Let $F : \mathcal{A}\rightarrow\mathcal{B}$ be an additive functor of abelian categories, such that $F$ has cohomological dimension $\le n$. Suppose $\mathcal{A}$ has enough injectives. Let $P\subset\text{Ob}(\mathcal{A})$ be the set of objects such that $R^iF(X) = 0$ for all $X\in P$ and $i\ne n$, and suppose every object of $\mathcal{A}$ is a quotient of an object in $P$.


Let $G : \mathcal{A}\rightarrow\mathcal{B}$ be the functor $R^nF$. Since $F$ has cohomological dimension $\le n$, $G$ is right exact, and sends exact sequences of objects in $P$ to exact sequences.


Now let $X^\bullet$ be a complex of objects in $P$. Let $I^{\bullet,\bullet}$ be a Cartan-Eilenberg resolution.


Let's fix our attention on a single "column" of the resolution, which we'll write:
$$0\rightarrow X\rightarrow I^0\stackrel{d^0}{\rightarrow} I^1\stackrel{d^1}{\rightarrow} I^2\stackrel{d^2}{\rightarrow} \cdots$$
where $X\in P$, and $I^i$ are injective objects of $\mathcal{A}$, and is an injective resolution of $X$. Consider the truncated exact sequence:


$$0\rightarrow X\rightarrow I^0\rightarrow I^1\rightarrow\cdots\rightarrow I^{n-1}\rightarrow\text{Ker}(d^n)\rightarrow 0$$


At the top of page 78 (second part of the proof of Proposition I.7.4), Hartshorne claims that because $X\in P$, and $I^i$ are all injective, then $\text{Ker}(d^n)$ is $F$-acyclic.


This must be obvious, but I can't see why this is true. Any attempts to prove it get stuck when I realize that $\text{Ker}(d^n)$ is just some random subobject of an injective object.



Question about the proof of Prop I.7.4 in Hartshorne's Residues and Duality
This is a standard dimension shifting argument (*décalage*): split the long exact sequence
$$0 \to X \to I^0 \to I^1 \to \ldots \to I^{n-1} \to \ker(d^n) \to 0$$
into short exact sequences
$$0 \to J^i \to I^i \to J^{i+1} \to 0.$$
Here, $J^0 = X$ and $J^n = \ker(d^n)$. For each $k$, the associated long exact sequence gives
$$\ldots \to R^kF(I^i) \to R^kF(J^{i+1}) \to R^{k+1}F(J^i) \to R^{k+1}F(I^i) \to \ldots.$$
For $k>0$, the outer terms vanish, giving an isomorphism $R^kF(J^{i+1}) \stackrel\sim\to R^{k+1}F(J^i)$. By induction,
$$R^kF(J^n) = R^{k+1}F(J^{n-1}) = \ldots = R^{k+n}F(J^0).$$
This vanishes when $k+n \neq n$, i.e. when $k > 0$. $\square$


**Remark.** The only thing we use is that $R^iF(I) = 0$ for $i > 0$ and $I$ injective. For left exact functors, this is well-known. You should convince yourself it's still true without that assumption (the way Hartshorne sets it up, this is a little hidden, but it's still true more or less by definition). Similarly, the long exact sequence is never stated explicitly, but it's definitely still fine.


