Let $S$ be a surface. $C$ be a curve on $S$ and $i:C\hookrightarrow S$ is the inclusion. Let $E$ be a rank 2 vector bundle on $S$ and $F$ a line bundle on $C$. Suppose we have a surjection $E\longrightarrow i_*F$, then the elementary transformation of $E$ along $F$ is the kernel $K$.
$0\longrightarrow K\longrightarrow E\longrightarrow i_*F\longrightarrow 0$
In Maruyama's paper - On a family of algebraic vector bundles, he calls a vector bundle of the form $K\otimes \mathcal{O}_s(C)$ a regular vector bundle.
Now $X=\mathbb{P}(E)$ is a $\mathbb{P}^1$-bundle over $S$, and $Y\hookrightarrow X$ is given by $Y=\mathbb{P}(F)\cong C$ since $F$ is a line bundle.
In the paper, families $R^2(S,C,D)$ of regular vector bundles of rank 2 are constructed, parametrized by $S,C$ and a divisor $D$ on $C$. The divisor $D$ is described as follows.
Because of Kunneth formula, Lemma 2.4 in the paper says that, $\mathcal{O}_{\mathbb{P}^1_C}(Y)\cong\mathcal{O}_{\mathbb{P}^1_C}(Z\times C)\otimes \pi_C^*\mathcal{O}_C(D)$. Here $Z$ is a hyperplane in $\mathbb{P}^1_k$, and hence is a point, and $\pi_C:\mathbb{P}^1_C\longrightarrow C$ is the natural projection. But this means that, the above translates to $\mathcal{O}_{\mathbb{P}^1_C}(C)\cong\mathcal{O}_{\mathbb{P}^1_C}( C)\otimes \pi_C^*\mathcal{O}_C(D)$ isn't it? So that $D$ is trivial always?
I almost sure that I am missing something somewhere. I don't know what or where. Any help will be appreciated.
