This lemma will follow in a straightforward manner from Lemmas \ref{lemma-resolution-property-perfect-complex}, \ref{lemma-resolution-property-map-perfect-complex}, and \ref{lemma-resolution-property-homotopy-map-perfect-complex} whose results we will use without further mention. Let us construct an inverse map $$ c : K_0(X) = K_0(D_{perf}(\mathcal{O}_X)) \longrightarrow K_0(\textit{Vect}(X)) $$ 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 $$ c([\mathcal{E}^\bullet]) = \sum (-1)^i[\mathcal{E}^i] $$ 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. \medskip\noindent 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 $$ 0 \to \mathcal{K}^n \to \mathcal{E}^n \to \mathcal{F}^n \to 0 $$ 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 $K_0(\textit{Vect}(X))$. \medskip\noindent 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 $$ \begin{matrix} 0 \to \mathcal{E}^a \to \mathcal{E}^{a + 1} \to \mathcal{F}^{a + 1} \to 0,\\ 0 \to \mathcal{F}^{a + 1} \to \mathcal{E}^{a + 2} \to \mathcal{F}^{a + 3} \to 0, \\ \ldots \\ 0 \to \mathcal{F}^{b - 3} \to \mathcal{E}^{b - 2} \to \mathcal{F}^{b - 2} \to 0, \\ 0 \to \mathcal{F}^{b - 2} \to \mathcal{E}^{b - 1} \to \mathcal{E}^b \to 0 \end{matrix} $$ 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 $$ c(\mathcal{E}^\bullet) = \sum (-1)[\mathcal{E}^n] = \sum (-1)^n([\mathcal{F}^{n - 1}] + [\mathcal{F}^n]) = 0 $$ Thus our construction gives zero on acyclic complexes. \medskip\noindent 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 $$ 0 \to \mathcal{E}^\bullet \to C(a)^\bullet \to \mathcal{G}^\bullet[1] \to 0 $$ see Derived Categories, Definition \ref{derived-definition-cone}. Since $a$ is a quasi-isomorphism, the cone $C(a)^\bullet$ is acyclic (this follows for example from the discussion in Derived Categories, Section \ref{derived-section-canonical-delta-functor}). Hence $$ 0 = c(C(f)^\bullet) = c(\mathcal{E}^\bullet) + c(\mathcal{G}^\bullet[1]) = c(\mathcal{E}^\bullet) - c(\mathcal{G}^\bullet) $$ 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. \medskip\noindent 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 $c : K_0(X) \to K_0(\textit{Vect}(X))$. \medskip\noindent 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 \ref{homology-section-truncations}) $$ \sigma_{\geq n}\mathcal{E}^\bullet \to \sigma_{\geq n - 1}\mathcal{E}^\bullet \to \mathcal{E}^{n - 1}[-n + 1] \to (\sigma_{\geq n}\mathcal{E}^\bullet)[1] $$ and induction show that $$ [\mathcal{E}^\bullet] = \sum (-1)^i[\mathcal{E}^i[0]] $$ 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.