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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