The question is basically the one outlined in the title. Let $\mathcal{T}$ be a triangulated category containing infinite direct sums (e.g. $D_{qc}(X)$ for some separated, finite type over a field $k$, scheme $X$) and consider the subcategory $\mathcal{E}$ generated by an object $E$ of $\mathcal{T}$. Here by subcategory generated I mean the smallest thick full triangulated subcategory containing all direct sums (and thus homotopy colimits). Is it true that $\mathcal{E}$ contains all the cohomology sheaves of $E$? Does it contain only some of them? It is clear that if I were to consider the subcategory generated in $\mathcal{T}^c$ (hence admitting only finite direct sums) this would necessarily be true. Indeed, considering the geometric example $\mathcal{T} = D_{qc}(X)$, then the cohomology sheaves of a perfect complex are not necessarily perfect. Thanks.


1 Answer 1



Let $j:\mathbb{A}^2_k\smallsetminus\{0\}\to \mathbb{A}^2_k$ be the canonical open embedding. Then the derived pushforward $Rj_*$ is fully faithful and colimit-preserving. In particular, the subcategory of $D_{qc}(\mathbb{A}^2)$ generated under colimits by $\mathscr{A}:=Rj_*\mathcal{O}$ is contained in this subcategory (in fact it coincides with the category of $\mathscr{A}$-modules, which is a subcategory because $\mathscr{A}$ is an idempotent algebra). However $H_0\mathscr{A}=k[x,y]$ is not.

  • $\begingroup$ Let me see if I understand you counterexample. You are saying that the subcategory generated by $\mathcal{A}$ is contained in the subcategory $Rj_{\ast} D_{qc}(\mathbb{A}^2_{k} \setminus \{0\})$. by fully faithfulness and colimit preservation. However, $H_0 \mathcal{A}$ is not in this larger subcategory because it is not a module over $k[x^{\pm 1}, y^{\pm 1}]$, right? $\endgroup$ Dec 15, 2019 at 17:55
  • 3
    $\begingroup$ @Federico almost: $\mathcal{A}$-modules are nothing to do with $k[x^{\pm1},y^{\pm1}]$-modules. $H_0\mathcal{A}$ is not in the subcategory because it generates everything under homotopy colimits (and so if it were, the subcategory would be everything but it's not). $\endgroup$ Dec 15, 2019 at 17:59
  • $\begingroup$ I see, thank you! $\endgroup$ Dec 15, 2019 at 18:01
  • $\begingroup$ Another way to see $H_0\mathcal{A}$ is not in the image of the pushforward is to check that it isn’t fixed by the operation “pull back, then push forward”. $\endgroup$ Dec 16, 2019 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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