Lemma 13.35.2. Let \mathcal{T} be a triangulated category. Given full subcategories \mathcal{A}, \mathcal{B} we have smd(\mathcal{A}) \star smd(\mathcal{B}) \subset smd(\mathcal{A} \star \mathcal{B}) and smd(smd(\mathcal{A}) \star smd(\mathcal{B})) = smd(\mathcal{A} \star \mathcal{B}).
Proof. Suppose we have a distinguished triangle A_1 \to X \to B_1 where A_1 \oplus A_2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) and B_1 \oplus B_2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B}). Then we obtain a distinguished triangle A_1 \oplus A_2 \to A_2 \oplus X \oplus B_2 \to B_1 \oplus B_2 which proves that X is in smd(\mathcal{A} \star \mathcal{B}). This proves the inclusion. The equality follows trivially from this. \square
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