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

by Derived Categories, Lemma 13.11.6. Thus this is a countable set as a countable colimit of $\square$

## Post a comment

Your email address will not be published.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.

## Comments (0)