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 for example by Lemma 13.16.4. 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 distinghuished 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.

Comment #5889 by on

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

Comment #5890 by on

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

I do not see is how to prove that $RF$ is defined at $A^\bullet$. The problem, I think, is the following. Let $s : A^\bullet \to M^\bullet$ be a quasi-isomorphism. Then I think the arguments in the proof and Lemma 13.5.9 allow us to conclude that given an integer $n$ there is a quasi-isomorphism $s' : M^\bullet \to N^\bullet$ such that the composition $F(A^\bullet) \to F(M^\bullet) \to F(N^\bullet)$ is an isomorphism on cohomology in degrees $\leq n$. But in order to show that $RF$is defined at $A^\bullet$ with value $F(A^\bullet)$ we would need to show that we can do this in all degrees simultaneously!

Thanks for the interesting question.

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