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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 s : 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. From Categories, Lemma 4.22.11 we deduce that RF is the value of the essentially constant ind-object F(L^\bullet ) for these s. This means that \text{id} : RF(K^\bullet ) \to RF(K^\bullet ) factors through F(L^\bullet ) for some complex L^\bullet with L^ n = 0 for n < a. 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.


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