For a projective variety $X$ and an ample line bundle $L$ on it, we consider the family of line bundles $L^{\otimes i}$ for $i\in \mathbb{Z}$. Let $\mathfrak{C}$ be the category generated by the aforementioned line bundles by taking successive extensions. The category $\mathfrak{C}$ is closed under extensions so it is an exact category. Does the higher $K$-groups of $\mathfrak{C}$ coincide with higher $K$-groups of $X$? or is the map $K_n(\mathfrak{C})\rightarrow K_n(X)$ surjective? (for $n>0$)
The reason for my question is that the line bundles mentioned above generate the category of perfect complexes on $X$. You can see the proof here theorem 4. The proof implies that every perfect complex is a direct summand of a complex with elements in $\mathfrak{C}$, which is very similar to the cofinality theorem. I'm not sure of a version of cofinality theorem that applies here or this has any implications on the $K$-groups.
Edit: Few things I want to add. First let's assume $X$ is a smooth projective variety of dimension $\geq 2$ over a finite field. I will call a morphisms of varieties a nice map if it is a composition of Frobenius and finite etale covers. Then you can prove that every object of $\mathfrak{C}$ can turn into a direct sum of line bundles of the form $L^{\otimes i}$ after pullback along a nice map. The reason for this is, you can kill the elements in the cohomology groups $H^{1}(X, L^{\otimes i})$ under pullback along nice maps. For $i>0$ it is just Serre vanishing, for $i<0$ it follows from Serre duality and for $i=0$ the proof is sketched here.
This also implies that you can split every short exact sequence in $\mathfrak{C}$ by pullback along nice maps. The Corollary of these facts is that $\mathfrak{C}$ is idempotent complete. In order to prove that we just need to prove that it is closed under direct summands. Consider a direct summand of an object in $\mathfrak{C}$, pullback so the object splits into direct sum of line bundles. Any direct summand of such a thing is just direct sum of a subset of line bundles. This implies that any direct summand is in $\mathfrak{C}$. (There are some details removed but we also need to note that pullback of vector bundles along nice maps are faithful functors.)
The algebraic $K$-groups $K_n(X)$, are the $K$-groups of the Waldhausen category of perfect complexes $Perf(X)$ with the usual notion of quasi-isomoprhisms defined in the category of complexes in coherent sheaves. By the second paragraph and the link provided there this category is cofinal in the Waldhausen category of bounded complexes in $\mathfrak{C}$ with the inherited notion of quasi-isomorphism. We denote the latter category by $(Ch^b(\mathfrak{C}), q_1)$. So for $n>0$ we have $K_n(X)\cong K_n((Ch^b(\mathfrak{C}), q_1))$. Now on $Ch^b(\mathfrak{C})$ we can consider other family of acyclic complexes that requires the kernels and cokernels of differentials to be in $\mathfrak{C}$ this leads to another notion of quasi-isomorphisms denoted by $q_2$. It is easy to see that $q_2$ is a subset of $q_1$. Since $\mathfrak{C}$ is idempotent complete by Gillet-Waldhausen $K_n(Ch^b(\mathfrak{C}),q_2)\cong K_n(\mathfrak{C})$.
So the problem is comparing $K_n(Ch^b(\mathfrak{C}),q_2)$ and $K_n(Ch^b(\mathfrak{C}),q_1)$. There is a localizing long exact sequence relating them to each other. In the following form: $$\ldots \rightarrow K_n(Ch^{b,q_1}(\mathfrak{C}),q_2)\rightarrow K_n(Ch^b(\mathfrak{C}),q_2)\rightarrow K_n(Ch^b(\mathfrak{C}),q_1)\rightarrow \ldots$$
Here $(Ch^{b,q_1}(\mathfrak{C}),q_2)$ is the Waldhausen category of acyclic bounded chain complexes in $\mathfrak{C}$, where by acyclic it means acyclic as a chain complex of coherent sheaves. The weak equivalences are induced by the acyclic chain complexes where the kernel and cokernel of differentials are in $\mathfrak{C}$. So the question is when these groups becomes trivial? Or the long exact sequence splits into short exact sequences?
Some conditions that makes this happen is that $\mathfrak{C}$ is closed under taking cokernels or kernels, but I'm not sure how common or rare this assumption is.
