The Stacks project

Lemma 13.16.7 (Leray's acyclicity lemma). Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Let $A^\bullet $ be a bounded below complex of right $F$-acyclic objects such that $RF$ is defined at $A^\bullet $1. The canonical map

\[ F(A^\bullet ) \longrightarrow RF(A^\bullet ) \]

is an isomorphism in $D^{+}(\mathcal{B})$, i.e., $A^\bullet $ computes $RF$.

Proof. Let $A^\bullet $ be a bounded complex of right $F$-acyclic objects. We claim that $RF$ is defined at $A^\bullet $ and that $F(A^\bullet ) \to RF(A^\bullet )$ is an isomorphism in $D^+(\mathcal{B})$. Namely, it holds for complexes with at most one nonzero right $F$-acyclic object by Definition 13.15.3. Next, suppose that $A^ n = 0$ for $n \not\in [a, b]$. Using the “stupid” truncations we obtain a termwise split short exact sequence of complexes

\[ 0 \to \sigma _{\geq a + 1} A^\bullet \to A^\bullet \to \sigma _{\leq a} A^\bullet \to 0 \]

see Homology, Section 12.15. Thus a distinguished triangle $(\sigma _{\geq a + 1} A^\bullet , A^\bullet , \sigma _{\leq a} A^\bullet )$. By induction hypothesis $RF$ is defined for the two outer complexes and these complexes compute $RF$. Then the same is true for the middle one by Lemma 13.14.12.

Suppose that $A^\bullet $ is a bounded below complex of acyclic objects such that $RF$ is defined at $A^\bullet $. To show that $F(A^\bullet ) \to RF(A^\bullet )$ is an isomorphism in $D^{+}(\mathcal{B})$ it suffices to show that $H^ i(F(A^\bullet )) \to H^ i(RF(A^\bullet ))$ is an isomorphism for all $i$. Pick $i$. Consider the termwise split short exact sequence of complexes

\[ 0 \to \sigma _{\geq i + 2} A^\bullet \to A^\bullet \to \sigma _{\leq i + 1} A^\bullet \to 0. \]

Note that this induces a termwise split short exact sequence

\[ 0 \to \sigma _{\geq i + 2} F(A^\bullet ) \to F(A^\bullet ) \to \sigma _{\leq i + 1} F(A^\bullet ) \to 0. \]

Hence we get distinguished triangles

\[ (\sigma _{\geq i + 2} A^\bullet , A^\bullet , \sigma _{\leq i + 1} A^\bullet ) \quad \text{and}\quad (\sigma _{\geq i + 2} F(A^\bullet ), F(A^\bullet ), \sigma _{\leq i + 1} F(A^\bullet )) \]

Since $RF$ is defined at $A^\bullet $ (by assumption) and at $\sigma _{\leq i + 1}A^\bullet $ (by the first paragraph) we see that $RF$ is defined at $\sigma _{\geq i + 1}A^\bullet $ and we get a distinguished triangle

\[ (RF(\sigma _{\geq i + 2} A^\bullet ), RF(A^\bullet ), RF(\sigma _{\leq i + 1} A^\bullet )) \]

See Lemma 13.14.6. Using these distinguished triangles we obtain a map of exact sequences

\[ \xymatrix{ H^ i(\sigma _{\geq i + 2} F(A^\bullet )) \ar[r] \ar[d] & H^ i(F(A^\bullet )) \ar[r] \ar[d]^\alpha & H^ i(\sigma _{\leq i + 1} F(A^\bullet )) \ar[r] \ar[d]^\beta & H^{i + 1}(\sigma _{\geq i + 2} F(A^\bullet )) \ar[d] \\ H^ i(RF(\sigma _{\geq i + 2} A^\bullet )) \ar[r] & H^ i(RF(A^\bullet )) \ar[r] & H^ i(RF(\sigma _{\leq i + 1} A^\bullet )) \ar[r] & H^{i + 1}(RF(\sigma _{\geq i + 2} A^\bullet )) } \]

By the results of the first paragraph the map $\beta $ is an isomorphism. By inspection the objects on the upper left and the upper right are zero. Hence to finish the proof it suffices to show that $H^ i(RF(\sigma _{\geq i + 2} A^\bullet )) = 0$ and $H^{i + 1}(RF(\sigma _{\geq i + 2} A^\bullet )) = 0$. This follows immediately from Lemma 13.16.1. $\square$

[1] For example this holds if $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined.

Comments (5)

Comment #5889 by on

Can we remove the hypothesis that is everywhere defined? At least in the bounded case it's obvious that we can.

Comment #5890 by on

Let be a bounded below complex of acyclic objects. I think we can replace the assumption that is everywhere defined by the assumption that is defined at . I will make this change the next time I go through all the comments.

I do not see is how to prove that is defined at . The problem, I think, is the following. Let be a quasi-isomorphism. Then I think the arguments in the proof and Lemma 13.5.10 allow us to conclude that given an integer there is a quasi-isomorphism such that the composition is an isomorphism on cohomology in degrees . But in order to show that is defined at with value we would need to show that we can do this in all degrees simultaneously!

Thanks for the interesting question.

Comment #8402 by on

In the first paragraph, when we say "by Lemma 13.16.4", wouldn't it suffice to say "by Definition 13.15.3"? (or simply "by definition"). Also, "distinguished" is misspelled just before expression .


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