Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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.

share|improve this answer
Thank you so much, I'm guessing that does it for now. –  L-A 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.

share|improve this answer
Thank you mate. –  L-A Apr 17 '13 at 2:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.