Processing math: 100%

The Stacks project

Lemma 13.17.4. Let \mathcal{A} be an abelian category. Let \mathcal{B} \subset \mathcal{A} be a Serre subcategory. Assume that for every surjection X \to Y with X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) and Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B}) there exists X' \subset X, X' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B}) which surjects onto Y. Then the functor D^-(\mathcal{B}) \to D^-_\mathcal {B}(\mathcal{A}) of (13.17.1.1) is an equivalence.

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


Comments (0)

There are also:

  • 9 comment(s) on Section 13.17: Triangulated subcategories of the derived category

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.