For a curve $C$ and its Jacobian $J$, is $\Gamma(J, \Omega) \to \Gamma(C, \Omega)$ an isomorphism?

Let $k$ be a field, $C$ a smooth complete $k$- curve, $J$ its Jacobian variety, $P \in C(k)$ a rational point, $f : C \to J$ the morphism associated with $P$. Then does $f$ induce an isomorphism $\Gamma(J, \Omega) \to \Gamma(C, \Omega)$?

This is trivial if $k = \mathbb{C}$. But I can't show this algebraically.

A text says that $\require{AMScd}$ \begin{CD} \Gamma(J, \Omega_J) @>{f^*}>> \Gamma(C, \Omega_C)\\ @V{h}VV @V{i}VV\\ T_{J,0}^\vee @>{j^\vee}>> H^1(C,\mathscr{O}_C)^\vee \end{CD} is commutative. ($j$ is the canonical isomorphism, $h$ is the evaluation map at $0$, and $i$ is the Serre's dual, which are all isomorphisms.) I know explicitly what the canonical isomorphism $T_{J,0} \to H^1(C, \mathscr{O}_C)$ is, however, I can't show the commutativity. And I think showing it is very complicated, because this diagram involves the Serre's dual isomoprhism, and because $j$ is already complicated.

So if there is, please show it without using this diagram.

I believe that this is no very difficult (although I can't) and there is a proof without using this diagram, because in Silverman's "Diophantine Geometry: an introduction", the author suggest this as an exercise. (I think this book is easier than my text, because this does not use scheme-theoritic language, and so of course does not use the Serre duality.)

Thank you very much!

• I imagine that the text you mention is Jacobian Varieties, J.S. Milne, proposition 2.2, that you can find here: jmilne.org/math/xnotes/JVs.pdf – Xarles Sep 14 at 20:24
• @Xarles Thank you for your comment. Yes, now I study Jacobian varieties with it. It leaves this commutativity as an exercise, and I couldn't find the proof of this in any sites. – k.j. Sep 14 at 21:19
• Serre, Algebraic Groups and Class Fields, V §2.10 Corollary 1 – Felipe Voloch Sep 14 at 22:58