Let $k$ be a finite field, and let $X$ and $Y$ be some curves ($k$-variety of dimension $1$ with all the goods properties we want), and $\pi : X \rightarrow Y$ be a morphism. Then, it induces a morphism : $\pi^\star : \textrm{Pic}^0(Y) \rightarrow \textrm{Pic}^0(X)$, and in general, for any extension $L$ of $k$, to a morphism : $\pi^\star : \textrm{Pic}^0(Y_L) \rightarrow \textrm{Pic}^0(X_L)$, with $Y_L$ and $X_L$ curves on $L$ obtained by base change. So, all of this lead to a morphism : $\alpha : \textrm{Jac}(Y) \rightarrow \textrm{Jac}(X)$.
Suppose now that $\pi$ is not constant, i.e $\pi$ is surjective. Then, why $\pi^\star : \textrm{Jac}(Y) \rightarrow \textrm{Jac}(X)$ has finite kernel (i.e $\textrm{dim}(\alpha^{-1}(0))=0)$ ?
I have : $\textrm{dim}(\alpha^{-1}(0)) = \textrm{dim}(\textrm{Jac}(Y)) - \textrm{dim}(\alpha(\textrm{Jac}(Y)))$ (stop me if I'm wrong) ; but what indicates that we have $\textrm{dim}(\textrm{Jac}(Y)) - \textrm{dim}(\alpha(\textrm{Jac}(Y)))$ ?
Thank you !
If $\pi$ is finite (for example, in your case, surjective), you also have a map $\pi_*:\operatorname{Pic}^0 X\to\operatorname{Pic}^0 Y$ so that $\pi_*\circ\pi^*$ is multiplication by $m=\deg\pi$ (You can see this in many books, an exercise in Hartshorne). This shows that the kernel of $\pi^*$ is $m$-torsion and thus finite.