The Stacks project

Remark 13.10.4. Let $\mathcal{A}$ be an additive category. Exactly the same proof as the proof of Proposition 13.10.3 shows that the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ are triangulated categories. Namely, the cone of a morphism between bounded (above, below) is bounded (above, below). But we prove below that these are triangulated subcategories of $K(\mathcal{A})$ which gives another proof.


Comments (2)

Comment #8866 by on

Let . I think that when proving is a triangulated category from (a) showing that it is a triangulated subcategory of we obtain less information than from (b) adapting the proof of Proposition 13.10.3. Namely, with (a) we learn that the triangulated structure of is the distinguished triangles in whose objects belong to , whereas with (b) we define the triangulated structure on to be the triangle isomorphism class in spawned by the triangles , with . With (a), all we can say is that a triangle in is distinguished iff it is isomorphic to a triangle with in , but not necessarily in . From here, I don't see how one can conclude that may be picked in . Note that is not strictly full in , since the map from the zero chain complex to is a chain homotopy equivalence.


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