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Let $X$ be a noetherian scheme (actually, I need the case where $X$ is proper over an affine scheme), $C$ is an object of the derived category $D_{coh}(X)$ of coherent sheaves on $X$. Under which assumptions on $C$ one can be sure that there exists a distinguished triangle (in the bigger derived category $D_{qc}(X)$ of quasi-coherent sheaves) $A\to B\to C\to A[1]$, where $A$ and $B$ are some coproducts of perfect complexes on $X$? Note that this is automatically the case if $C$ is a countable homotopy colimit of perfect complexes.

Recall that homotopy colimits of this sort were defined by Bokstedt and Neeman; it appears that they are sometimes called Milnor colimits and also coincide with "true" countable filtered homotopy colimits. Also, if $X$ is countable (in the easily defined sense) then one can easily apply certain results of Neeman to prove that a triangle as above exists for any object of $D_{qc}(X)$; hence this case is not interesting.

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  • $\begingroup$ I don't quite understand what do you mean by a countable scheme, but even if you take $X = \mathrm{Spec}(\mathbb{F}_2)$, the category $D_{qc}(X)$ does not have the property that you want. Indeed, just take any vector space over $\mathbb{F}_2$ of uncountable dimension. $\endgroup$
    – Sasha
    Commented Feb 8, 2022 at 18:28
  • $\begingroup$ It is not a countable hcl of perfect of complexes; yet one can take $B=C$ in the triangle I want. $\endgroup$
    – Mikhail Bondarko
    Commented Feb 8, 2022 at 19:06
  • $\begingroup$ I think this is true if $C$ can be represented by a complex of coherent modules (for example this is true if $C$ is bounded above). Namely, write $X$ as a limit of a directed inverse system $\{X_i\}$ of "countable" schemes with affine transition maps. Any bounded complex of coherent modules comes from one of the $X_i$. It follows the whole complex comes from a countable limit of $X_i$'s which is also countable. Then apply the statement you have in your question. $\endgroup$
    – Johan
    Commented Feb 9, 2022 at 2:03
  • $\begingroup$ Thank you for this interesting comment, Johan! Does there exist any reference for the fact that any bounded (above?) complex of coherent modules comes from one of the $X_i$? $\endgroup$
    – Mikhail Bondarko
    Commented Feb 9, 2022 at 7:26
  • $\begingroup$ Coherent modules are finitely presented so they descend. $\endgroup$
    – Johan
    Commented Feb 10, 2022 at 14:30

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In the case where C is bounded above this statement was established in Neeman's https://arxiv.org/abs/1804.02240v4. Now I will try to extend his "approximation" statements to the case where $C$ is an arbitrary object of $D_{coh}(X)$ (and put the result into the arxiv:)).

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