Proof that derived functors don't depend on choice of resolution. Can somebody help me out with this? Let $X$ be an object in an abelian category $A$ with enough injectives, let $0 \rightarrow X \rightarrow M^{\bullet}$ be an injective resolution , let $0 \rightarrow X \rightarrow N^{\bullet}$ be another injective resolution , let $F:A \rightarrow B$ (where $B$ is an abelian category) be a left exact functor, how does one go about proving that both resolutions induce the same cohomology groups $R^iF(X)$?
 A: The proof uses a lemma known as "The Fundamental Lemma of Homological Algebra" or "The Comparison Theorem for Resolutions."

Lemma. Let $\mathcal{A}$ be an abelian category with enough injectives, $X \longrightarrow I^\bullet$ an injective resolution of $X \in \mathrm{Ob}(\mathcal{A})$, and $f \in \mathrm{Hom}_{\mathcal{A}}(Y, X)$. Then for every resolution $Y \longrightarrow J^\bullet$, there is a cochain map $\tilde{f}: J^\bullet \longrightarrow I^\bullet$ lifting $f$. Furthermore, $\tilde{f}$ is unique up to chain homotopy equivalence.

It follows that if $X \longrightarrow I^\bullet$ and $X \longrightarrow J^\bullet$ are two injective resoutions of the same object, $I^\bullet$ and $J^\bullet$ are chain homotopy equivalent. Hence the cohomologies $H^i(F(I^\bullet))$ and $H^i(F(J^\bullet))$ are canonically isomorphic.
The proof of the fundamental lemma is slightly lengthy (but not too difficult). See Theorem 2.2.6 of Weibel's An Introduction to Homological Algebra or Theorem 2.22 of Davis and Kirk's Lecture Notes in Algebraic Topology for a proof.
