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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