Lemma 13.36.2. Let \mathcal{D} be a triangulated category. Let E be an object of \mathcal{D}. The subcategory
\langle E \rangle = \bigcup \nolimits _ n \langle E \rangle _ n = \bigcup \nolimits _{n, m \geq 1} smd(add(E[-m, m])^{\star n})
is a strictly full, saturated, triangulated subcategory of \mathcal{D} and it is the smallest such subcategory of \mathcal{D} containing the object E.
Proof.
The equality on the right follows from Lemma 13.36.1. It is clear that \langle E \rangle = \bigcup \langle E \rangle _ n contains E, is preserved under shifts, direct sums, direct summands. If A \in \langle E \rangle _ a and B \in \langle E \rangle _ b and if A \to X \to B \to A[1] is a distinguished triangle, then X \in \langle E \rangle _{a + b} by Lemma 13.36.1. Hence \bigcup \langle E \rangle _ n is also preserved under extensions and it follows that it is a triangulated subcategory.
Finally, let \mathcal{D}' \subset \mathcal{D} be a strictly full, saturated, triangulated subcategory of \mathcal{D} containing E. Then \mathcal{D}'[-\infty , \infty ] \subset \mathcal{D}', add(\mathcal{D}) \subset \mathcal{D}', smd(\mathcal{D}') \subset \mathcal{D}', and \mathcal{D}' \star \mathcal{D}' \subset \mathcal{D}'. In other words, all the operations we used to construct \langle E \rangle out of E preserve \mathcal{D}'. Hence \langle E \rangle \subset \mathcal{D}' and this finishes the proof.
\square
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