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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