The Stacks project

Lemma 13.16.1. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Let $K^\bullet $ be a complex of $\mathcal{A}$ and $a \in \mathbf{Z}$.

  1. If $H^ i(K^\bullet ) = 0$ for all $i < a$ and $RF$ is defined at $K^\bullet $, then $H^ i(RF(K^\bullet )) = 0$ for all $i < a$.

  2. If $RF$ is defined at $K^\bullet $ and $\tau _{\leq a}K^\bullet $, then $H^ i(RF(\tau _{\leq a}K^\bullet )) = H^ i(RF(K^\bullet ))$ for all $i \leq a$.

Proof. Assume $K^\bullet $ satisfies the assumptions of (1). Let $K^\bullet \to L^\bullet $ be any quasi-isomorphism. Then it is also true that $K^\bullet \to \tau _{\geq a}L^\bullet $ is a quasi-isomorphism by our assumption on $K^\bullet $. Hence in the category $K^\bullet /\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : K^\bullet \to L^\bullet $ with $L^ n = 0$ for $n < a$ are cofinal. Thus $RF$ is the value of the essentially constant ind-object $F(L^\bullet )$ for these $s$ it follows that $H^ i(RF(K^\bullet )) = 0$ for $i < a$.

To prove (2) we use the distinguished triangle

\[ \tau _{\leq a}K^\bullet \to K^\bullet \to \tau _{\geq a + 1}K^\bullet \to (\tau _{\leq a}K^\bullet )[1] \]

of Remark 13.12.4 to conclude via Lemma 13.14.6 that $RF$ is defined at $\tau _{\geq a + 1}K^\bullet $ as well and that we have a distinguished triangle

\[ RF(\tau _{\leq a}K^\bullet ) \to RF(K^\bullet ) \to RF(\tau _{\geq a + 1}K^\bullet ) \to RF(\tau _{\leq a}K^\bullet )[1] \]

in $D(\mathcal{B})$. By part (1) we see that $RF(\tau _{\geq a + 1}K^\bullet )$ has vanishing cohomology in degrees $< a + 1$. The long exact cohomology sequence of this distinguished triangle then shows what we want. $\square$

Comments (4)

Comment #5949 by Dylan on

I believe it should be "for i<a" in the last sentence of the first paragraph.

Comment #8399 by on

Typos: instead of «Thus is» I think it should be «Thus is». After, instead of «for these it follows» it should be «for these . It follows» On the other hand, in «it follows that for », maybe one could give more detail? I guess saying «there is a quasi-isomorphism with for and a map as in Categories, Definition 4.22.1, (1). In particular is monic, so for some by Lemma 13.4.12, and the claim follows.»

Comment #9011 by on

Very good. Yes, I think this was an incomplete arguement. For example, if you look at the essentially constant system given following Definition 4.22.2, then the system lives in a subcategory (namely free -modules of even rank) but the value does not! Fixed here.

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 05TC. Beware of the difference between the letter 'O' and the digit '0'.