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Nov 16, 2020 at 18:26 comment added Maxime Ramzi @Denis-CharlesCisinski : that seems like it is, thanks a lot !
Nov 16, 2020 at 18:25 comment added D.-C. Cisinski @MaximeRamzi I think this is this paper : mathematik.uni-regensburg.de/cisinski/unik.pdf (also arXiv:1911.02338)
Nov 16, 2020 at 18:08 comment added Maxime Ramzi @Denis-CharlesCisinski : Something seems to have happened since the time you wrote this comment, and I can't see the paper you're referring to. Could you indicate which one that was ?
May 19, 2020 at 15:07 vote accept Amos Kaminski
May 19, 2020 at 15:07 vote accept Amos Kaminski
May 19, 2020 at 15:07
May 19, 2020 at 15:07 vote accept Amos Kaminski
May 19, 2020 at 15:07
May 19, 2020 at 14:54 history edited Martin Sleziak
added top-level tag; https://meta.mathoverflow.net/questions/1457/why-are-mo-tags-formatted-as-they-are
May 19, 2020 at 14:07 comment added D.-C. Cisinski For the case of the bounded derived category, this follows right away from Corollary 7.59 in this paper (for E an abelian category); see Remark 7.60 in loc. cit. For bounded above complexes or unbounded complexes, similar universal properties may be found in Lurie's Higher Algebra, aber you need some nice properties for the heart (being cocomplete and having enough projectives or being Grothendieck).
May 19, 2020 at 12:27 answer added Dan Petersen timeline score: 6
May 19, 2020 at 10:21 comment added Simone Virili I have studied some related questions in a general (strong and stable) Grothendieck derivator in a recent preprint: arxiv.org/abs/1807.01505
May 19, 2020 at 8:25 answer added Maxime Ramzi timeline score: 10
May 18, 2020 at 12:34 comment added Maxime Ramzi Also : see Dustin Clausen's second comment under his answer here : mathoverflow.net/questions/112412/…
May 18, 2020 at 11:43 comment added Maxime Ramzi You might be interested in Higher Algebra, theorem 1.3.3.2. and remark 1.3.3.3.. Depending on what you call "derived category", and with suitable hypotheses on $\mathcal D$ there will be a canonical functor $\mathcal D^{-}(\mathcal A)\to \mathcal D$. I think under more suitable hypotheses, this should extend to $\mathcal{D(A)\to D}$
May 18, 2020 at 10:00 history asked Amos Kaminski CC BY-SA 4.0