This is on some confusion on the proof of lemma $1.6$ of the paper titled Special divisors on curves on a $K3$ surface(For convenience I am attaching the link here: https://link.springer.com/article/10.1007/BF01389083)
The statement of the lemma is as follows : If $E$ is a vector bundle on a $K3$ surface $X$ which is generated away from finitely many points, then there exists a globally generated vector bundle $F$ on $X$ with $det(F) =det(E)$,$h^{0}(F) \geq h^{0}(E)$ and $h^{i}(F) = h^{i}(E)$ for $i \geq 1$.
Sketch of the proof : At the beginning they worked under the hypothesis $H^{0}(E^*)=0$.We then automatically have a short exact sequence of the following form:
$0 \to V \to H^{0}(E) \otimes \mathcal O_X \to E \to S_E \to 0$....$(1)$,
where $V$ and $S_E$ are respectively kernel and cokernel of the evaluation map.
Confusion $1$ : After taking sheaf Hom I got the the short exact sequence $0 \to E^{*} \to H^{0}(E)^* \otimes \mathcal O_X \to V^{*} \to \mathcal {E}xt^1(S_E, \mathcal O_X) \to 0$.....$(2)$. But in the paper they got the following exact sequence : $0 \to E^{*} \to H^{0}(E)^* \otimes \mathcal O_X \to V^{*} \to Ext^2(S_E, \mathcal O_X) \to 0$...$(3)$.But, I don't see how $Ext^2(S_E, \mathcal O_X)$ is there in the S.E.S $(3)$.Is there a way to connect the terms $\mathcal{E}xt^1(S_E,\mathcal O_X)$and $Ext^2(S_E, \mathcal O_X)$?
At this point starting from S.E.S $(2)$ looking at this long cohomology sequence I got,$0 \to H^{0}(E)^* \to H^{0}(V^*) \to H^{0}(R_E) \to 0$, where $R_E := \mathcal {E}xt^1(S_E, \mathcal O_X)$.
Then in the paper it's stated that we may thus choose a subspace $W \subset H^{0}(V^*)$ such that $0 \to H^{0}(E)^* \to W \to H^{0}(R_E) \to 0$ and $W$ generates $V^*$ as a vector bundle (i.e the evaluation map $ W \otimes \mathcal O_X \to V^*$ is surjective)
Doubt 1: I can see only two candidates for $W$ namely $ H^{0}(E)^* ,H^{0}(V^*) $.But in any case the evaluation maps are evidently not surjective. So what ensures existence of such a $W$?
Any help from anyone is welcome
