Lemma 13.17.7 (Leray's acyclicity lemma). Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined. Let $A^\bullet$ be a bounded below complex of $F$-acyclic objects. The canonical map

$F(A^\bullet ) \longrightarrow RF(A^\bullet )$

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

Proof. First we claim the lemma holds for a bounded complex of acyclic objects. Namely, it holds for complexes with at most one nonzero object by definition. Suppose that $A^\bullet$ is a complex with $A^ n = 0$ 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 12.14. Thus a distinguished triangle $(\sigma _{\geq a + 1} A^\bullet , A^\bullet , \sigma _{\leq a} A^\bullet )$. By induction hypothesis the two outer complexes compute $RF$. Then the middle one does too by Lemma 13.15.12.

Suppose that $A^\bullet$ is a bounded below complex of acyclic objects. To show that $F(A) \to RF(A)$ is an isomorphism in $D^{+}(\mathcal{B})$ it suffices to show that $H^ i(F(A)) \to H^ i(RF(A))$ 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

$\begin{matrix} (\sigma _{\geq i + 2} A^\bullet , A^\bullet , \sigma _{\leq i + 1} A^\bullet ) \\ (\sigma _{\geq i + 2} F(A^\bullet ), F(A^\bullet ), \sigma _{\leq i + 1} F(A^\bullet )) \\ (RF(\sigma _{\geq i + 2} A^\bullet ), RF(A^\bullet ), RF(\sigma _{\leq i + 1} A^\bullet )) \end{matrix}$

Using the last two we obtain a map of exact sequences

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

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 $R^ iF(\sigma _{\geq i + 2} A^\bullet ) = 0$ and $R^{i + 1}F(\sigma _{\geq i + 2} A^\bullet ) = 0$. This follows immediately from Lemma 13.17.1. $\square$

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