Elementary transform of a ruled surface - description of the sheaf I struggling with the problem III.24 (2) from Beauville's Complex Algebraic Surfaces:

Let:
  
  
*
  
*$E$ be a rank two vector bundle on a smooth projective curve $C$ over $\mathbb C$,
  
*$S := \mathbb P(E)$, $p : S \to C$,
  
*$s \in S$ be a point over $x \in X$, corresponding to a morphism $f : E \to i_{x, *} \mathbb C$.
  
  
  Show that the elementary transform of $S$ at $s$ is $S' = \mathbb P(E')$, where $E' = \ker f$. Moreover, show that $S \dashrightarrow S'$ is the map corresponding to the inclusion $E' \to E$.

Recall that elementary transform arises in the following way:


*

*we blow $S$ at $s$, obtaining $\widetilde S$, 

*the strict transform of the fiber $F$ containing $s$ is (as easily seen) a $(-1)$-rational curve and thus we can blow it down, obtaining $S'$.



My problem nr 1: what does it mean that $S \dashrightarrow S'$ corresponds to the inclusion $E' \to E$?

I see that there the inclusion induces a morphism $S' = \mathbb P E' \to \mathbb P E = S$. Is it supposed to be the inverse of $S \dashrightarrow S'$? I guess no - the inverse of an elementary transform is an elementary transform and thus it should not be a morphism, but a rational map.
I tried to find the map $S \dashrightarrow S'$ that should somehow correspond to the inclusion $E' \subset E$ by considering $S$ as pairs $(x, \xi \textrm{ - one dim. subspace of } \mathbb P(E_x \otimes \kappa(x)))$, but I didn't find any reasonable map.

My problem nr 2: it is straightforward that $S'$ is geometrically ruled over $C$ and thus $S' = \mathbb P E''$ for some $E''$. But how to show that there is an inclusion $E'' \to E$?

My thoughts: if $U := C \setminus \{ x \}$, then $S$, $\widetilde S$ and $S'$ are isomorphic over $U$ and thus we may assume that $E''|_U \cong E|_U$. I guess that the image of $F$ on $S'$ will be the point corresponding to the morphism:
$$E'_x \to \ker (f : E_x \otimes \kappa(x) \to \mathbb C) \cong \mathbb C.$$
 A: I believe the problem in Beauville contains a mistake - the author mixes two conventions for projective bundles:


*

*Convention nr 1: $\mathbb P(E) =  \mathrm{Proj} \, \mathrm{Sym} E^{\vee}$ - in this case the fiber over a point $x \in C$ is $\mathbb P(E_x \otimes \kappa(x))$. In other words points of $\mathbb P(E)$ over a point $x \in C$ correspond equivalently to:


*

*lines in the $\mathbb C$-vector space $E_x \otimes \kappa(x)$,

*morpshisms $E^{\vee} \to i_{x, *} \mathbb C \to 0$.


*Convention nr 2: $\mathbb P(E) =  \mathrm{Proj} \, \mathrm{Sym} E$ - in this case points of $\mathbb P(E)$ over a point $x \in C$ correspond to
one dimensional quotients of the vector space $E_x \otimes \kappa(x)$, etc.
I will stick to the first convention, since Beauville does this for the most part of his book. The problem should be corrected as follows:

Let $s \in S$ be a point over $x \in C$ corresponding to a morphism:
  $$  E^{\vee} \to i_{x, *} \mathbb C \to 0. $$ Define $E'$ by the exact
  sequence: $$  0 \to (E')^{\vee} \to E^{\vee} \to i_{x, *} \mathbb C
\to 0. $$ Show that the elementary transform of $S$ at $s$ is $\mathbb
 P(E')$ and that the map $S \to S'$ is induced by the map $E \to E'$.

Now we have a morphism $(E')^{\vee} \to E^{\vee}$, which yields after dualizing a morphism $h : E \to E'$. This induces a rational map $\mathbb P(h) : \mathbb P(E) \to \mathbb P(E')$. It has the following properties:


*

*$\mathbb P(h)$ is an isomorphism out of $F := p^{-1}(x)$,

*$\mathbb P(h) : \mathbb P(E) \to \mathbb P(E')$ is defined out of $s$ and contracts $F$ to a point,

*$\mathbb P(E')$ contains an "additional" rational line $p'^{-1}(x)$.


I believe those properties show that $S' = \mathbb P(E')$. Let me show the proof of the second property, which seems to be crucial. We have the exact sequence:
$$ 0 \to \xi_s \to E_x \otimes \kappa(x) \stackrel{h}{\rightarrow} E_x' \otimes \kappa(x), $$
where $\xi_s$ is the line corresponding to $s$. Note that the image of $h$ is one dimensional, denote it by $I$. Under $h$, every line in $E_x \otimes \kappa(x)$ goes either to $0$ (if this line is $\xi_s$, i.e. if it is a point corresponding to $s$) or to $I$ (if this line is not $\xi_s$). Thus $\mathbb P(h)$ is not defined at $s$ and the image of $F \setminus \{ s \}$ under $h$ is the point $s' \in S'$ corresponding to the line:
$$0 \to I \to E'_x \otimes \kappa(x). $$
