Remark 38.40.4. Let $X$ be a scheme. Let $E \in D(\mathcal{O}_ X)$ be a perfect object such that $H^ i(E)$ is a perfect $\mathcal{O}_ X$-module of tor dimension $\leq 1$ for all $i \in \mathbf{Z}$. This property sometimes allows one to reduce questions about $E$ to questions about $H^ i(E)$. For example, suppose

$\mathcal{E}^ a \xrightarrow {d^ a} \ldots \xrightarrow {d^{b - 2}} \mathcal{E}^{b - 1} \xrightarrow {d^{b - 1}} \mathcal{E}^ b$

is a bounded complex of finite locally free $\mathcal{O}_ X$-modules representing $E$. Then $\mathop{\mathrm{Im}}(d^ i)$ and $\mathop{\mathrm{Ker}}(d^ i)$ are finite locally free $\mathcal{O}_ X$-modules for all $i$. Namely, suppose by induction we know this for all indices bigger than $i$. Then we can first use the short exact sequence

$0 \to \mathop{\mathrm{Im}}(d^ i) \to \mathop{\mathrm{Ker}}(d^{i + 1}) \to H^{i + 1}(E) \to 0$

and the assumption that $H^{i + 1}(E)$ is perfect of tor dimension $\leq 1$ to conclude that $\mathop{\mathrm{Im}}(d^ i)$ is finite locally free. The same argument used again for the short exact sequence

$0 \to \mathop{\mathrm{Ker}}(d^ i) \to \mathcal{E}^ i \to \mathop{\mathrm{Im}}(d^ i) \to 0$

then gives that $\mathop{\mathrm{Ker}}(d^ i)$ is finite locally free. It follows that the distinguished triangles

$\tau _{\leq k - 1}E \to \tau _{\leq k}E \to H^ k(E)[-k] \to (\tau _{\leq k - 1}E)[1]$

are represented by the following short exact sequences of bounded complexes of finite locally free modules

$\begin{matrix} & & & & & & 0 \\ & & & & & & \downarrow \\ \mathcal{E}^ a & \to & \ldots & \to & \mathcal{E}^{k - 2} & \to & \mathop{\mathrm{Ker}}(d^{k - 1}) \\ \downarrow & & & & \downarrow & & \downarrow \\ \mathcal{E}^ a & \to & \ldots & \to & \mathcal{E}^{k - 2} & \to & \mathcal{E}^{k - 1} & \to & \mathop{\mathrm{Ker}}(d^ k) \\ & & & & & & \downarrow & & \downarrow \\ & & & & & & \mathop{\mathrm{Im}}(d^{k - 1}) & \to & \mathop{\mathrm{Ker}}(d^ k) \\ & & & & & & \downarrow \\ & & & & & & 0 \end{matrix}$

Here the complexes are the rows and the “obvious” zeros are omitted from the display.

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).