Lemma 13.35.3. Let \mathcal{T} be a triangulated category. Given full subcategories \mathcal{A}, \mathcal{B} the full subcategories add(\mathcal{A}) \star add(\mathcal{B}) and smd(add(\mathcal{A})) are closed under direct sums.
Proof. Namely, if A \to X \to B and A' \to X' \to B' are distinguished triangles and A, A' \in add(\mathcal{A}) and B, B' \in add(\mathcal{B}) then A \oplus A' \to X \oplus X' \to B \oplus B' is a distinguished triangle with A \oplus A' \in add(\mathcal{A}) and B \oplus B' \in add(\mathcal{B}). The result for smd(add(\mathcal{A})) is trivial. \square
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