Lemma 13.6.4. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor of a pre-triangulated category into an abelian category. Let $\mathcal{D}_ H^{+}, \mathcal{D}_ H^{-}, \mathcal{D}_ H^ b$ be the full subcategory of $\mathcal{D}$ with objects

$\begin{matrix} \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ H^{+}) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \ll 0\} \\ \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ H^{-}) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }n \gg 0\} \\ \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ H^ b) = \{ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid H(X[n]) = 0\text{ for all }|n| \gg 0\} \end{matrix}$

Each of these is a strictly full saturated pre-triangulated subcategory of $\mathcal{D}$. If $\mathcal{D}$ is a triangulated category, then each is a triangulated subcategory.

Proof. Let us prove this for $\mathcal{D}_ H^{+}$. It is clear that it is preserved under $$ and $[-1]$. If $(X, Y, Z, f, g, h)$ is a distinguished triangle of $\mathcal{D}$ and $H(X[n]) = H(Y[n]) = 0$ for all $n \ll 0$, then also $H(Z[n]) = 0$ for all $n \ll 0$ by the long exact sequence (13.3.5.1). Hence we may apply Lemma 13.4.16 to see that $\mathcal{D}_ H^{+}$ is a pre-triangulated subcategory (respectively a triangulated subcategory if $\mathcal{D}$ is a triangulated category). The assertion of being saturated follows from

\begin{align*} H((X \oplus Y)[n]) = 0 & \Rightarrow H(X[n] \oplus Y[n]) = 0 \\ & \Rightarrow H(X[n]) \oplus H(Y[n]) = 0 \\ & \Rightarrow H(X[n]) = H(Y[n]) = 0 \end{align*}

for all $n \in \mathbf{Z}$. $\square$

There are also:

• 2 comment(s) on Section 13.6: Quotients of triangulated categories

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).