# Derived Category of strictly simplicial algebraic space vs. systems of objects in the derived categories

Let $$X_{\bullet}^+$$ be a strictly simplicial algebraic space and for a morphism $$\delta:[m]\to[n]$$ in $$\Delta^+$$, let $$\delta:X_n\to X_m$$ also denote the associated map (by abuse of notation). Then one can consider the category $$\operatorname{Mod}(\mathcal{O}_{X_{\bullet}^+})$$ of $$\mathcal{O}_{X_{\bullet}^+}$$-modules in the étale topos of $$X_{\bullet}^+$$. Let $$D(\mathcal{O}_{X_{\bullet}^+})$$ be the associated derived category. Let $$D'$$ be the category whose objects are pairs $$(E_n,\varphi_{\delta})$$, where for each $$n$$, $$E_n$$ is an object of $$D(\mathcal{O}_{X_n})$$, and for $$\delta:[m]\to[n]$$ we have a morphism $$\varphi_{\delta}:\delta^*E_{m}\to E_{n}$$, such that the various $$\varphi_{\delta}$$ are compatible with composition.

It appears that we have a morphism of (triangulated?) categories $$D(\mathcal{O}_{X^+_{\bullet}})\to D'$$ given by restriction. Is this morphism an equivalence?

I was trying to construct a quasi-inverse by replacing a system $$(E_n,\varphi_{\delta})$$ with an isomorphic one in which the $$\varphi_{\delta}$$ are given by actual complex maps, and the compatibility holds on the level of complexes. This way, one should get an object in $$D(\mathcal{O}_{X^+_{\bullet}})$$. However, I can only get the compatibility up to homotopy, so it seems like this doesn't work.

• This sort of thing is typically false-- in fact I doubt that D' is triangulated. What happens when all of the $X_n$ are just points? – Phil Tosteson Jan 16 at 1:38
• The answer to your question is a plain no (except in trivial cases, where each $X_n$ is empty for instance). This is one of the many reasons we work with $\infty$-categories. If you pass to derived $\infty$-categories (inverting quasi-isomorphisms in the setting of $\infty$-categories), then what you ask becomes true. – Denis-Charles Cisinski Jan 16 at 9:10

This kind of functor is flat-out never faithful, because it is very easy for two natural transformations to have components that are equivalent in the derived category without being equivalent in a natural way leading to equality in $$D$$. For extremely simple diagram shapes it can be full and essentially surjective, but the simplex category is not very simple at all, so this won’t happen here either. I would give explicit counterexamples but your framework is quite complex. It’s just never going to be anywhere close to working.