# Can distinct morphisms between curves induce the same morphism on singular cohomology?

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 $$f=g$$?

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 $$H^1(-, \mathbf{C})$$.
• It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition). – naf Apr 10 '19 at 4:15
• @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). – rj7k8 Apr 10 '19 at 16:22
• @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. – Piotr Achinger Apr 10 '19 at 16:45