Processing math: 100%

The Stacks project

Lemma 13.17.3. Let \mathcal{A} be an abelian category. Let \mathcal{B} \subset \mathcal{A} be a Serre subcategory. Suppose that the functor v : \mathcal{A} \to \mathcal{A}/\mathcal{B} has a left adjoint u : \mathcal{A}/\mathcal{B} \to \mathcal{A} such that vu \cong \text{id}. Then

D(\mathcal{A})/D_\mathcal {B}(\mathcal{A}) = D(\mathcal{A}/\mathcal{B})

and similarly for the bounded versions.

Proof. The functor D(v) : D(\mathcal{A}) \to D(\mathcal{A}/\mathcal{B}) is essentially surjective by Lemma 13.17.2. For an object X of D(\mathcal{A}) the adjunction mapping c_ X : uvX \to X maps to an isomorphism in D(\mathcal{A}/\mathcal{B}) because vuv \cong v by the assumption that vu \cong \text{id}. Thus in a distinguished triangle (uvX, X, Z, c_ X, g, h) the object Z is an object of D_\mathcal {B}(\mathcal{A}) as we see by looking at the long exact cohomology sequence. Hence c_ X is an element of the multiplicative system used to define the quotient category D(\mathcal{A})/D_\mathcal {B}(\mathcal{A}). Thus uvX \cong X in D(\mathcal{A})/D_\mathcal {B}(\mathcal{A}). For X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})) the map

\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})/D_\mathcal {B}(\mathcal{A})}(X, Y) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A}/\mathcal{B})}(vX, vY)

is bijective because u gives an inverse (by the remarks above). \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Your email address will not be published.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.