I am trying to prove the following statement:
Let $F: \mathcal{A} \to \mathcal{B}$ be a left-exact functor between abelian categories. Suppose $\mathcal{A}$ has enough injectives. Then the right derived functors $R^iF : \mathcal{A} \to \mathcal{B}$ for all $i$ are additive functors.
I am totally lost and do not know where to start. The main problem is in two parts:
First, I'm sure I have to induce some kind of morphism between the injective resolutions given some $f: A \to A^\prime$ in $\mathcal{A}$, say $f^\bullet: I^\bullet(A) \to I^\bullet(A^\prime)$ between injective resolutions. For the first term $f^0 : I^0(A) \to I^0(A^\prime)$, I was able to induce a morphism using the fact that $\epsilon : A \to I^0(A)$ is injective, and $\epsilon^\prime \circ f : A \to I^0(A^\prime)$ is a morphism, and $I^0(A^\prime)$ is by definition an injective object in $\mathcal{A}$ so there must exists some $f^0 : I^0(A) \to I^0(A^\prime)$ that makes the first square commute. But how do I extend this to $f^i: I^i(A)\to I^i(A^\prime)$ for $i \geq 1$?
The second problem is, that I think I have to use the additive properties of the functor $F$, but I am not sure whether functors between abelian categories have to be additive.
Thanks in advance!
For the first part, you can construct $f^\bullet$ by induction. The proof is more natural with left derived functors and projectif resolution, but it is dual, so let's do this.
You have $f: A \to A'$ and $\epsilon' \circ f: A \to I^0(A')$. You also have a monomorphism $\epsilon: A \to I^0 (A)$, so $\epsilon' \circ f$ factorises through $\epsilon$, i.e. there exists $f^0: I^0(A) \to I^0(A')$ such that $$f^0 \circ \epsilon = \epsilon' \circ f$$
This first factorisation is easy because we have zeroes on the left, let us now suppose we have $f^0, \dots ,f^n$ well defined, how do we define $f^{n+1}$?
We have a factorisation of $\epsilon: I^n(A) \to I^{n+1}(A)$ given by $I^n(A) \to \operatorname{coIm} \epsilon = \operatorname{Im} \epsilon \to I^{n+1} (A)$ where the morphism on the left is epic (p) and on the right is monic (i). By properties of the coimage, $\epsilon' \circ f^{n}$ factorises through $\operatorname{coIm} \epsilon$ so we have $$\tilde{f^{n}} \circ p = \epsilon' \circ f^n: I^n(A) \to I^{n+1}(A')$$ Then since $i: \operatorname{coIm} \epsilon \to I^{n+1}(A)$ is monic and its target is injective, we can factorize $\tilde{f^n}$ to get $$\tilde{f^n} = f^{n+1} \circ i$$ This $f^{n+1}$ is the suitable candidate and this proves our recursion.
For your second problem, feel free to observe that if we have $f$ and $g$ and one candidate for each $f^\bullet$ and $g^\bullet$ then $f^\bullet + g^\bullet$ will be a suitable candidate for $f+g$.
Furthermore, if you have two suitable candidates $f^\bullet$ and $g^\bullet$ for $f$, both will be homotopic.
From now on you can construct an additive functor $$I_R : \mathcal{A} \to \mathcal{K}^{\geq 0} ( \operatorname{Inj} \mathcal{A})$$ that associates to $A$ its injective resolution and to $f$ the class of the chain complex morphism $[f^\bullet]$.
Then if you have an additive functor $F: \mathcal{A} \to \mathcal{B}$, you can extend it to $\bar{F}:\mathcal{K}^{\geq 0} ( \operatorname{Inj} \mathcal{A}) \to \mathcal{K}^{\geq 0} ( \mathcal{B})$
Your then have $R^k F = H^k \circ \bar{F} \circ I_R$
Which is a composition of additive functors!! (Hence additive)