Strict Transform of a Line Bundle? Assume I am blowing up an algebraic variety $X$ in an ideal sheaf $\mathcal{I}$, write $Y:=\mathrm{Bl}_\mathcal{I}(X)$. Now, also assume I have a globally generated line bundle $\mathcal{O}_X^k\twoheadrightarrow\mathcal{L}$. Denote by $h_i\in\mathcal{L}(X)$ the images of the canonical base vectors, i.e. the global generators of $\mathcal{L}$. I can consider the strict transform of the vanishing set $Z_i:=Z(h_i)$ under the blowing-up $Y\to X$. 
Edit: We assume that there is a connected component $Y_i$ of $Z(\mathcal{I})$ such that $Y_i\subseteq Z_i$ for all $i$. 
I am now wondering if anything like the "strict transform" of $\mathcal{L}$ exists - I would want this to be a globally generated line bundle $\mathcal{O}_Y^k\twoheadrightarrow\mathcal{M}$ with generators $g_i\in\mathcal{M}(X)$ such that $Z(g_i)$ is the strict transform of $Z_i$.
The question is, does such a construction exist and is it well-known? If no, before I start and try to construct it, am I missing some obvious example where this does not even work?
 A: There are a number of issues with what you want mostly along the lines of the comments.
The main problem is that the strict transform is an operation on a divisor and not on a divisor class. A line bundle corresponds to a divisor class. So, basically you want to figure out a way to define the strict transform for a divisor class hoping that for a general choice in a basepoint-free system you would get the same.
I think that actually happens, but it will simply be the pull-back. 
The condition that $Z(h)\cap Z(\mathscr I)\neq\emptyset$ does not mean that the strict transform of $Z(h)$ is different from its pull-back. As long as $Y\not\subseteq Z(h)$ for any irreducible component $Y$ of $Z(\mathscr I)$, the strict transform of $Z(h)$ is the same as its pull-back. Of course, if $Z(\mathscr I)$ is a point, then $Z(\mathscr I)\subseteq Z(h)$ if and only if $Z(h)\cap Z(\mathscr I)\neq\emptyset$. So, to see the difference, try blowing up a curve in a threefold and see what happens to a surface that intersects the curve, but does not contain it. Its pull-back cannot contain any exceptional divisor, so it will be the same as its strict transform.
Assuming that for all $h$ there exists an irreducible component $Y$ of $Z(\mathscr I)$ such that  $Y\subseteq Z(h)$ is still not enough. I see at least two problems with this approach.
1) You would have to guarantee that the strict transforms are linearly equivalent. In other words, the pre-image of $Y$ in the pre-image of $Z(h)$ would have to be equivalent to the pre-image of $Y'$ in  the pre-image of $Z(h')$ including multiplicities. However, their pre-images would be different exceptional divisors and hence not comparable in the Picard group.
2) A perhaps even bigger problem is that you don't have enough irreducible components to do this for the whole linear system. Since you want your line bundle to be generated by global sections, you can't have one irreducible component contained in all elements of the linear system, but then if you take a general member, it will contain none of them. You're back to the previous case: the "strict transform" of the general element of the linear system is the same as its pull-back. 
I don't think you can escape this. If you have a basepoint free linear system, then the general member will be transversal to your $Z(\mathscr I)$ and hence its strict transform will be the same as its pull-back.
So, I think this still stands:
Conclusion: If your requirement is that you want to define the strict transform of a globally generated line bundle such that this strict transform is the line bundle corresponding to the strict transform of a general member of the linear system associated to the original line bundle, then the only way to do this is to take the pull-back. However, that already exists and is easier to define, so there is no point to bend over backwards to define the strict transform. Then again, it satisfies your goal requirement: the pull back will also be globally generated.
