, and this holds for every $k>0$. Therefore $X_n$ is acyclic.

Another proof closer to what you had in mind, but requiring the use of hyperderived functors :
Again, let $k>0$, $K$ be the kernel of $X_0\rightarrow X_1$ and $X_*$ be the complex $X_0\rightarrow X_1\rightarrow...\rightarrow X_{n-1}$. You have a short exact sequence of complexes :
$$ 0\rightarrow K[n-1]\rightarrow X_*\rightarrow X_n[0]\rightarrow 0$$
Thus, you have a long exact sequence 
$$ R^kF(X_*)\rightarrow R^k(X_n)\rightarrow R^{k+1}F(K[n-1])\rightarrow R^{k+1}F(X_*)$$
As you said in your post, $R^kF(X_*)=H^kF(X_*)=0$ since $k>0$ and $X_*$ is a complex of acyclics in non positive (cohomological) degree. The same holds for $R^{k+1}F(X_*)=0$. Thus, you have an isomorphism $$R^k F(X_n)\overset{\sim}\longrightarrow R^{k+1}F(K[n-1])=R^{k+n}F(K)$$
But this last object is zero because $F$ is of cohomological dimension $\leq n$.
