# Can homotopy colimits recover cohomology sheaves?

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

No.

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.

• 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? – Federico Barbacovi Dec 15 '19 at 17:55
• @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). – Denis Nardin Dec 15 '19 at 17:59
• I see, thank you! – Federico Barbacovi Dec 15 '19 at 18:01
• 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”. – Peter LeFanu Lumsdaine Dec 16 '19 at 2:23