Thanks for the answer. Could you also explain why $\Phi$ is a linear map?

Sorry, I mean equivariant.

Maybe I am overlooking something but it seems to me that one needs that $iE$ is reflexive or something in that direction. Is that true?

In the case of finite groups, induction is both left and right adjoint to restriction. Is that true in general? In the book I am looking at, they only prove one of these two. In particular, I only get a natural map $iE\to E$ from that theorem. What is the natural map $E\to iE$?

You are right. What I meant is that there can be fibers that are not geometrically irreducible (I am not working over an algebraically closed field).

Thanks, but I this is not exactly what I need. Note that $\pi$ can have some reducible fibers whereas in Lazarsfeld all fibers are a $\mathbb{P}^1$.

okay, I got it!

Thanks! So this shows that any homomorphism $B/b\to A/a$ comes from a homomorphismm $B\to A$ which maps $(b)$ to $(a)$. Is it clear that every such can be obtained by multiplication with $c$ as above?

Thanks! So the situation is even better than I have expected. Am I right in assuming that the reflexive sheaf corresponding to a Weil divisor is just the subsheaf of the quotient field of all functions with the prescribed pole and zero loci?