Timeline for Topology of the preimage of a point for degree one holomorphic maps
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 1, 2011 at 15:16 | vote | accept | Dmitri Panov | ||
| Jan 1, 2011 at 14:45 | answer | added | BS. | timeline score: 6 | |
| Dec 31, 2010 at 17:19 | comment | added | Dmitri Panov | BS, I think, it works, thanks! Would you be kind to write down these comments as the answer to the question? | |
| Dec 31, 2010 at 16:13 | comment | added | BS. | To conclude the argument, it would suffice to know that $f^{-1}(x)$ is a retract (not necessarily by deformation) of some neighborhood in $M$. But this surely results from sharp results on the structure of analytic sets (triangulability - in a "relative" situation), | |
| Dec 31, 2010 at 15:55 | comment | added | BS. | but this seems easy : $t\mapsto d_x^2(f(\gamma(t)))$ has to be constant, which is absurd. | |
| Dec 31, 2010 at 15:47 | comment | added | BS. | Isn't Milnor's curve selection lemma adapted to this situation ? Namely, if $(z_n)$ is a sequence of critical points in $f^{-1}(U\setminus x)$, with $f(z_n)\to x$, one may assume by properness that $z_n\to z\in f^{-1}(x)$, and by selection lemma there is an analytic curve $\gamma:[0,\delta[\to M$ of critical points with $\gamma(0)=z$ and $f(\gamma(t))\neq x$ for $t>0$. It remains to rule this out... | |
| Dec 31, 2010 at 14:27 | comment | added | Dmitri Panov | S, nice suggestion. Though one needs to show that the function will not have critical points in a punctured neighbourhood of $f^{-1}(x)$. I don't see straight away how to prove this. (obviously if we blow up a point $y$ close to $x$ your function will have a critical point somewhere close to the exceptional divisor over $y$). | |
| Dec 31, 2010 at 10:20 | comment | added | BS. | A suggestion about the second statement : follow the gradient of $d_x^2\circ f$, with $d_x$ the distance to $x\in N$ with respect to a suitable (real analytic, hermitian, kaehler?) metric on $N$, the gradient being taken with respect to another such metric on $M$. | |
| Dec 30, 2010 at 22:18 | comment | added | Dustin Clausen | Sorry, maybe not any proper map of manifolds -- the "general grounds" might require more. But I imagine complex analytic is enough, though I know neither a reference nor an argument. | |
| Dec 30, 2010 at 22:07 | comment | added | Dustin Clausen | Dmitri, I think your second statement will hold for any proper map f between manifolds... with some argument like, on general grounds there is an open subset U of f^{-1}(x) with the inclusion a homotopy equivalence; then if U_x is an epsilon-neighborhood of x in N-f(M-U) we will have f^{-1}(x) inside U_x inside U giving the deformation retraction up to homotopy. | |
| Dec 30, 2010 at 21:09 | history | edited | Dmitri Panov | CC BY-SA 2.5 |
added 426 characters in body; edited tags
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| Dec 30, 2010 at 19:15 | answer | added | Sándor Kovács | timeline score: 3 | |
| Dec 30, 2010 at 13:18 | history | asked | Dmitri Panov | CC BY-SA 2.5 |