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Don't know if this kind of a dumb question but let $A$ and $B$ be abelian categories and suppose they're equivalent: there are two functors $P: A \rightarrow B$ and $Q: B \rightarrow A$ satisfying the equivalence conditions. Let $F:A \rightarrow D$ and $G: B \rightarrow D$ be left exact functors, I can choose an object $X$ in $A$, put it in an injective resolution

$0 \rightarrow X \rightarrow I^0 \rightarrow I^1 \rightarrow I^2 \rightarrow I^3 \rightarrow \cdots$

and compute the derived functors $R^nF(X)$, does the equivalance of categories help at all compute $R^nGP(X)$? Is there something to be gained from knowing $A$ and $B$ are equivalent in terms of computing derived functors?

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2 Answers 2

up vote 5 down vote accepted

Since $F,G$ are not related at all, the answer is of course: No. But the answer is Yes if $F=GP$.

More generally, let $P : A \to B$ be any exact functor which preserves injectives. The latter property holds for example when $P$ has an exact left adjoint (nice exercise). In particular, this applies to the case that $P$ is an equivalence of categories.

Then for every left exact functor $F : B \to C$, the functors $(R^q F) \circ P$ and $R^q(F \circ P)$ are canonically isomorphic. The proof is straight forward, using the properties of $P$ and the definitions of a derived functor.

Remark: If $P$ is not exact, but still maps injective objects to $F$-acyclic objects, then one can try to "approximate" $R^{p+q}(F \circ P)$ by $R^p F \circ R^q P$. This is made precise by the Grothendieck spectral sequence.

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Thank you so much, I'm guessing that does it for now. –  Luis Almaguer Apr 17 '13 at 2:30

I see nothing in the definition of derived functors that wouldn't be preserved by equivalence of categories. So $R^nGP(X)=PR^nG(X)$. (Derived functors are, as far as I know, defined only up to isomorphism, so "$=$" here means "canonically isomorphic".)

EDIT: Martin Brandenburg is right. I assumed $F=GP$, but this is not stated in the question.

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Thank you mate. –  Luis Almaguer Apr 17 '13 at 2:38

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