Remark 42.56.8. Let $X$, $Z$, $D_{Z, perf}(\mathcal{O}_ X)$ be as in Remark 42.56.7. Assume $X$ is regular. Then there is a canonical map
defined as follows. For any coherent $\mathcal{O}_ Z$-module $\mathcal{F}$ denote $\mathcal{F}[0]$ the object of $D(\mathcal{O}_ X)$ which has $\mathcal{F}$ in degree $0$ and is zero in other degrees. Then $\mathcal{F}[0]$ is a perfect complex on $X$ by Derived Categories of Schemes, Lemma 36.11.8. Hence $\mathcal{F}[0]$ is an object of $D_{Z, perf}(\mathcal{O}_ X)$. On the other hand, given a short exact sequence $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to 0$ of coherent $\mathcal{O}_ Z$-modules we obtain a distinguished triangle $\mathcal{F}[0] \to \mathcal{F}'[0] \to \mathcal{F}''[0] \to \mathcal{F}[1]$, see Derived Categories, Section 13.12. This shows that we obtain a map $K_0(\textit{Coh}(Z)) \to K_0(D_{Z, perf}(\mathcal{O}_ X))$ by sending $[\mathcal{F}]$ to $[\mathcal{F}[0]]$ with apologies for the horrendous notation.
Comments (2)
Comment #7964 by WhyFiniteDimension on
Comment #8192 by Stacks Project on