Lemma 42.56.9. Let $X$ be a Noetherian regular scheme. Let $Z \subset X$ be a closed subscheme. The maps constructed in Remarks 42.56.7 and 42.56.8 are mutually inverse and we get $K'_0(Z) = K_0(D_{Z, perf}(\mathcal{O}_ X))$.

Proof. Clearly the composition

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

is the identity map. Thus it suffices to show the first arrow is surjective. Let $E$ be an object of $D_{Z, perf}(\mathcal{O}_ X)$. Recall that $D_{perf}(\mathcal{O}_ X) = D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.8. Hence the cohomologies $H^ i(E)$ are coherent, can be viewed as objects of $D_{Z, perf}(\mathcal{O}_ X)$, and only a finite number are nonzero. Using the distinguished triangles of canonical truncations the reader sees that

$[E] = \sum (-1)^ i[H^ i(E)[0]]$

in $K_0(D_{Z, perf}(\mathcal{O}_ X))$. Then it suffices to show that $[\mathcal{F}[0]]$ is in the image of the map for any coherent $\mathcal{O}_ X$-module set theoretically supported on $Z$. Since we can find a finite filtration on $\mathcal{F}$ whose subquotients are $\mathcal{O}_ Z$-modules, the proof is complete. $\square$

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