Suppose $f,g:X \rightarrow Y$ are finite morphisms between connected smooth curves over $\mathbb{C}$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,\mathbb{C}) \rightarrow H^*(X,\mathbb{C})$, does $f=g$?
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Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, g\colon X\to B$ sending a base point $x\in X$ to $0\in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, g\colon A\to B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $\pi_1 = H_1$, or on $H^1(-, \mathbf{C})$.
answered Apr 9, 2019 at 23:01
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3$\begingroup$ It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition). $\endgroup$– nafCommented Apr 10, 2019 at 4:15
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1$\begingroup$ @ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=\alpha \circ g$ where $\alpha:B \rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $\alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind). $\endgroup$– rj7k8Commented Apr 10, 2019 at 16:22
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$\begingroup$ @ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $b\in B$, $Y\cap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch. $\endgroup$ Commented Apr 10, 2019 at 16:45