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Let $X$ be an integral scheme, proper over an algebraically closed field $k$. Let $\mathrm{Vect}(X)$ be the exact category of finite locally free $O_X$-modules. Let $K_0(\mathrm{Vect}(X))$ be its Grothendieck group. Let $K_0'(X)$ be the Grothendieck group of the abelian category $\mathrm{Coh}(O_X)$ of coherent sheaves on $X$. The inclusion functor $\mathrm{Vect}(X)\to \mathrm{Coh}(O_X)$ induces a map $\epsilon:K_0(\mathrm{Vect}(X))\to K_0'(X)$.

Question: Is $\epsilon$ always injective?

The map $\epsilon$ appears in p.105 of Le théorème de Riemann-Roch by Borel and Serre. In Thm. 2, they show that if $X$ is smooth projective, then $\epsilon$ is bijective. I am interested in the case that $X$ is not regular. One can show that for every $E\in \mathrm{Vect}(X)$ with $\epsilon(E)=0$, one has $E=0$.

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    $\begingroup$ When the scheme $X$ does not have resolution property (it is probably not known whether proper schemes have resolution property), the $K_0$ of vector bundles might be bad, and should be replaced by $K_0$ of perfect complexes. In the quasi-projective case (where there is resolution property) this seems to be addressed in [Pavic–Shinder, $K$-theory and the singularity category of quotient singularities]. $\endgroup$
    – Z. M
    Commented Oct 29 at 16:37
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    $\begingroup$ No, this map is not always injective. If $X$ is a nodal or cuspidal curve, for example, then the map has a big kernel. $\endgroup$
    – Eoin
    Commented Oct 29 at 17:09

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