Ruled surfaces and sections given by base point free divisors (Hartshorne exercise V.2.11) Let $C$ be a smooth projective curve over an algebraically closed field. Let $\mathcal{E}$ be a locally free sheaf on $C$ of rank two, which is normalized in the sense that $H^0(\mathcal{E})\neq 0$ but $H^0(\mathcal{E}\otimes\mathcal{O}_C(D))=0$ for any $D$ of degree $<0$. Let $C_0$ be a section of $\pi:X\cong\Bbb P(\mathcal{E})\to C$ such that $\mathcal{O}_X(C_0)\cong\mathcal{O}_X(1)$. Let $\mathfrak{e}$ be a divisor on $C$ so that $\det\mathcal{E}\cong\mathcal{O}_C(\mathfrak{e})$, and let $\mathfrak{b}$ be a divisor on $C$.
I am trying to prove that if $|\mathfrak{b}|$ and $|\mathfrak{b}+\mathfrak{e}|$ have no base points and $\mathfrak{b}$ is nonspecial, then there is a section $D\sim C_0+\mathfrak{b}f$. (This is one part of exercise V.2.11 in Hartshorne.)
The main tool that jumps out at me for this situation is Harshorne proposition V.2.9:

If $D$ is a section of $\pi:X\to C$, corresponding to a surjection $\mathcal{E}\to\mathcal{L}\to 0$, and if $\mathcal{L}\cong\mathcal{O}_C(\mathfrak{d})$ for some divisor $\mathfrak{d}$ on $C$, then $\deg\mathfrak{d}=C_0.D$ and $D\sim C_0+(\mathfrak{d}-\mathfrak{e})f$.

With this in mind, I think I should exhibit a surjection $\mathcal{E}\to \mathcal{O}_C(\mathfrak{b}+\mathfrak{e})$, which by the above proposition will give me a section $D$ which is linearly equivalent to $C_0+(\mathfrak{b}+\mathfrak{e}-\mathfrak{e})f=C_0+\mathfrak{b}f$. Unfortunately, I am a little lost on how to construct this surjection. Can you help?
I've thought about trying to define an injective morphism $\mathcal{O}_C(-\mathfrak{b})\to\mathcal{E}$ and proving the quotient is a line bundle (and thus $\mathcal{O}_C(\mathfrak{b}+\mathfrak{e})$ by taking determinants), but I'm not sure this works and it seems to only use base point freeness of $\mathfrak{b}$ - so I think this probably isn't the solution.

Edit: a duplicate has been suggested, but it does not construct a surjection $\mathcal{E}\to \mathcal{O}_C(\mathfrak{b}+\mathfrak{e})$. I would really like to understand how to construct such a surjection, so please don't close this as a duplicate of a post that does not do that.
 A: $\def\Ext{\operatorname{Ext}}\def\Hom{\operatorname{Hom}}$
After some time studying other material, I came back to this and figured it out. From previous work, we know that $\mathcal{E}$ fits in to an exact sequence $$0\to \mathcal{O}_C\to\mathcal{E}\to\mathcal{O}_C(\mathfrak{e})\to 0.$$ Apply $\Hom(-,\mathcal{O}_C(\mathfrak{b+e}))$ to this exact sequence to get $$0 \to \Hom(\mathcal{O}_C(\mathfrak{e}),\mathcal{O}_C(\mathfrak{b+e})) \to \Hom(\mathcal{E},\mathcal{O}_C(\mathfrak{b+e})) \to \Hom(\mathcal{O}_C,\mathcal{O}_C(\mathfrak{b+e})) \to $$ $$\to \Ext^1(\mathcal{O}_C(\mathfrak{e}),\mathcal{O}_C(\mathfrak{b+e})) \to \cdots.$$
We may identify $\Ext^1(\mathcal{O}_C(\mathfrak{e}),\mathcal{O}_C(\mathfrak{b+e}))\cong \Ext^1(\mathcal{O}_C,\mathcal{O}_C(\mathfrak{b})) \cong H^1(\mathcal{O}_C(\mathfrak{b}))$, which is zero by the assumption that $\mathfrak{b}$ is nonspecial. Therefore a map $\mathcal{E}\to\mathcal{O}_C(\mathfrak{b+e})$ is equivalent to the data of two maps $\mathcal{O}_C(\mathfrak{e})\to\mathcal{O}_C(\mathfrak{b+e})$ and $\mathcal{O}_C\to\mathcal{O}_C(\mathfrak{b+e})$ which can be arranged as $\mathcal{E}\to\mathcal{O}_C(\mathfrak{e})\to\mathcal{O}_C(\mathfrak{b+e})$ and $\mathcal{O}_C\to\mathcal{E}\to\mathcal{O}_C(\mathfrak{b+e})$.
Thus to find a map $\mathcal{E}\to\mathcal{O}_C(\mathfrak{b+e})$ which is surjective, it suffices to find two maps $\mathcal{O}_C(\mathfrak{e})\to\mathcal{O}_C(\mathfrak{b+e})$ and $\mathcal{O}_C\to\mathcal{O}_C(\mathfrak{b+e})$ so that the locii where these maps fail to be surjective are disjoint.
A map $\mathcal{O}_C(\mathfrak{e})\to\mathcal{O}_C(\mathfrak{b+e})$ is equivalent to a map $\mathcal{O}_C\to\mathcal{O}_C(\mathfrak{b})$ by twisting by $\mathcal{O}_C(-\mathfrak{e})$, and this twisting preserves surjectivity: if $c\in C$, then $\mathcal{O}_C(\mathfrak{e})_c\to\mathcal{O}_C(\mathfrak{b+e})_c$ is surjective iff $\mathcal{O}_{C_c}\to\mathcal{O}_C(\mathfrak{b})_c$ is surjective. So it suffices to show that we can choose effective divisors linearly equivalent to $\mathfrak{b}$ and $\mathfrak{b+e}$ which are disjoint.
As $l(\mathfrak{b+e})\geq 1$, there is an effective divisor $E$ linearly equivalent to $\mathfrak{b+e}$. Now consider the map $\varphi_{\mathfrak{b}}:C\to\Bbb P^N$ determined by $|\mathfrak{b}|$: since $\mathfrak{b}$ is base-point free and nonspecial, by exercise IV.6.8, it has $l(\mathfrak{b})\geq g+1+1-g=2$ and therefore $N\geq 1$ and the map $\varphi_{\mathfrak{b}}$ is nonconstant. We can choose a hyperplane in $\Bbb P^N$ which avoids the image $\varphi_{\mathfrak{b}}(E)$, producing an effective divisor $E'$ linearly equivalent to $\mathfrak{b}$ which is disjoint from $E$. Therefore we have shown that there is a surjection $\mathcal{E}\to\mathcal{O}_C(\mathfrak{b+e})$.
