Lemma 13.31.5. 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. Observe that the complex
\[ C : \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I^{b - 1}) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I^ b) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I^{b + 1}) \]
has cohomology $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$ in the middle. Similarly, the complex
\[ C_ t : \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I_ t^{b - 1}) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I_ t^ b) \to \prod \nolimits _ b \mathop{\mathrm{Hom}}\nolimits (K^{-b}, I_ t^{b + 1}) \]
computes $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet )$. Next, observe that we have
\[ C = \prod \nolimits _{t \in T} C_ t \]
as complexes of abelian groups by our choice of $I$. Taking products is an exact functor on the category of abelian groups. Hence if $K^\bullet $ is acyclic, then $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I_ t^\bullet ) = 0$, hence $C_ t$ is acyclic, hence $C$ is acyclic, hence we get $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) = 0$. Thus we find that $I^\bullet $ is K-injective. Having said this, we can use Lemma 13.31.2 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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