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visits member for 2 years, 10 months
seen Sep 22 at 22:03

Nov
22
asked Set of generators of the commutator subgroup of a surface group
Oct
31
accepted Range of a holomorphic function on the disc
Oct
31
comment Range of a holomorphic function on the disc
I have edited, I hope it is clearer.
Oct
31
revised Range of a holomorphic function on the disc
added 51 characters in body
Oct
31
asked Range of a holomorphic function on the disc
Sep
7
accepted Holomorphic function constant on a lattice.
Sep
7
asked Holomorphic function constant on a lattice.
Jul
19
comment Local homeomorphisms which are not covering map?
Ok but without including non-Haussdorff cases, can you think of a counterexample ?
Jul
19
comment Local homeomorphisms which are not covering map?
Of course there exists diffeomorphisms between manifolds, i meant local diffeomorphisms between manifolds which are not covering map. I edited
Jul
19
revised Local homeomorphisms which are not covering map?
added 34 characters in body
Jul
19
comment Local homeomorphisms which are not covering map?
Ok that's a good remark. Now can one think of couterexamples which are not a covering map with points from the domain removed ?
Jul
19
comment Local homeomorphisms which are not covering map?
Yes of course, I edited.
Jul
19
revised Local homeomorphisms which are not covering map?
edited body
Jul
19
asked Local homeomorphisms which are not covering map?
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\}) $
Mar
12
accepted Universal covering of $SO(3,\mathbb{R})$
Mar
11
comment A problem on Residue Theorem
I'm sure $C$ is note a discrete set of point
Mar
10
comment 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
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 $S^2$.
Mar
10
comment 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