Timeline for Why does a homologically trivial cycle have trivial projections?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2012 at 8:53 | comment | added | jmc | I see, I actually should have been able to think that up myself. Thanks for helping me! | |
| Nov 6, 2012 at 4:48 | comment | added | Angelo | To Johan: Well, if $X$ is a smooth irreducible complete variety, the only cycles in codimension 0 are the multiples of the fundamental class of $X$. Since the top cohomology is also isomorphic to $\mathbb Z$, and is generated by the cohomology class of the fundamental class of $X$, you see that the cycle map is an isomorphism. | |
| Nov 5, 2012 at 21:58 | comment | added | jmc | Thanks for the edit. Do you mean that the cycle map is injective on codimension $0$ cycles? That is a fact I did not know yet. | |
| Nov 5, 2012 at 21:06 | vote | accept | jmc | ||
| Nov 5, 2012 at 15:42 | history | edited | Angelo | CC BY-SA 3.0 |
added 540 characters in body
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| Nov 5, 2012 at 11:33 | comment | added | jmc | Thanks for your answer. However, I interpreted Zhangs claim as if $\pi_{i,*}(\gamma)$ is rationally trivial. I should have been more clear about that in my question. I really need rational equivalence to $0$. | |
| Nov 5, 2012 at 11:20 | history | answered | Angelo | CC BY-SA 3.0 |