Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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:

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?

2 - Can you define filtrations, and a spectral sequence to compute the derived functors?

Any good references on this?

share|improve this question
2  
You can find the construction in Spaltenstein's original paper [N. Spaltenstein, Resolutions of unbounded complexes, Compositio Math. 65 (1988). 121–154.] which is freely available heere numdam.org/item?id=CM_1988__65_2_121_0 –  Mariano Suárez-Alvarez Sep 17 '12 at 17:58
    
Will look into it, thanks –  Mario Carrasco Sep 18 '12 at 0:08

1 Answer 1

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].

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.

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 $L_\ast Ch(F)$.

A discussion of projective resolutions in $Ch(\mathcal{A})$ can be found in my answer to this question:

On the difference between a projective chain complex and a level-wise projective chain complex

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.

share|improve this answer
    
Thank you, I'm reading Spaltenstein's paper too, I'm trying to digest all the info so I guess I'll get back to you later, thanks! –  Mario Carrasco Sep 18 '12 at 0:11

Your Answer

 
discard

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.