Suppose $X$ is any quasi-projective variety. let $K^0(X)$ denote the Grothendieck group of locally free sheaves.
Suppose $U$ is an open subset of $X$. Is there a localization sequence:
$$ K^0(X)\rightarrow K^0(U)\rightarrow 0. $$
I saw that this exists for group of coherent sheaves $K_0(X)$. If $X$ is not smooth then do we still have the localization sequence for $K^0$.
thanks.
An element of $K_0(U)$ is represented by a perfect complex $F^.$ in the derived category of $U$-modules. As long as $X$ and $U$ are quasi-compact and quasi-separated, the class $[F^.]$ lifts to $K_0(X)$ if and only if $F^.$ is the restriction (in the derived category) of a perfect complex on $X$. (This is the "key proposition" of Thomason and Trobaugh's paper on "Higher Algebraic K-Theory of Schemes and of Derived Categories".) So any perfect complex that doesn't lift gives a counterexample to the surjectivity on $K_0$.
There is a localization sequence, as given in the reference mentioned in Angelo's comments. However the map is not always surjective. Perhaps an easy example is $X=Spec(R)$ where $R=k[x,y]_{(x,y)}/(xy)$ and $U=X$ minus the closed point. Then $K^0(X)=\mathbb Z$ since $R$ is local, but $K^0(U)=\mathbb Z^2$ since $U$ is two disjoint points.
