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Fix an abelian variety $A$ of dimension at least $3$ and a (smooth, projective, irreducible) curve $C$ inside $A$ that generates $A$ as an algebraic group, over an algebraically closed field. Now, fix a curve $X$ (also smooth, projective, irreducible) and consider the set $S$ of morphisms $f:X \to A$ with $f(X) \ne C$. Is $\# (f(X) \cap C)$ bounded or unbounded as $f$ varies in $S$?

Edit: Brendan Creutz pointed out to me that if $X=C$ is defined over $\mathbb{F}_q$ and we take $f$ to be multiplication by $1 + \# A(\mathbb{F}_{q^n})$ for growing $n$ we get unbounded intersection. So, the question is answered over the algebraic closure of finite fields but I am still interested in the answer over $\mathbb{C}$, for instance.

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    $\begingroup$ If the Manin-Mumford conjecture were false, a similar technique would work in characteristic zero, using multiplication by $1+n!$. So any technique to prove boundedness would have to be strong enough to prove Manin-Mumford. Don't existing proofs of Manin-Mumford use heavily the Galois action on the torsion points, which seems not usefully available here? $\endgroup$
    – Will Sawin
    Commented Oct 5, 2017 at 3:19
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    $\begingroup$ I believe boundedness should be expected in characteristic zero, but proving it might be difficult. There is a simpler problem in linear tori: If $X, Y \subset \mathbb{G}_{m/\mathbb{C}}^3$ are fixed curves, prove that there is an upper bound on the number of points that belong to both $X$ and $[n]Y+Q$, as $n \in \mathbb{Z}$ and $Q \in \mathbb{G}_m^3(\mathbb{C})$ vary in such a way that these curves have no component in common. The original problem may be reduced to this kind of statement inside an abelian variety, replacing $[n]$ by the isogenies of $A$. $\endgroup$
    – Vesselin Dimitrov
    Commented Oct 5, 2017 at 3:38

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