Lemma 13.15.6. In Situation 13.15.1. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset with the following properties:

1. every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$,

2. for any short exact sequence $0 \to P \to Q \to R \to 0$ of $\mathcal{A}$ with $P, Q \in \mathcal{I}$, then $R \in \mathcal{I}$, and $0 \to F(P) \to F(Q) \to F(R) \to 0$ is exact.

Then every object of $\mathcal{I}$ is acyclic for $RF$.

Proof. We may add $0$ to $\mathcal{I}$ if necessary. Pick $A \in \mathcal{I}$. Let $A[0] \to K^\bullet$ be a quasi-isomorphism with $K^\bullet$ bounded below. Then we can find a quasi-isomorphism $K^\bullet \to I^\bullet$ with $I^\bullet$ bounded below and each $I^ n \in \mathcal{I}$, see Lemma 13.15.5. Hence we see that these resolutions are cofinal in the category $A[0]/\text{Qis}^{+}(\mathcal{A})$. To finish the proof it therefore suffices to show that for any quasi-isomorphism $A[0] \to I^\bullet$ with $I^\bullet$ bounded below and $I^ n \in \mathcal{I}$ we have $F(A)[0] \to F(I^\bullet )$ is a quasi-isomorphism. To see this suppose that $I^ n = 0$ for $n < n_0$. Of course we may assume that $n_0 < 0$. Starting with $n = n_0$ we prove inductively that $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n)$ and $\mathop{\mathrm{Im}}(d^{-1})$ are elements of $\mathcal{I}$ using property (2) and the exact sequences

$0 \to \mathop{\mathrm{Ker}}(d^ n) \to I^ n \to \mathop{\mathrm{Im}}(d^ n) \to 0.$

Moreover, property (2) also guarantees that the complex

$0 \to F(I^{n_0}) \to F(I^{n_0 + 1}) \to \ldots \to F(I^{-1}) \to F(\mathop{\mathrm{Im}}(d^{-1})) \to 0$

is exact. The exact sequence $0 \to \mathop{\mathrm{Im}}(d^{-1}) \to I^0 \to I^0/\mathop{\mathrm{Im}}(d^{-1}) \to 0$ implies that $I^0/\mathop{\mathrm{Im}}(d^{-1})$ is an element of $\mathcal{I}$. The exact sequence $0 \to A \to I^0/\mathop{\mathrm{Im}}(d^{-1}) \to \mathop{\mathrm{Im}}(d^0) \to 0$ then implies that $\mathop{\mathrm{Im}}(d^0) = \mathop{\mathrm{Ker}}(d^1)$ is an elements of $\mathcal{I}$ and from then on one continues as before to show that $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n)$ is an element of $\mathcal{I}$ for all $n > 0$. Applying $F$ to each of the short exact sequences mentioned above and using (2) we observe that $F(A)[0] \to F(I^\bullet )$ is an isomorphism as desired. $\square$

There are also:

• 7 comment(s) on Section 13.15: Derived functors on derived categories

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).