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.
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 colimitpreserving. 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? $\endgroup$ – Federico Dec 15 '19 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). $\endgroup$ – Denis Nardin Dec 15 '19 at 17:59


$\begingroup$ 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”. $\endgroup$ – Peter LeFanu Lumsdaine Dec 16 '19 at 2:23