Unbounded complexes, resolutions and computation of derived functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:13:51Z http://mathoverflow.net/feeds/question/107405 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107405/unbounded-complexes-resolutions-and-computation-of-derived-functors Unbounded complexes, resolutions and computation of derived functors Mario Carrasco 2012-09-17T17:38:07Z 2012-09-17T23:06:19Z <p>Hey guys, let $F: \mathcal{A} \rightarrow \mathcal{B}$ be a left exact functor between abelian categories with enough injectives, let $K \in Kom(\mathcal{A})$ be an unbounded complex, I've heard that you can construct resolutions and define a derived functor $RF$ for $K$ but I was wondering:</p> <p>1 - Are the resolutions similar to Cartan-Eilenberg resolutions for bounded complexes, where you find an injective (projective) resolution for each member $K^i$ of $K$ and check it satisfies certain properties?</p> <p>2 - Can you define filtrations, and a spectral sequence to compute the derived functors?</p> <p>Any good references on this?</p> http://mathoverflow.net/questions/107405/unbounded-complexes-resolutions-and-computation-of-derived-functors/107421#107421 Answer by Ralph for Unbounded complexes, resolutions and computation of derived functors Ralph 2012-09-17T21:00:05Z 2012-09-17T23:06:19Z <p>All references below (unless otherwise stated) refer to Weibel (We): An Introduction to Hom. Algebra. Futhermore I consider projective resolutions and assume $\mathcal{A}$ has enough projectives and $F$ is right exact since this case is treated in Weibel. The case of injective resolutions can be easily adapted by switching to the opposite category [We, 2.3.4]. </p> <p>Let $Ch(\mathcal{A})$ be the category of (unbounded) chain complexes in $\mathcal{A}$. Since $\mathcal{A}$ is abelian, $Ch(\mathcal{A})$ is abelian as well [We, Th. 1.2.3] and has enough projectives [We, 2.2.2]. The functor $F: \mathcal{A} \to \mathcal{B}$ induces a functor $Ch(F): Ch(\mathcal{A}) \to Ch(\mathcal{B})$. A morphism $h: C \to D$ of chain complees in $\mathcal{A}$ is epi, iff each $h_i:C_i \to D_i$ is epi [We, Proof of 1.2.3]. Hence, the right exactness of $F$ implies that $Ch(F)$ is also right exact. </p> <p>In summary, we have shown: $Ch(F)$ is a right exact functor between abelian categories and $Ch(\mathcal{A})$ has enough projectives. Consequently, $Ch(F)$ has a left derived functor and everything what can be done for $L_\ast F$ (i.e. filtations, spectral sequences, etc.) can also be done for $L_\ast Ch(F)$. </p> <p>A discussion of projective resolutions in $Ch(\mathcal{A})$ can be found in my answer to this question: </p> <p><a href="http://mathoverflow.net/questions/103584/on-the-difference-between-a-projective-chain-complex-and-a-level-wise-projective/103591#103591" rel="nofollow">http://mathoverflow.net/questions/103584/on-the-difference-between-a-projective-chain-complex-and-a-level-wise-projective/103591#103591</a></p> <p>Similar, a chain complex $I$ in $\mathcal{A}$ (which is now supposed to have enough injectives) is an injective object in $Ch(\mathcal{A})$, iff $I$ consists of injective objects $I_i\in \mathcal{A}$ such that all short sequences $$0 \to \ker(d_i) \to I_i \to \operatorname{im}(d_i) \to 0$$ are exact and do split. </p>