I got a question about a proof I found in Gelfand-Manin's "Methods of homological algebra" (Page 200):
Theorem 1. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be three abelian categories, $F: \mathcal{A} \rightarrow \mathcal{B}$, $G: \mathcal{B} \rightarrow \mathcal{C}$ be two additive left exact functors. Let $\mathcal{R_\mathcal{A}} \subset Ob \ \mathcal{A}$ (resp. $\mathcal{R}_\mathcal{B} \subset Ob \ \mathcal{B}$) be a class of objects adapted to $F$ (resp. to $G$). Assume that $F(\mathcal{R}_\mathcal{A}) \subset \mathcal{R}_\mathcal{B}$. Then the derived functors $RF$, $RG$, $R(G \circ F): \mathcal{D}^+(\bullet) \rightarrow \mathcal{D}^+(\bullet)$ exist and the natural morphism of functors $R(G \circ F) \rightarrow RG \circ RF$ is an isomorphism.
Proof: The definition of an adapted class in III.6.3 and the conditions of the theorem show that $\mathcal{R}_\mathcal{A}$ is adapted not only to $F$, but to $G \circ F$ as well. Hence, $RF$, $RG$ and $R(G \circ F)$ exist and to compute them we can use the construction from III.6.6 (Universal property of derived functors).
Next, $RF$ and $RG$ are exact functors. Hence $R(G \circ F)$ is also exact and the morphism $E: R(G \circ F) \rightarrow RG \circ RF$ is defined by the universal property.
For $K^\bullet \in Ob\, Kom^+(\mathcal{R}_\mathcal{A})$ the morphism $E(K^\bullet): R(G \circ F)(K^\bullet) \rightarrow RG \circ RF(K^\bullet)$ is an isomoprhism. Since any object of $D^+(\mathcal{A})$ is isomorphic to such an object $K^\bullet$, $E$ is an isomorphism of functors. $\square$
That last part was where I got lost. From the universal property I understand that there's a morphism of functors $R(G \circ F) \rightarrow RG \circ RF$, the part I don't get is the reason why it is an isomorphism, specifically this:
Since any object of $D^+(\mathcal{A})$ is isomorphic to such an object $K^\bullet$, $E$ is an isomorphism of functors
That's where I got completely lost, I was reading another book and picked up Gelfand-Manin's book just to look at the proof and don't know where that came from, Does anyone know?
Best Answer
To show that $E$ is an isomorphism of functors, it suffices to show that $E(A)$ is an isomorphism for each object $A$ of $D^+(\mathcal{A})$. This has been shown for each $K^\bullet$ an object of $\operatorname{Kom}^+(\mathcal{R}_\mathcal{A})$. For an arbitrary object $A$, choose a quasi-isomorphism $f:A\to K^\bullet$ for such a $K^\bullet$, which then becomes an isomorphism in the derived category. Then $R(G\circ F)(f):R(G\circ F)(K)\to R(G\circ F)(A)$ and $(RG\circ RF)(f):(RG\circ RF)(K^\bullet)\to(RG\circ RF)(A)$ are isomorphisms. We then have that $E(A) = (RG\circ RF)(f)\circ E(K^\bullet) \circ R(G\circ F)(f)^{-1}$ is an isomorphism, being a composition of isomorphisms.