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Dec 16, 2019 at 2:23 comment added Peter LeFanu Lumsdaine 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”.
Dec 15, 2019 at 18:02 vote accept Federico Barbacovi
Dec 15, 2019 at 18:01 comment added Federico Barbacovi I see, thank you!
Dec 15, 2019 at 17:59 comment added Denis Nardin @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).
Dec 15, 2019 at 17:55 comment added Federico Barbacovi 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?
Dec 15, 2019 at 17:43 history answered Denis Nardin CC BY-SA 4.0