The Stacks project

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

Moreover, property (2) also guarantees that the complex

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. $\square$