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I'm not sure how I can show the following:

If F is a left exact functor from an abelian category A to an abelian category B, whose derived functor RF in the sense of derived categories exists, then the following holds:

if $Z^{.}$ is a complex consisting of F-acyclic objects in A, then $RF(Z^{.})$ is equal to $KF(Z^{.})$; with the last symbol I just mean: apply F to the complex $Z^{.}$ and understand the result as belonging to $D^{+}(B)$.

I don't want to assume the existence of F-adapted classes or enough Injectives, just the Existence of RF.

Thanks a lot!

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Please do not cross-post: mathoverflow.net/questions/72196/derived-functor-and-acyclics –  Qiaochu Yuan Aug 5 '11 at 21:15

1 Answer 1

up vote 7 down vote accepted

With the proper definitions you have the following facts. I'll be using Keller's terminology in Derived Categories and their uses:

  1. For any functor $F: \mathscr{A} \to \mathscr{B}$ (between abelian or even exact categories), the class of (right) $F$-acyclic objects are a subcategory $Ac \subset \mathscr{A}$ closed under extensions.
  2. The restriction of $F$ to $Ac$, is exact. [Keller, Lemma 15.3]
  3. For every short exact sequence $0 \to Z' \to A \to A''$ with $Z'$ in $Ac$ there exists a commutative diagram $$\begin{array}{ccccccccc} 0 & \to & Z' & \to & A & \to & A'' & \to & 0 \\ & & \parallel & & \downarrow & & \downarrow \\ 0 & \to & Z' & \to & Z & \to & Z'' & \to & 0 \end{array}$$ where the second row is exact and $Z, Z''$ are in $Ac$. [Keller, Lemma 15.3].

This implies the following:

  1. $Ac$ is an exact category and the inclusion functor $I: Ac \to \mathscr{A}$ is exact. [Keller, §4].
  2. The functor $I = D^{+}I: D^{+}(Ac) \to D^{+}(\mathscr{A})$ is fully faithful because $K^{+}(Ac) \subset K^{+}(\mathscr{A})$ is right cofinal with respect to quasi-isomorphisms. [Keller, Theorem 12.1 (b)]
  3. By exactness of $FI$ and $I$ we have $FI = D^{+}(FI) = \mathbf{R}(FI)$ and $I = D^{+}I= \mathbf{R}I$
  4. By [Keller, Lemma 13.5], we have that for $Z \in K^{+}(Ac)$ the derived functor $\mathbf{R}F$ is defined at $Z$ if and only if $\mathbf{R}(FI)$ is defined at $Z$. The canonical morphism $\mathbf{R}(FI)(Z) \to \mathbf{R}F \mathbf{R}I (Z)$ is invertible.

Piecing 3. and 4. together we get that $\mathbf{R}F$ is defined at every $Z \in K^{+}(Ac)$ and $FI(Z) = \mathbf{R}(FI)(Z) = \mathbf{R}F \mathbf{R}I(Z) = \mathbf{R}F(IZ)$ with a slight abuse of notation that you'll hopefully allow me to make. Neglecting the inclusion functor $I$ we get $F(Z) = \mathbf{R}F(Z)$ as you wanted.

So that's the argument. Now Keller's text does not provide too many details but it gives sufficiently many for following through the entire argument. Sections 6.7 and 6.6 in Derived Categories and Universal Problems may be helpful when verifying Proposition 13.1. and Lemma 13.6, as well as Deligne's text Cohomologie à supports propres, SGA 4 Vol. 3, Exposé XVII, Springer LNM 305 (1973).


Some Further References:

  • You'll find many references in Keller's article mentioned above. The standard texts treating derived categories on a rather basic level are of course Weibel's and Gelfand–Manin's books on homological algebra.
  • If you want to learn about exact categories, you may find most of the references I'm aware of in my survey article on them, T. Bühler, Exact Categories, Volume 28(1), (2010), 1-69. MR2606234
  • A rather detailed treatment of derived functors in the setting of derived categories can be found in Chapter 13 of M. Kashiwara–P. Schapira, Categories and Sheaves, Springer Grundlehren 332, (2006). MR2182076
  • A nice article comparing various notions of derived functors in many different settings (although mostly homotopical) is Maltsiniotis–Kahn, Structures de derivabilité, Adv. Math. Volume 218 (4), (2008), 1286–1318. MR2419385
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Sorry for the previous slightly over the top answer, in case you've seen it. –  t.b. Aug 6 '11 at 0:55
    
great comment, thanks a lot! –  Descartes Aug 6 '11 at 7:08

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