Timeline for Proper morphisms
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2011 at 16:27 | comment | added | Gunnar Þór Magnússon | @Qing Liu, it seems I do, good example. | |
| Nov 5, 2011 at 13:04 | comment | added | Qing Liu | @Gunnar, I guess you want to suppose the fibers of $f$ are connected. In my counterexample below, $f$ is a submersion. | |
| Nov 5, 2011 at 7:54 | comment | added | Gunnar Þór Magnússon | @Inkspot: This is true under the additional (strong) hypothesis that $f$ is a submersion. The proof is by standard techniques in point set topology. | |
| Nov 4, 2011 at 23:49 | answer | added | Sándor Kovács | timeline score: 6 | |
| Nov 4, 2011 at 21:58 | comment | added | Angelo | You could take a point in the boundary of $U$; then the inverse image is compact, since it is empty, but the function is not proper in a neighborhood of the point. | |
| Nov 4, 2011 at 21:50 | history | edited | Qing Liu |
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| Nov 4, 2011 at 21:49 | answer | added | Qing Liu | timeline score: 5 | |
| Nov 4, 2011 at 21:34 | comment | added | Qing Liu | @Tom: if $f$ is an open immersion, the answer is yes (take $U=f(X)$). Inkspot doesn't ask $f$ to be proper over $S$. | |
| Nov 4, 2011 at 11:24 | comment | added | inkspot | Thank you. Wanting something to be true is, yet again, not enough. | |
| Nov 4, 2011 at 10:56 | comment | added | Tom Goodwillie | No. Let $f$ be the inclusion of an open subset. | |
| Nov 4, 2011 at 10:31 | history | asked | inkspot | CC BY-SA 3.0 |