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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 3 months |
| seen | Apr 24 at 19:51 | |
| stats | profile views | 125 |
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Mar 12 |
comment |
Shows that $M=\{(x,y,z):xy=0, x^2+y^2+z^2=1, z\ne +1 \ and -1\} $ is a 1- manifold. yes but you also have the condition $xy = 0$ so the good function to consider is $\Phi(x,y,z) = (x^2 + y^2 + z^2 -1, xy) $, and so $M = \Phi^{-1}(\{0,0\}) $ |
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Mar 12 |
accepted | Universal covering of $SO(3,\mathbb{R})$ |
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Mar 11 |
comment |
A problem on Residue Theorem I'm sure $C$ is note a discrete set of point |
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Mar 10 |
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If radial projection is bijective then is it a homeomorphism? I don't think this approach would lead to a counter example. The problem here is that you consider surfaces (or curves) with boundary, which seems proscribed. In that case I can prove there is no such counter-example in the one-dimensional case |
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Mar 10 |
comment |
If radial projection is bijective then is it a homeomorphism? The intersting question raised up by this remark is to find out if there exists a non-compact surface $S$ admiting a continuous bijection from $S$ to $S^2$. |
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Mar 10 |
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If radial projection is bijective then is it a homeomorphism? If $S$ is compact, it is true. Exercise : every continuous bijective map from a compact space to a Haussdorf space is a homeomorphism on its image |
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Mar 9 |
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$f: \mathbb{Q} \rightarrow \mathbb{R} \ \ \lim _{q \rightarrow t, \ q\in \mathbb{Q}} f(q) =g$ "I know that if a function is continuous on rational points, then it's continuous on whole R, but that isn't relevant to the problem, is it?" What about $\tan$ restricted to $\mathbb{Q}$ ? |
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Mar 9 |
comment |
Hatcher - simplicial and quotient representations of spheres The sterographic projection gives you an homeomorphism between $\mathbb{R}^n$ and $S^n - \{x\}$. But it is not really difficult to prove that $\mathbb{R}^n$ is homeomorphic to $D^n$. |
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Mar 9 |
comment |
Hatcher - simplicial and quotient representations of spheres If you admit that $S^n - \{x\}$ is homeomorphic to $D^n$ with $\varphi : D^n \longrightarrow S^n - \{x\}$, put $\tilde{\varphi} : \overline{D^n} \longrightarrow S^n$ that equals $\varphi$ on $D^n$ and $x$ on $\partial D^n$. You now have to check that $\tilde{\varphi}$ factors to a continuous function from $D^n / \partial D^n$ with respect to the quotient topology |
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Mar 9 |
comment |
Hatcher - simplicial and quotient representations of spheres It depends on how explicit you want the homeomorphism to be. |
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Mar 8 |
answered | nth derivative of an exterior conformal mapping in complex analysis |
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Mar 8 |
awarded | Custodian |
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Mar 8 |
reviewed | Approve suggested edit on How does Thurston's geometrisation conjecture imply Poincaré's conjecture? |
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Mar 8 |
accepted | How does Thurston's geometrisation conjecture imply Poincaré's conjecture? |
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Mar 8 |
asked | How does Thurston's geometrisation conjecture imply Poincaré's conjecture? |
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Mar 6 |
accepted | Measure on a quotient |
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Mar 6 |
comment |
Measure on a quotient Such a $\Pi$ is an infinite covering map, this would imply that the measure of the image of any set of positive measure would be infinite, and that is not really good, isn't it ? |
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Mar 5 |
awarded | Tumbleweed |
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Mar 5 |
comment |
A question on a dense subspace What do you call weight ? |
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Mar 1 |
answered | Homeomorphisms of the 3D sphere |
