# 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*).