The Stacks project

Proof. This lemma will follow in a straightforward manner from Lemmas 36.37.2, 36.37.3, and 36.37.4 whose results we will use without further mention. Let us construct an inverse map

Namely, any object of $D_{perf}(\mathcal{O}_ X)$ can be represented by a bounded complex $\mathcal{E}^\bullet$ of finite locally free $\mathcal{O}_ X$-modules. Then we set

Of course we have to show that this is well defined. For the moment we view $c$ as a map defined on bounded complexes of finite locally free $\mathcal{O}_ X$-modules.

Suppose that $\mathcal{E}^\bullet \to \mathcal{F}^\bullet$ is a surjective map of bounded complexes of finite locally free $\mathcal{O}_ X$-modules. Let $\mathcal{K}^\bullet$ be the kernel. Then we obtain short exact sequences of $\mathcal{O}_ X$-modules

which are locally split because $\mathcal{F}^ n$ is finite locally free. Hence $\mathcal{K}^\bullet$ is also a bounded complex of finite locally free $\mathcal{O}_ X$-modules and we have $c(\mathcal{E}^\bullet ) = c(\mathcal{K}^\bullet ) + c(\mathcal{F}^\bullet )$ in $K_0(\textit{Vect}(X))$.

Suppose given a bounded complex $\mathcal{E}^\bullet$ of finite locally free $\mathcal{O}_ X$-modules which is acyclic. Say $\mathcal{E}^ n = 0$ for $n \not\in [a, b]$. Then we can break $\mathcal{E}^\bullet$ into short exact sequences

Arguing by descending induction we see that $\mathcal{F}^{b - 2}, \ldots , \mathcal{F}^{a + 1}$ are finite locally free $\mathcal{O}_ X$-modules, and

Thus our construction gives zero on acyclic complexes.

It follows from the results of the preceding two paragraphs that $c$ is well defined. Namely, suppose the bounded complexes $\mathcal{E}^\bullet$ and $\mathcal{F}^\bullet$ of finite locally free $\mathcal{O}_ X$-modules represent the same object of $D(\mathcal{O}_ X)$. Then we can find quasi-isomorphisms $a : \mathcal{G}^\bullet \to \mathcal{E}^\bullet$ and $b : \mathcal{G}^\bullet \to \mathcal{F}^\bullet$ with $\mathcal{G}^\bullet$ bounded complex of finite locally free $\mathcal{O}_ X$-modules. We obtain a short exact sequence of complexes

see Derived Categories, Definition 13.9.1. Since $a$ is a quasi-isomorphism, the cone $C(a)^\bullet$ is acyclic (this follows for example from the discussion in Derived Categories, Section 13.12). Hence

as desired. The same argument using $b$ shows that $0 = c(\mathcal{F}^\bullet ) - c(\mathcal{G}^\bullet )$. Hence we find that $c(\mathcal{E}^\bullet ) = c(\mathcal{F}^\bullet )$ and $c$ is well defined.

A similar argument using the cone on a map $\mathcal{E}^\bullet \to \mathcal{F}^\bullet$ of bounded complexes of finite locally free $\mathcal{O}_ X$-modules shows that $c(Y) = c(X) + c(Z)$ if $X \to Y \to Z$ is a distinguished triangle in $D_{perf}(\mathcal{O}_ X)$. Details omitted. Thus we get the desired homomorphism of abelian groups $c : K_0(X) \to K_0(\textit{Vect}(X))$.

It is clear that the composition $K_0(\textit{Vect}(X)) \to K_0(X) \to K_0(\textit{Vect}(X))$ is the identity. On the other hand, let $\mathcal{E}^\bullet$ be a bounded complex of finite locally free $\mathcal{O}_ X$-modules. Then the the existence of the distinguished triangles of “stupid truncations” (see Homology, Section 12.15)

and induction show that

in $K_0(X) = K_0(D_{perf}(\mathcal{O}_ X))$ with apologies for the notation. Hence the map $K_0(\textit{Vect}(X)) \to K_0(D_{perf}(\mathcal{O}_ X)) = K_0(X)$ is surjective which finishes the proof. $\square$