Remark 13.35.6. Let \mathcal{T} be a triangulated category. Given full subcategories \mathcal{A}_1 \subset \mathcal{A}_2 \subset \mathcal{A}_3 \subset \ldots and \mathcal{B} of \mathcal{T} we have
\left(\bigcup \mathcal{A}_ i\right)[a, b] = \bigcup \mathcal{A}_ i[a, b]
smd\left(\bigcup \mathcal{A}_ i\right) = \bigcup smd(\mathcal{A}_ i),
add\left(\bigcup \mathcal{A}_ i\right) = \bigcup add(\mathcal{A}_ i),
\left(\bigcup \mathcal{A}_ i\right) \star \mathcal{B} = \bigcup \mathcal{A}_ i \star \mathcal{B},
\mathcal{B} \star \left(\bigcup \mathcal{A}_ i\right) = \bigcup \mathcal{B} \star \mathcal{A}_ i,
\left(\bigcup \mathcal{A}_ i\right)^{\star n} = \bigcup \mathcal{A}_ i^{\star n}.
We omit the trivial verifications.
Comments (0)