Elementary transformations of ruled surfaces as maps of vector bundles This comes as a question in Beauville's Algebraic surfaces book (III.24 (2)). We work over $\mathbb{C}$.
All geometrically ruled surfaces (grs) $p:S\longrightarrow C$ over a curve $C$ can be seen as $S=\mathbb{P}(E)$ where $E$ is a vector bundle of rank $2$ over $C$, i.e. a locally free sheaf of rank $2$ over $C$. They are always minimal, or relatively minimal, depending on your book.
An elementary transformation $S\dashrightarrow S'$ with centre $s\in S$ corresponds to blowing up $s$ and contracting the strict transform of a fibre. This gives another grs $S'$ over $C$.
That point of view which is explicit and algebraic is the one I understand and usually work with. But you can also think in the following way. Take $s\in S$ and consider the pushforward of the skyscraper sheaf $\mathbb{C}(s)$, $F=p_*\mathbb{C}(S)$. Since $s$ can be seen as $(p(s),D_s)$ where $D_s\in E_{p(s)}$ is a line at the stalk of $E$ at $p(s)$, we get a map
$$u_s:E\longrightarrow F$$
which I understand it can be defined on the stalks and sends $(c,v)$ to $(c,0)$ if $c\neq p(s)$ and to $(c,v+D_s)$ otherwise. Here we identify $F_s$ with $E_s/(D_s)$ or some similar quotient. Beauville did not write $F$ as a pushforward but as the skyscraper sheaf itself. I think that is an unimportant typo. Define $E'=ker(u_s)$.
The question is:
1) $E'$ is a vector bundle of rank $2$.
2) $S'=\mathbb{P}(E')$ corresponds to the elementary transformation with centre $s$ $S\dashrightarrow S'$.
3) That transformation corresponds to the inclusion $E\rightarrow E'$.
I find hard to believe 1) is true although it must be because it is written in the book and an article by Hartshorne says it is 'easy' to see. I fail to see how the rank of $E'$ is going to be $2$ over $p(s)$. It seems to me that in the short exact sequence 
$$E'\longrightarrow E \longrightarrow F,$$
if you restrict to stalks you necessarily have $E'_s\cong \mathbb{C}$, but I must be being somewhat naive.
Probably the problem is that I do not understand $u_s$. I probably can do 2) and 3) once I solve 1). Any ideas?
 A: The kernel $E'$ of $u_s \colon E \to F$ is a torsion free sheaf over the curve $C$, hence it is necessarily a vector bundle (torsion free sheaves over curves are locally free).
Since the generic rank of $E'$ is $2$, it follows that  $E'$ is a rank $2$ vector bundle.
The point you are missing is that the inclusion $E' \to E$ is not an inclusion of vector bundles, but only an inclusion of sheaves. 
In fact, we can have a similar situation in rank one, for instance
$0 \to \mathcal{O}_C(-s) \to \mathcal{O}_C \to \mathbb{C}_s \to 0.$
Also in this case, the kernel of $v_s \colon \mathcal{O}_C \to \mathbb{C}_s$ is locally free.
A: 1) To see that the rank of $E'_s$ is 2 you can tensor the short exact sequence 
$$
0 \to E' \to E \to O_s \to 0
$$
by $O_s$. You will get
$$
0 \to Tor_1(O_s,O_s) \to E'_s \to E_s \to O_s\otimes O_s \to 0.
$$
Now it is easy to see that $Tor_1(O_s,O_s) = O_s\otimes O_s = O_s$ (for example you can use the locally free resolution $0 \to O_C(-s) \to O_C \to  O_s \to 0$ to compute the Tor and the tensor product), so one obtains
$$
0 \to O_S \to E'_s \to E_s \to O_s \to 0,
$$
whereof the rank of $E'_s$ is 2.
2,3) Both answers are "yes". 
