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Let $X$ be an algebraic curve defined over the complex numbers $\mathbb{C}$ of genus $g > 1$. A theorem of De Franchis states that there exist only finitely many (isomorphism classes of) curves $Y$ defined over $\mathbb{C}$ with genus exceeding one such that there exists a non-constant map $f: X \rightarrow Y$. For a given curve $X$, put $S(X)$ for the set of (isomorphism classes of) curves $Y$ with genus exceeding one such that there exists a non-constant map $f: X \rightarrow Y$.

Now suppose that for two curves $X_1, X_2$ with equal genus $g > 1$, that $S(X_1) = S(X_2)$. What can be said about $X_1, X_2$?

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  • $\begingroup$ A very general curve $X/\mathbb{C}$ (eg with simple Jacobian) only maps to $\mathbb{P}^1$. $\endgroup$
    – Raju
    Jan 6, 2019 at 2:03
  • $\begingroup$ You require the maps $f$ to be holomorphic, right? $\endgroup$
    – Alex Suciu
    Jan 6, 2019 at 2:58
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    $\begingroup$ $X$ is the only curve in $S(X)$ of genus $g$, so if $X_1,X_2$ have the same genus $g$ and $S(X_1)=S(X_2)$ then $X_1=X_2$. $\endgroup$ Jan 6, 2019 at 4:24
  • $\begingroup$ If, as in Felipe's comment X is in S(X), you lose. If you require S(X) to only include curves of lower genus, then consider Y to be any curve g>1, and let X be, say a ramified double cover. Lots of those. $\endgroup$
    – meh
    Jan 7, 2019 at 14:06

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