Lemma 13.35.1. Let $\mathcal{T}$ be a triangulated category. Given full subcategories $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ we have $(\mathcal{A} \star \mathcal{B}) \star \mathcal{C} = \mathcal{A} \star (\mathcal{B} \star \mathcal{C})$.
13.35 Operations on full subcategories
Let $\mathcal{T}$ be a triangulated category. We will identify full subcategories of $\mathcal{T}$ with subsets of $\mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$. Given full subcategories $\mathcal{A}, \mathcal{B}, \ldots $ we let
$\mathcal{A}[a, b]$ for $-\infty \leq a \leq b \leq \infty $ be the full subcategory of $\mathcal{T}$ consisting of all objects $A[-i]$ with $i \in [a, b] \cap \mathbf{Z}$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ (note the minus sign!),
$smd(\mathcal{A})$ be the full subcategory of $\mathcal{T}$ consisting of all objects which are isomorphic to direct summands of objects of $\mathcal{A}$,
$add(\mathcal{A})$ be the full subcategory of $\mathcal{T}$ consisting of all objects which are isomorphic to finite direct sums of objects of $\mathcal{A}$,
$\mathcal{A} \star \mathcal{B}$ be the full subcategory of $\mathcal{T}$ consisting of all objects $X$ of $\mathcal{T}$ which fit into a distinguished triangle $A \to X \to B$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$,
$\mathcal{A}^{\star n} = \mathcal{A} \star \ldots \star \mathcal{A}$ with $n \geq 1$ factors (we will see $\star $ is associative below),
$smd(add(\mathcal{A})^{\star n}) = smd(add(\mathcal{A}) \star \ldots \star add(\mathcal{A}))$ with $n \geq 1$ factors.
If $E$ is an object of $\mathcal{T}$, then we think of $E$ sometimes also as the full subcategory of $\mathcal{T}$ whose single object is $E$. Then we can consider things like $add(E[-1, 2])$ and so on and so forth. We warn the reader that this notation is not universally accepted.
Proof. If we have distinguished triangles $A \to X \to B$ and $X \to Y \to C$ then by Axiom TR4 we have distinguished triangles $A \to Y \to Z$ and $B \to Z \to C$. $\square$
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. $\square$
Lemma 13.35.3. Let $\mathcal{T}$ be a triangulated category. Given full subcategories $\mathcal{A}$, $\mathcal{B}$ the full subcategories $add(\mathcal{A}) \star add(\mathcal{B})$ and $smd(add(\mathcal{A}))$ are closed under direct sums.
Proof. Namely, if $A \to X \to B$ and $A' \to X' \to B'$ are distinguished triangles and $A, A' \in add(\mathcal{A})$ and $B, B' \in add(\mathcal{B})$ then $A \oplus A' \to X \oplus X' \to B \oplus B'$ is a distinguished triangle with $A \oplus A' \in add(\mathcal{A})$ and $B \oplus B' \in add(\mathcal{B})$. The result for $smd(add(\mathcal{A}))$ is trivial. $\square$
Lemma 13.35.4. Let $\mathcal{T}$ be a triangulated category. Given a full subcategory $\mathcal{A}$ for $n \geq 1$ the subcategory defined above is a strictly full subcategory of $\mathcal{T}$ closed under direct sums and direct summands and $\mathcal{C}_{n + m} = smd(\mathcal{C}_ n \star \mathcal{C}_ m)$ for all $n, m \geq 1$.
\[ \mathcal{C}_ n = smd(add(\mathcal{A})^{\star n}) = smd(add(\mathcal{A}) \star \ldots \star add(\mathcal{A})) \]
Proof. Immediate from Lemmas 13.35.1, 13.35.2, and 13.35.3. $\square$
Remark 13.35.5. Let $F : \mathcal{T} \to \mathcal{T}'$ be an exact functor of triangulated categories. Given a full subcategory $\mathcal{A}$ of $\mathcal{T}$ we denote $F(\mathcal{A})$ the full subcategory of $\mathcal{T}'$ whose objects consists of all objects $F(A)$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. We have We omit the trivial verifications.
\[ F(\mathcal{A}[a, b]) = F(\mathcal{A})[a, b] \]
\[ F(smd(\mathcal{A})) \subset smd(F(\mathcal{A})), \]
\[ F(add(\mathcal{A})) \subset add(F(\mathcal{A})), \]
\[ F(\mathcal{A} \star \mathcal{B}) \subset F(\mathcal{A}) \star F(\mathcal{B}), \]
\[ F(\mathcal{A}^{\star n}) \subset F(\mathcal{A})^{\star n}. \]
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 We omit the trivial verifications.
\[ \left(\bigcup \mathcal{A}_ i\right)[a, b] = \bigcup \mathcal{A}_ i[a, b] \]
\[ smd\left(\bigcup \mathcal{A}_ i\right) = \bigcup smd(\mathcal{A}_ i), \]
\[ add\left(\bigcup \mathcal{A}_ i\right) = \bigcup add(\mathcal{A}_ i), \]
\[ \left(\bigcup \mathcal{A}_ i\right) \star \mathcal{B} = \bigcup \mathcal{A}_ i \star \mathcal{B}, \]
\[ \mathcal{B} \star \left(\bigcup \mathcal{A}_ i\right) = \bigcup \mathcal{B} \star \mathcal{A}_ i, \]
\[ \left(\bigcup \mathcal{A}_ i\right)^{\star n} = \bigcup \mathcal{A}_ i^{\star n}. \]
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 $\mathcal{D}$.
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 $smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)})$.
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 $smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)})$.
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 $smd(add(\mathcal{E}[a, b])^{\star (b - a + 1)})$.
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). $\square$
Lemma 13.35.8. Let $\mathcal{T}$ be a triangulated category. Let $H : \mathcal{T} \to \mathcal{A}$ be a homological functor to an abelian category $\mathcal{A}$. Let $a \leq b$ and $\mathcal{E} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$ be a subset such that $H^ i(E) = 0$ for $E \in \mathcal{E}$ and $i \not\in [a, b]$. Then for $X \in smd(add(\mathcal{E}[-m, m])^{\star n})$ we have $H^ i(X) = 0$ for $i \not\in [-m + na, m + nb]$.
Proof. Omitted. Pleasant exercise in the definitions. $\square$
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