Lemma 13.10.6. Let $\mathcal{A}$, $\mathcal{B}$ be additive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor. The induced functors

are exact functors of triangulated categories.

Lemma 13.10.6. Let $\mathcal{A}$, $\mathcal{B}$ be additive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor. The induced functors

\[ \begin{matrix} F : K(\mathcal{A}) \longrightarrow K(\mathcal{B})
\\ F : K^{+}(\mathcal{A}) \longrightarrow K^{+}(\mathcal{B})
\\ F : K^{-}(\mathcal{A}) \longrightarrow K^{-}(\mathcal{B})
\\ F : K^ b(\mathcal{A}) \longrightarrow K^ b(\mathcal{B})
\end{matrix} \]

are exact functors of triangulated categories.

**Proof.**
Suppose $A^\bullet \to B^\bullet \to C^\bullet $ is a termwise split sequence of complexes of $\mathcal{A}$ with splittings $(s^ n, \pi ^ n)$ and associated morphism $\delta : C^\bullet \to A^\bullet [1]$, see Definition 13.9.9. Then $F(A^\bullet ) \to F(B^\bullet ) \to F(C^\bullet )$ is a termwise split sequence of complexes with splittings $(F(s^ n), F(\pi ^ n))$ and associated morphism $F(\delta ) : F(C^\bullet ) \to F(A^\bullet )[1]$. Thus $F$ transforms distinguished triangles into distinguished triangles.
$\square$

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