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

$K_0(\textit{Coh}(Z)) \longrightarrow K_0(D_{Z, perf}(\mathcal{O}_ X))$

defined as follows. For any coherent $\mathcal{O}_ Z$-module $\mathcal{F}$ denote $\mathcal{F}$ the object of $D(\mathcal{O}_ X)$ which has $\mathcal{F}$ in degree $0$ and is zero in other degrees. Then $\mathcal{F}$ is a perfect complex on $X$ by Derived Categories of Schemes, Lemma 36.11.8. Hence $\mathcal{F}$ 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 $\mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to \mathcal{F}$, 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 $[\mathcal{F}]$ with apologies for the horrendous notation.

Comment #7964 by WhyFiniteDimension on

It seems the construction does not need $X$ to be finite dimensional.

Comment #8192 by on

WhyFiniteDimension indeed! This led to some slight improvements also in later results, for example Proposition 42.57.1. Thanks! Changes are here.

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