The Stacks project

Proof. Let $X^\bullet $ be a bounded above complex of $\mathcal{A}$ such that $H^ i(X^\bullet ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ for all $i \in \mathbf{Z}$. Moreover, suppose we are given $B^ i \subset X^ i$, $B^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ for all $i \in \mathbf{Z}$. Claim: there exists a subcomplex $Y^\bullet \subset X^\bullet $ such that

1. $Y^\bullet \to X^\bullet $ is a quasi-isomorphism,

2. $Y^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ for all $i \in \mathbf{Z}$, and

3. $B^ i \subset Y^ i$ for all $i \in \mathbf{Z}$.

To prove the claim, using the assumption of the lemma we first choose $C^ i \subset \mathop{\mathrm{Ker}}(d^ i : X^ i \to X^{i + 1})$, $C^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ surjecting onto $H^ i(X^\bullet )$. Setting $D^ i = C^ i + d^{i - 1}(B^{i - 1}) + B^ i$ we find a subcomplex $D^\bullet $ satisfying (2) and (3) such that $H^ i(D^\bullet ) \to H^ i(X^\bullet )$ is surjective for all $i \in \mathbf{Z}$. For any choice of $E^ i \subset X^ i$ with $E^ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ and $d^ i(E^ i) \subset D^{i + 1} + E^{i + 1}$ we see that setting $Y^ i = D^ i + E^ i$ gives a subcomplex whose terms are in $\mathcal{B}$ and whose cohomology surjects onto the cohomology of $X^\bullet $. Clearly, if $d^ i(E^ i) = (D^{i + 1} + E^{i + 1}) \cap \mathop{\mathrm{Im}}(d^ i)$ then we see that the map on cohomology is also injective. For $n \gg 0$ we can take $E^ n$ equal to $0$. By descending induction we can choose $E^ i$ for all $i$ with the desired property. Namely, given $E^{i + 1}, E^{i + 2}, \ldots $ we choose $E^ i \subset X^ i$ such that $d^ i(E^ i) = (D^{i + 1} + E^{i + 1}) \cap \mathop{\mathrm{Im}}(d^ i)$. This is possible by our assumption in the lemma combined with the fact that $(D^{i + 1} + E^{i + 1}) \cap \mathop{\mathrm{Im}}(d^ i)$ is in $\mathcal{B}$ as $\mathcal{B}$ is a Serre subcategory of $\mathcal{A}$.

The claim above implies the lemma. Essential surjectivity is immediate from the claim. Let us prove faithfulness. Namely, suppose we have a morphism $f : U^\bullet \to V^\bullet $ of bounded above complexes of $\mathcal{B}$ whose image in $D(\mathcal{A})$ is zero. Then there exists a quasi-isomorphism $s : V^\bullet \to X^\bullet $ into a bounded above complex of $\mathcal{A}$ such that $s \circ f$ is homotopic to zero. Choose a homotopy $h^ i : U^ i \to X^{i - 1}$ between $0$ and $s \circ f$. Apply the claim with $B^ i = h^{i + 1}(U^{i + 1}) + s^ i(V^ i)$. The resulting map $s' : V^\bullet \to Y^\bullet $ is a quasi-isomorphism as well and $s' \circ f$ is homotopic to zero as is clear from the fact that $h^ i$ factors through $Y^{i - 1}$. This proves faithfulness. Fully faithfulness is proved in the exact same manner. $\square$