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?



Elementary transformations of ruled surfaces as maps of vector bundles
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.


