Timeline for Can distinct morphisms between curves induce the same morphism on singular cohomology?
Current License: CC BY-SA 4.0
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| Apr 10, 2019 at 16:45 | comment | added | Piotr Achinger | @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. | |
| Apr 10, 2019 at 16:22 | comment | added | rj7k8 | @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). | |
| Apr 10, 2019 at 4:15 | comment | added | naf | It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition). | |
| Apr 9, 2019 at 23:39 | vote | accept | rj7k8 | ||
| Apr 9, 2019 at 23:01 | history | answered | Piotr Achinger | CC BY-SA 4.0 |