Lemma 19.13.4. Let $\mathcal{A}$ be a Grothendieck abelian category. Then

1. $D(\mathcal{A})$ has both direct sums and products,

2. direct sums are obtained by taking termwise direct sums of any complexes,

3. products are obtained by taking termwise products of K-injective complexes.

Proof. Let $K^\bullet _ i$, $i \in I$ be a family of objects of $D(\mathcal{A})$ indexed by a set $I$. We claim that the termwise direct sum $\bigoplus _{i \in I} K^\bullet _ i$ is a direct sum in $D(\mathcal{A})$. Namely, let $I^\bullet$ be a K-injective complex. Then we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(\bigoplus \nolimits _{i \in I} K^\bullet _ i, I^\bullet ) & = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(\bigoplus \nolimits _{i \in I} K^\bullet _ i, I^\bullet ) \\ & = \prod \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet _ i, I^\bullet ) \\ & = \prod \nolimits _{i \in I} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet _ i, I^\bullet ) \end{align*}

as desired. This is sufficient since any complex can be represented by a K-injective complex by Theorem 19.12.6. To construct the product, choose a K-injective resolution $K_ i^\bullet \to I_ i^\bullet$ for each $i$. Then we claim that $\prod _{i \in I} I_ i^\bullet$ is a product in $D(\mathcal{A})$. This follows from Derived Categories, Lemma 13.31.5. $\square$

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