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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