All varieties are defined over $\mathbb{C}$. Let $\pi : X \to C$ be an elliptic surface with $X$ and $C$ smooth. Then there is a Jacobian surface $\overline{\pi}: J(X) \to C$ (with a section) associated to $X$. Do we also have a morphism $\varphi : X \to J(X)$ such that $\overline{\pi}\circ \varphi = \pi$? (If yes, a precise reference for the construction of that morphism would be greatly appreciated!)
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1$\begingroup$ A stupid answer is to compose $\pi$ with the zero section of $\overline\pi$, I assume you want to rule that out... $\endgroup$– Torsten EkedahlCommented Apr 28, 2010 at 7:35
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$\begingroup$ yeah, I want to rule that out. I guess I want $\varphi$ to be surjective. $\endgroup$– TuanCommented Apr 28, 2010 at 16:39
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I think the following at least gives a rational map: Choose a curve $D$ in $X$ which maps dominantly to $C$ with degree $n$. For a smooth fibre $F$ of $\pi$ define a map to the correspoding fibre of $\bar{\pi}$ by sending a point $p$ to the class of $n[p] - [F\cdot D]$. (This is a cycle of degree 0.) With a little work one can check that this gives a well defined rational map which is a morphism over the smooth locus.
It seems unlikely though that the above actually gives a morphism in general.
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1$\begingroup$ Just a note -- if you think of X as Pic^1(X) and J as Pic^0(X), this is (over the generic fiber) just the composition of the isomorphism Pic^n(X) = Pic^(X) coming from your choice of D, with multiplication-by-n from Pic^1(X) to Pic^n(X). $\endgroup$– JSECommented Apr 28, 2010 at 16:39