# Theorem on composition of derived functors, question about proof

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?

• This theorem is a brain twister and I struggled with it myself. I have written some notes, nd.edu/~lnicolae/Verdier-ams.pdf and at pages 11-12 of I describe in "human language" the meaning of the theorem. Maybe these explanations will help you better absorb Gelfand-Manin. – Liviu Nicolaescu Oct 3 '12 at 21:18
• Thank you so much! I've been actually reading your notes, super well explained, they've been of great help, I just couldn't understand that last line – Richard Jennings Oct 3 '12 at 22:46

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.
• Thank you so much, don't you mean $f: A \rightarrow K^\bullet$ is a quasi-isomorphism?? – Richard Jennings Oct 3 '12 at 22:36
If you want to check that a morphism of functors is an isomorphism, it suffices to verify this only on one object in each isomorphism class. Since each isomorphism class in $D^+(A)$ contains a bounded below complex of adapted objects (this is the so-called "adapted resolution", it exists by definition of an adapted class of objects), it suffices to consider only those. But on such complexes the derived functors are just termwise applications of $G\circ F$ and the morphism of functors is the identity morphism. This completes the proof.