The Stacks project

Lemma 13.32.2. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor of abelian categories. Assume

  1. every object of $\mathcal{A}$ is a subobject of an object which is right acyclic for $F$,

  2. there exists an integer $n \geq 0$ such that $R^ nF = 0$,

Then

  1. $RF : D(\mathcal{A}) \to D(\mathcal{B})$ exists,

  2. any complex consisting of right acyclic objects for $F$ computes $RF$,

  3. any complex is the source of a quasi-isomorphism into a complex consisting of right acyclic objects for $F$,

  4. for $E \in D(\mathcal{A})$

    1. $H^ i(RF(\tau _{\leq a}E) \to H^ i(RF(E))$ is an isomorphism for $i \leq a$,

    2. $H^ i(RF(E)) \to H^ i(RF(\tau _{\geq b - n + 1}E))$ is an isomorphism for $i \geq b$,

    3. if $H^ i(E) = 0$ for $i \not\in [a, b]$ for some $-\infty \leq a \leq b \leq \infty $, then $H^ i(RF(E)) = 0$ for $i \not\in [a, b + n - 1]$.

Proof. Note that the first assumption implies that $RF : D^+(\mathcal{A}) \to D^+(\mathcal{B})$ exists, see Proposition 13.16.8. Let $A$ be an object of $\mathcal{A}$. Choose an injection $A \to A'$ with $A'$ acyclic. Then we see that $R^{n + 1}F(A) = R^ nF(A'/A) = 0$ by the long exact cohomology sequence. Hence we conclude that $R^{n + 1}F = 0$. Continuing like this using induction we find that $R^ mF = 0$ for all $m \geq n$.

We are going to use Lemma 13.32.1 with the function $d : \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) \to \{ 0, 1, 2, \ldots \} $ given by $d(A) = \max \{ 0\} \cup \{ i \mid R^ iF(A) \not= 0\} $. The first assumption of Lemma 13.32.1 is our assumption (1). The second assumption of Lemma 13.32.1 follows from the fact that $RF(A \oplus B) = RF(A) \oplus RF(B)$. The third assumption of Lemma 13.32.1 follows from the long exact cohomology sequence. Hence for every complex $K^\bullet $ there exists a quasi-isomorphism $K^\bullet \to L^\bullet $ into a complex of objects right acyclic for $F$. This proves statement (3).

We claim that if $L^\bullet \to M^\bullet $ is a quasi-isomorphism of complexes of right acyclic objects for $F$, then $F(L^\bullet ) \to F(M^\bullet )$ is a quasi-isomorphism. If we prove this claim then we get statements (1) and (2) of the lemma by Lemma 13.14.15. To prove the claim pick an integer $i \in \mathbf{Z}$. Consider the distinguished triangle

\[ \sigma _{\geq i - n - 1}L^\bullet \to \sigma _{\geq i - n - 1}M^\bullet \to Q^\bullet , \]

i.e., let $Q^\bullet $ be the cone of the first map. Note that $Q^\bullet $ is bounded below and that $H^ j(Q^\bullet )$ is zero except possibly for $j = i - n - 1$ or $j = i - n - 2$. We may apply $RF$ to $Q^\bullet $. Using the second spectral sequence of Lemma 13.21.3 and the assumed vanishing of cohomology (2) we conclude that $H^ j(RF(Q^\bullet ))$ is zero except possibly for $j \in \{ i - n - 2, \ldots , i - 1\} $. Hence we see that $RF(\sigma _{\geq i - n - 1}L^\bullet ) \to RF(\sigma _{\geq i - n - 1}M^\bullet )$ induces an isomorphism of cohomology objects in degrees $\geq i$. By Proposition 13.16.8 we know that $RF(\sigma _{\geq i - n - 1}L^\bullet ) = \sigma _{\geq i - n - 1}F(L^\bullet )$ and $RF(\sigma _{\geq i - n - 1}M^\bullet ) = \sigma _{\geq i - n - 1}F(M^\bullet )$. We conclude that $F(L^\bullet ) \to F(M^\bullet )$ is an isomorphism in degree $i$ as desired.

Part (4)(a) follows from Lemma 13.16.1.

For part (4)(b) let $E$ be represented by the complex $L^\bullet $ of objects right acyclic for $F$. By part (2) $RF(E)$ is represented by the complex $F(L^\bullet )$ and $RF(\sigma _{\geq c}L^\bullet )$ is represented by $\sigma _{\geq c}F(L^\bullet )$. Consider the distinguished triangle

\[ H^{b - n}(L^\bullet )[n - b] \to \tau _{\geq b - n}L^\bullet \to \tau _{\geq b - n + 1}L^\bullet \]

of Remark 13.12.4. The vanishing established above gives that $H^ i(RF(\tau _{\geq b - n}L^\bullet ))$ agrees with $H^ i(RF(\tau _{\geq b - n + 1}L^\bullet ))$ for $i \geq b$. Consider the short exact sequence of complexes

\[ 0 \to \mathop{\mathrm{Im}}(L^{b - n - 1} \to L^{b - n})[n - b] \to \sigma _{\geq b - n}L^\bullet \to \tau _{\geq b - n}L^\bullet \to 0 \]

Using the distinguished triangle associated to this (see Section 13.12) and the vanishing as before we conclude that $H^ i(RF(\tau _{\geq b - n}L^\bullet ))$ agrees with $H^ i(RF(\sigma _{\geq b - n}L^\bullet ))$ for $i \geq b$. Since the map $RF(\sigma _{\geq b - n}L^\bullet ) \to RF(L^\bullet )$ is represented by $\sigma _{\geq b - n}F(L^\bullet ) \to F(L^\bullet )$ we conclude that this in turn agrees with $H^ i(RF(L^\bullet ))$ for $i \geq b$ as desired.

Proof of (4)(c). Under the assumption on $E$ we have $\tau _{\leq a - 1}E = 0$ and we get the vanishing of $H^ i(RF(E))$ for $i \leq a - 1$ from part (4)(a). Similarly, we have $\tau _{\geq b + 1}E = 0$ and hence we get the vanishing of $H^ i(RF(E))$ for $i \geq b + n$ from part (4)(b). $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07K7. Beware of the difference between the letter 'O' and the digit '0'.