I may get grilled for this but here I go: Let $\mathcal{A}$ be an abelian category with enough injectives. What I want to know is VERY VERY specific. Let's say I have a complex in $\mathcal{A}$

$0 \rightarrow A^0 \rightarrow A^1 \rightarrow A^2 \rightarrow \cdots, \:\:\: (1)$

which happens to be exact and where all the objects $A^i$ are acyclic. Furthermore, let $F$ be a left exact functor $F: \mathcal{A} \rightarrow \mathcal{B}$, now let's say (in this particular case) that the complex

$0 \rightarrow F(A^0) \rightarrow F(A^1) \rightarrow F(A^2) \rightarrow \cdots \:\:\: (2)$

is exact too.

I want to know:

QUESTION 1: Can I take (1) and put it as an acyclic resolution of $0$? That is:

$A^\bullet_0: 0 \rightarrow 0 \rightarrow A^0 \rightarrow A^1 \rightarrow A^2 \rightarrow \cdots$,

such that I can compute the right derived functors $R^nF(0)$ with this acyclic resolution? This way

$F(A^\bullet_0): 0 \rightarrow F(A^0) \rightarrow F(A^1) \rightarrow F(A^2) \rightarrow \cdots $

coincides with (2) and $R^nF(0) = $ for all $n$?

I want to do it this way SPECIFICALLY because it encodes some information I need.

Now check out (or skip through if you're familiar with this topic) right quick the following 2 theorems by my homeboys Gelfand and Manin:

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.

Theorem 2. Let under the conditions of the above theorem the class $\mathcal{R}_\mathcal{A}$ coincides with the class $\mathcal{I}_\mathcal{A}$ of all injective objects in $\mathcal{A}$ and let the class $\mathcal{I}_\mathcal{B}$ be sufficiently large. Then for any $X \in Ob \mathcal{A}$ there exists a spectral sequence with

$E^{p,q}_r = R^pG(R^qF(X))$

converging to $R^n(G \circ F) (X)$. It is functorial in $X$.

The proof of the last theorem uses a Cartan-Eilenberg resolution of a complex $K^\bullet$, where $K^\bullet = F(I^\bullet_X)$, where $I^\bullet_X$ is an injective resolution of $X$. Now what I want to do basically is compute in this case the groups $R^n(G \circ F) (0)$ using SPECIFICALLY the acyclic resolution $A^\bullet_0$ I wrote above, so I'd have $K^\bullet = F(A^\bullet_0)$. My other question is:

QUESTION 2: Can I apply these theorems and the Cartan-Eilenberg resolution if I use an acyclic resolution $A^\bullet_X$, so $K^\bullet = F(A^\bullet_X)$, instead of an injective resolution in $K^\bullet = F(I^\bullet_X)$ to compute the $R^n(G \circ F) (X)$'s in the above theorem??

  • 3
    $\begingroup$ The answer to your first question: yes. The quick way to see this is that $R^n F(0) = 0$ and the cohomology of $F(A_0^\bullet)$ is also $0$ by the exactness hypothesis. $\endgroup$ – Zhen Lin Sep 11 '12 at 5:07

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