Lemma 20.46.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)_{n \in \mathbf{N}}$ be a system of perfect objects of $D(\mathcal{O}_ X)$. Let $K = \text{hocolim} K_ n$ be the derived colimit (Derived Categories, Definition 13.33.1). Then for any object $E$ of $D(\mathcal{O}_ X)$ we have

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, E) = R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee$

where $(K_ n^\vee )$ is the inverse system of dual perfect complexes.

Proof. By Lemma 20.46.4 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

$R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (20.38.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

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).