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The Stacks project

Lemma 13.35.7. Let \mathcal{A} be an abelian category. Let \mathcal{D} = D(\mathcal{A}). Let \mathcal{E} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) be a subset which we view as a subset of \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) also. Let K be an object of \mathcal{D}.

  1. Let b \geq a and assume H^ i(K) is zero for i \not\in [a, b] and H^ i(K) \in \mathcal{E} if i \in [a, b]. Then K is in smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)}).

  2. Let b \geq a and assume H^ i(K) is zero for i \not\in [a, b] and H^ i(K) \in smd(add(\mathcal{E})) if i \in [a, b]. Then K is in smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)}).

  3. Let b \geq a and assume K can be represented by a complex K^\bullet with K^ i = 0 for i \not\in [a, b] and K^ i \in \mathcal{E} for i \in [a, b]. Then K is in smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)}).

  4. Let b \geq a and assume K can be represented by a complex K^\bullet with K^ i = 0 for i \not\in [a, b] and K^ i \in smd(add(\mathcal{E})) for i \in [a, b]. Then K is in smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)}).

Proof. We will use Lemma 13.35.4 without further mention. We will prove (2) which trivially implies (1). We use induction on b - a. If b - a = 0, then K is isomorphic to H^ i(K)[-a] in \mathcal{D} and the result is immediate. If b - a > 0, then we consider the distinguished triangle

\tau _{\leq b - 1}K^\bullet \to K^\bullet \to K^ b[-b]

and we conclude by induction on b - a. We omit the proof of (3) and (4). \square


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