Timeline for Surjective morphism from $X$ to itself is finite
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2014 at 14:10 | comment | added | Donu Arapura | OK, I see………….. | |
| Aug 7, 2014 at 14:00 | comment | added | user5117 | @DonuArapura: I think one can prove that $f_*$ is surjective by elementary means; all you need to do is, given a curve in $X$, find a 1-cycle mapping finitely to that curve via $f$. In particular there's no need to talk about $f^*$. Am I being too simple-minded? | |
| Aug 7, 2014 at 13:27 | comment | added | Donu Arapura | DCV: but this begs the question what is $f^*$ on $N_1$? I think you need some form of duality here. | |
| Aug 7, 2014 at 11:58 | comment | added | Davide Cesare Veniani | One can define $f_*$ on $N_1(X)$, the (real) vector space of 1-cycles up to numerical equivalence, which is finite-dimensional. Surjectivity then implies injectivity. | |
| Aug 7, 2014 at 11:57 | vote | accept | Davide Cesare Veniani | ||
| Aug 7, 2014 at 10:53 | comment | added | Donu Arapura | Since you're using $f_*$, you need to assume that $X$ is smooth, or perhaps use intersection homology. | |
| Aug 7, 2014 at 10:23 | history | answered | abx | CC BY-SA 3.0 |