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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Phil Tosteson Jan 16 at 1:38
  • 5
    $\begingroup$ 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. $\endgroup$ – 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.

  • 2
    $\begingroup$ Basically it only works when your indexing category has homological dimension less than or equal to 1. The simplex category has infinite homological dimension. $\endgroup$ – Phil Tosteson Jan 16 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.