Lemma 13.29.5. Let $\mathcal{A}$ be an abelian category. Let $T$ be a set and for each $t \in T$ let $I_ t^\bullet$ be a K-injective complex. If $I^ n = \prod _ t I_ t^ n$ exists for all $n$, then $I^\bullet$ is a K-injective complex. Moreover, $I^\bullet$ represents the product of the objects $I_ t^\bullet$ in $D(\mathcal{A})$.

Proof. Let $K^\bullet$ be an complex. Then we have

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) = \prod \nolimits _{t \in T} \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet )$

Since taking products is an exact functor on the category of abelian groups we see that if $K^\bullet$ is acyclic, then $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$ is acyclic because this is true for each of the complexes $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet )$. Having said this, we can use Lemma 13.29.2 to conclude that

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , I^\bullet ) = \prod \nolimits _{t \in T} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K^\bullet , I_ t^\bullet )$

and indeed $I^\bullet$ represents the product in the derived category. $\square$

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