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Dec 30, 2015 at 1:49 comment added Ben Wieland Your premise is false: the map on $K^0$ is not a rational isomorphism. Just take the subcategory of objects with trivial class in $K^0$. See Thomason's Classification of triangulated subcategories.
Dec 26, 2015 at 18:00 comment added Mikhail Bondarko I doubt that any property of pre-triangulated smooth DG categories (that is not valid in some more general context) may be classically known.:) So, could you give a reference?
Dec 26, 2015 at 17:30 comment added Dmitry Vaintrob Sure. Take complexes of vector spaces of even total dimension.
Dec 26, 2015 at 7:45 comment added მამუკა ჯიბლაძე Could you give an example when the map on $K^0$ is not an isomorphism?
Dec 26, 2015 at 4:43 history asked Dmitry Vaintrob CC BY-SA 3.0