Let $F$ be a stable torsion free sheaves on a smooth projective variety $X$ over $\mathbb{C}$. Assume that $F$ is generated by global sections, i.e. the evaluation map $ev: H^0(F)\otimes \mathcal{O}_X\to F$ is surjective. Let $K$ be the kernel sheaf of $ev$.
My question is: Is $K$ a simple sheaf, i.e. $Hom(K,K)=\mathbb{C}$?
If we assume $F$ is not generated by global sections, and consider the cone $C$ of map $ev: H^0(F)\otimes \mathcal{O}_X\to F$. Do we also have $Hom(C,C)=\mathbb{C}$? Here $Hom(C,C)$ is taken in the category of bounded complex of $Coh(X)$.
No. For instance, let $X = \mathbb{P}^2$ and $F = T$, the tangent bundle. Then $F$ is globally generated but it is easy to see that the kernel of the evaluation morphism is isomorphic to $\Omega(1)^{\oplus 3}$, hence is not simple.