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.



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


