Lemma 13.10.5. Let $\mathcal{A}$ be an additive category. The categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ are full triangulated subcategories of $K(\mathcal{A})$.

Proof. Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma 13.4.16 to show that they are triangulated subcategories. It is clear that each of the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ is preserved under the shift functors $[1], [-1]$. Finally, suppose that $f : A^\bullet \to B^\bullet$ is a morphism in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$. Then $(A^\bullet , B^\bullet , C(f)^\bullet , f, i, -p)$ is a distinguished triangle of $K(\mathcal{A})$ with $C(f)^\bullet \in K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$ as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, $K^\bullet \to L^\bullet$ is isomorphic to an termwise split injection of complexes in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, see Lemma 13.9.6 and then one can directly take the associated distinguished triangle.) $\square$

## Comments (2)

Comment #1706 by Keenan Kidwell on

In the statement of the lemma, the first instance of "subcategory" should be "category."

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