I'm reading through the section on derived functors in Lang's Algebra and I came across this explanation of the natural isomorphism $H^i(F(I_M)) = H^i(F(J_M))$ for two different injective resolutions $0\rightarrow M\rightarrow I_M$ and $0\rightarrow M\rightarrow J_M$:
Given an object $M$, let $$ 0 \to M \to I^0 \to I^1 \to I^2 \to $$ be an injective resolution, which we abbreviate by $$ 0 \to M \to I_M, $$ where $I_M$ is the complex $I^0 \to I^1 \to I^2 \to$. We let $I$ be the complex $$ 0 \to I^0 \to I^1 \to I^2 \to $$ We define the right-derived functor $R^n F$ by $$ R^n F(M) = H^n(F(I)), $$ in other words, the $n$-th homology of the complex $$ 0 \to F(I^0) \to F(I^1) \to F(I^2) \to $$ Directly from the definitions and the monomorphism $M \to I^0$, we see that there is an isomorphism $$ R^0F(M) = F(M). $$ This isomorphism seems at first to depend on the injective resolution, and so do the functors $R^n F(M)$ for other $n$. However, from Lemmas 5.1 and 5.2 we see that given two injective resolutions of $M$, there is a homomorphism between them, and that any two homomorphisms are homotopic. If we apply the functor $F$ to these homomorphisms and to the homotopy, then we see that the homology of the complex $F(I)$ is in fact determined up to unique isomorphism. One therefore omits the resolution from the notation and from the language.
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The two lemmas mentioned are as follows:
Lemma 5.1 If $f, g$ are homotopic, then $f, g$ induce the same homomorphism on the homology $H(E)$, that is $$ H(f_n) = H(g_n) \colon H^n(E) \to H^n(E'). $$
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Lemma 5.2 Consider two complexes: $$ \require{AMScd} \begin{CD} 0 @>>> M @>>> E^0 @>>> E^1 @>>> E^2 @>>> \cdots \\ {} @V \varphi VV {} {} {} {} \\ 0 @>>> M' @>>> I^0 @>>> I^1 @>>> I^2 @>>> \cdots \end{CD} $$ Suppose that the top row is exact, and that each $I^n$ ($n \geq 0$) is injective. Let $\varphi \colon M \to M'$ be a given homomorphism. Then there exists a morphism $f$ of complexes such that $f_{-1} = \varphi$; any any two such are homotopic.
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I don't really understand this proof sketch—I can't see where the natural isomorphism of functors comes from.
EDIT: $F$ is a left exact functor between abelian categories.