# Acyclic Objects and cohomologically finite functors

let's start with a left exact functor $F: A\longrightarrow B$ of abelian categories, where the derived functor $RF: D^{+}(A)\longrightarrow D^{+}(B)$ exists. Furthermore the class of F-acyclic objects be adapted to F and furthermore F of finite cohomological dimension.

My question: how can I find a natural number n with the following property: if

$X_n\rightarrow X_{n-1}\rightarrow ...\rightarrow X_{o}$ is exact in $A$ with $X_n,...,X_1$ F-acyclic, then also $X_{o}$ is F-acyclic.

Regards!

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Crossposted from MO: mathoverflow.net/questions/72261 Please don't do that. –  t.b. Aug 6 '11 at 23:51

I shall assume that the last map $X_1 \to X_0$ is surjective (or the answer is no because of the trivial resolution $0 \to 0 \to 0 \to \dots \to X$). The claim is that we can do this with $n$ the cohomological dimension of $F$. Namely, if we have a resolution $$X_n \to X_{n-1} \to \dots X_1 \to X_0 \to 0,$$ with the first $n$ objects $F$-acyclic, then the last one is too. We can see this using dimension-shifting. That is, draw the kernels and cokernels at each stage. We find from the long exact sequence in cohomology (the derived functors form a $\delta$-functor even in this case, by definition essentially) that $R^1 F(X_0) = R^{2} F(K_1)$, for $K_1$ the kernel of $X_1 \to X_0$. Similarly we find that $R^2 F(K_1) = R^3 F(K_3)$, and so on. Repeating, we get that $R^1 F(X_0)$ is $R^{n+1} F$ of some object (the kernel of $X_n \to X_{n-1}$), which is zero by hypothesis. So $R^1 F(X_0) = 0$, and by the same argument $R^k F(X_0) = 0$ for each $k>1$, too.

This is a useful fact in sheaf theory: it implies that, in general, soft resolutions on the left won't exist (soft resolutions on the right exist, because there are enough injectives in the category of sheaves). If there is an arbitrarily large soft resolution on the left, the sheaf is in fact soft: one corollary of this is that the tensor product of a soft, flat sheaf with any sheaf(!) is itself soft (by resolving the other sheaf with direct sums of objects of the form $j_!(\mathbb{Z}_U)$ for $U$ an open subset, and $j$ an open inclusion). Moreover, (and perhaps from your original post this is the context you encountered the claim), it shows that the derived functors of a cohomologically bounded functor extend to the total derived category (not just the bounded one*).

*There is a caveat here: using fancier technology, it is possible to talk about the total derived functor on the entire derived category without assuming that the functor has bounded cohomological dimension. These techniques were first developed by Spaltenstein for sheaves, and apparently has been extended to Grothendieck abelian categories. I have not read this paper in any detail, and cannot comment further, but anyway this is far less elementary.

Edit: (T.B.) The extension of Spaltenstein's work to Grothendieck abelian categories is relatively straightforward. It can be found e.g. in chapter 14 of Kashiwara-Schapira, Categories and Sheaves together with the most important references. Maybe Henning Krause's work, explained beautifully and in a very accessible way in his notes on Localization theory in triangulated categories deserves special mention.

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I took the liberty and added some references to the footnote, I hope you don't mind. –  t.b. Aug 7 '11 at 0:31
@Theo: Thanks for doing that! I hadn't realized it was in K-S, and should perhaps have a look at ch. 14. –  Akhil Mathew Aug 7 '11 at 0:36
Please do also have a look at these notes of Henning. I found them immensely enlightening (and I believe that they tie in well with views you expressed in your blog post on quasi-coherent sheaves). –  t.b. Aug 7 '11 at 0:47
Hi Akhil, thank you for your completely satisfying answer. Sorry for the crossposting, Theo, but after posting it in overflow I had the feeling that it should be quite elementary and posted it here. But I then didn't find out how to delete my own question in overflow and had to leave it doubled. Perhaps you can tell me how I do that. –  Descartes Aug 7 '11 at 7:18
@Descartes: That's probably because you're an unregistered user. As a registered user you have a "delete" link at the bottom left of your question (as long as there is no answer). You should consider registering your accounts because that makes it much easier for the software to recognize you (you'll lose your accounts if you erase your browser cookies and in order to recover them you'll have to bother the moderators). Registering is easy, all you need is an openid and link your account to that. Concerning the level of the questions I think that it would be better to ask them here first. –  t.b. Aug 7 '11 at 10:14