Consider the punctured affine plane $\mathbb{A}^2\setminus \{(0,0)\}$ over a field. Every vector bundle on this is trivial. Especially every vector bundle on this extends to a vector bundle on $\mathbb{A}^2$. Every morphism of vector bundles on this also extends uniquely to a morphism of vector bundles on $\mathbb{A}^2$. (You can look at the morphism of vector bundles as a section of the $\text{Hom}$ vector bundle and the section of vector bundles extends uniquely to a section of the extended vector bundle. This is a property that holds for reflexive sheaves on an open that its complement has codimension greater than or equal to 2.).
More generally If $X$ is a smooth variety and $Y$ a closed subvariety of codimension $\geq 2$. Every reflexive sheaf on $X\setminus Y$ extends uniquely to a reflexive sheaf on $X$. Every section of a reflexive sheaf on $X\setminus Y$ extends uniquely to a section of the extended reflexive sheaves. You can see here. The extension functor is simply just the pushforward and restriction is just the pullback to the open.
Now back to our situation, on the punctured plane and affine plane, reflexive sheaves coincide with trivial vector bundles. So pushforward and pullback functors induce mutually inverse functors between the category of vector bundles on the plane and the punctured plane. This implies an equivalence of categories between those two. However this cannot be true because if it was then this implies the algebraic $K$-theory of the two are isomorphic. Looking at the localization sequence for a point, $\mathbb{A}^2$ and $\mathbb{A}^2\setminus \{(0,0)\}$ this implies that algebraic $K$-theory of point is trivial, which cannot be true.
What mistake am I making in my logic?
The pushforward functor probably is not exact but how it is possible for a functor that induces equivalence of categories to not to be exact? (its inverse pullback is exact)
Best Answer
The categories of vector bundles are exact categories, and you need to use the structure of exact categories to define the $K$-theory. Now the point is that the functor you are considering is not compatible with the exact structures; indeed, the morphism $$ \mathcal{O} \oplus \mathcal{O} \stackrel{(x,y)}\to \mathcal{O} $$ is an admissible epimorphism on the punctured plane, but not on the plane itself.