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.)
$\square$

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## Comments (2)

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