Lemma 57.17.2. Let \mathcal{A} be a countable abelian category. Then D^ b(\mathcal{A}) is countable.
Proof. It suffices to prove the statement for D(\mathcal{A}) as the others are full subcategories of this one. Since every object in D(\mathcal{A}) is a complex of objects of \mathcal{A} it is immediate that the set of isomorphism classes of objects of D^ b(\mathcal{A}) is countable. Moreover, for bounded complexes A^\bullet and B^\bullet of \mathcal{A} it is clear that \mathop{\mathrm{Hom}}\nolimits _{K^ b(\mathcal{A})}(A^\bullet , B^\bullet ) is countable. We have
\mathop{\mathrm{Hom}}\nolimits _{D^ b(\mathcal{A})}(A^\bullet , B^\bullet ) = \mathop{\mathrm{colim}}\nolimits _{s : (A')^\bullet \to A^\bullet \text{ qis and }(A')^\bullet \text{ bounded}} \mathop{\mathrm{Hom}}\nolimits _{K^ b(\mathcal{A})}((A')^\bullet , B^\bullet )
by Derived Categories, Lemma 13.11.6. Thus this is a countable set as a countable colimit of \square
Comments (0)