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: