Lemma 13.34.2. Let $\mathcal{A}$ be an abelian category with exact countable products. Then
$D(\mathcal{A})$ has countable products,
countable products $\prod K_ i$ in $D(\mathcal{A})$ are obtained by taking termwise products of any complexes representing the $K_ i$, and
$H^ p(\prod K_ i) = \prod H^ p(K_ i)$.
Proof. Let $K_ i^\bullet $ be a complex representing $K_ i$ in $D(\mathcal{A})$. Let $L^\bullet $ be a complex. Suppose given maps $\alpha _ i : L^\bullet \to K_ i^\bullet $ in $D(\mathcal{A})$. This means there exist quasi-isomorphisms $s_ i : K_ i^\bullet \to M_ i^\bullet $ of complexes and maps of complexes $f_ i : L^\bullet \to M_ i^\bullet $ such that $\alpha _ i = s_ i^{-1}f_ i$. By assumption the map of complexes
is a quasi-isomorphism. Hence setting $f = \prod f_ i$ we see that $\alpha = s^{-1}f$ is a map in $D(\mathcal{A})$ whose composition with the projection $\prod K_ i^\bullet \to K_ i^\bullet $ is $\alpha _ i$. We omit the verification that $\alpha $ is unique. $\square$
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