The Stacks project

Lemma 13.33.5. Let $\mathcal{A}$ be an abelian category. If $\mathcal{A}$ has exact countable direct sums, then $D(\mathcal{A})$ has countable direct sums. In fact given a collection of complexes $K_ i^\bullet $ indexed by a countable index set $I$ the termwise direct sum $\bigoplus K_ i^\bullet $ is the direct sum of $K_ i^\bullet $ in $D(\mathcal{A})$.

Proof. Let $L^\bullet $ be a complex. Suppose given maps $\alpha _ i : K_ i^\bullet \to L^\bullet $ in $D(\mathcal{A})$. This means there exist quasi-isomorphisms $s_ i : M_ i^\bullet \to K_ i^\bullet $ of complexes and maps of complexes $f_ i : M_ i^\bullet \to L^\bullet $ such that $\alpha _ i = f_ is_ i^{-1}$. By assumption the map of complexes

\[ s : \bigoplus M_ i^\bullet \longrightarrow \bigoplus K_ i^\bullet \]

is a quasi-isomorphism. Hence setting $f = \bigoplus f_ i$ we see that $\alpha = fs^{-1}$ is a map in $D(\mathcal{A})$ whose composition with the coprojection $K_ i^\bullet \to \bigoplus K_ i^\bullet $ is $\alpha _ i$. We omit the verification that $\alpha $ is unique. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A5L. Beware of the difference between the letter 'O' and the digit '0'.