**Proof.**
Assume every object of $\mathcal{A}$ injects into an object acyclic for $RF$. Let $\mathcal{I}$ be the set of objects acyclic for $RF$. Let $K^\bullet $ be a bounded below complex in $\mathcal{A}$. By Lemma 13.16.4 there exists a quasi-isomorphism $\alpha : K^\bullet \to I^\bullet $ with $I^\bullet $ bounded below and $I^ n \in \mathcal{I}$. Hence in order to prove (1) it suffices to show that $F(I^\bullet ) \to F((I')^\bullet )$ is a quasi-isomorphism when $s : I^\bullet \to (I')^\bullet $ is a quasi-isomorphism of bounded below complexes of objects from $\mathcal{I}$, see Lemma 13.15.15. Note that the cone $C(s)^\bullet $ is an acyclic bounded below complex all of whose terms are in $\mathcal{I}$. Hence it suffices to show: given an acyclic bounded below complex $I^\bullet $ all of whose terms are in $\mathcal{I}$ the complex $F(I^\bullet )$ is acyclic.

Say $I^ n = 0$ for $n < n_0$. Setting $J^ n = \mathop{\mathrm{Im}}(d^ n)$ we break $I^\bullet $ into short exact sequences $0 \to J^ n \to I^{n + 1} \to J^{n + 1} \to 0$ for $n \geq n_0$. These sequences induce distinguished triangles $(J^ n, I^{n + 1}, J^{n + 1})$ in $D^+(\mathcal{A})$ by Lemma 13.12.1. For each $k \in \mathbf{Z}$ denote $H_ k$ the assertion: For all $n \leq k$ the right derived functor $RF$ is defined at $J^ n$ and $R^ iF(J^ n) = 0$ for $i \not= 0$. Then $H_ k$ holds trivially for $k \leq n_0$. If $H_ n$ holds, then, using Proposition 13.15.8, we see that $RF$ is defined at $J^{n + 1}$ and $(RF(J^ n), RF(I^{n + 1}), RF(J^{n + 1}))$ is a distinguished triangle of $D^+(\mathcal{B})$. Thus the long exact cohomology sequence (13.11.1.1) associated to this triangle gives an exact sequence

\[ 0 \to R^{-1}F(J^{n + 1}) \to R^0F(J^ n) \to F(I^{n + 1}) \to R^0F(J^{n + 1}) \to 0 \]

and gives that $R^ iF(J^{n + 1}) = 0$ for $i \not\in \{ -1, 0\} $. By Lemma 13.17.1 we see that $R^{-1}F(J^{n + 1}) = 0$. This proves that $H_{n + 1}$ is true hence $H_ k$ holds for all $k$. We also conclude that

\[ 0 \to R^0F(J^ n) \to F(I^{n + 1}) \to R^0F(J^{n + 1}) \to 0 \]

is short exact for all $n$. This in turn proves that $F(I^\bullet )$ is exact.

The proof in the case of $LF$ is dual.
$\square$

## Comments (6)

Comment #1274 by JuanPablo on

Comment #1275 by JuanPablo on

Comment #1276 by JuanPablo on

Comment #1301 by Johan on

Comment #2601 by Rogier Brussee on

Comment #2626 by Johan on