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.

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    $\begingroup$ 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? $\endgroup$ Jan 16, 2021 at 1:38
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    $\begingroup$ 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. $\endgroup$ Jan 16, 2021 at 9:10

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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.

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    $\begingroup$ Basically it only works when your indexing category has homological dimension less than or equal to 1. The simplex category has infinite homological dimension. $\endgroup$ Jan 16, 2021 at 1:56

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