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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: