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visits member for 2 years
seen Feb 19 '13 at 5:01

Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
20
awarded  Yearling
Jun
23
awarded  Popular Question
Jun
20
awarded  Yearling
Feb
19
asked preimage of a connected under a covering map has unique representation into slices
Feb
15
comment a proof concerning fundamental group and lifting of paths
you are right xd thanks
Feb
15
accepted a proof concerning fundamental group and lifting of paths
Feb
15
asked a proof concerning fundamental group and lifting of paths
Feb
11
comment The Direct Product of a finite number of cyclic groups, is again a cyclic group?
It is false, for example take $\mathbb{Z}_2^2 $ it's a group of order 4, but all the non identity elements have order 2
Feb
8
comment is this set a regular surface?
I think that you are wrong, because it is now a contradiction that $V\cap S $ is open with respect to $S$ (in fact all the open sets with respect to $S$ have that form)
Feb
3
accepted prove that this formula defines a measure on $(X,\mathfrak{A})$
Feb
2
asked prove that this formula defines a measure on $(X,\mathfrak{A})$
Jan
28
comment the differential of a regular map between varieties
Yes... It's similar but it's not the same, here we are working over any kind of algebraically closed field, and not necessary over the real numbers. We are working over Varieties, in Algebraic Geometry and not manifolds.
Jan
28
asked the differential of a regular map between varieties
Jan
26
comment is this set a regular surface?
that's what I mean, so I can't see the contradiction
Jan
25
comment is this set a regular surface?
For example, take the sphere $S^1$ , we now that this set it's a regular surface, and also we know that has empty interior. Let's take $p\in S^1$ and take a neighborhood $V$ of $p$ and then consider $V\cap S^1$ clearly this set it's no open in $\Bbb R^3$ (in fact it has empty interior), so under that conclusion the sphere would not be a regular surface. At least that it's how I understood your answer.
Jan
25
comment is this set a regular surface?
I think that your idea is good, but not your conclusion, because you are saying that only because $V\cap S$ is homeomorphic to $U$ then $V\cap S$ is open in $\Bbb R^3$, but that it's not true, because the homeomorphism it's between the sets $U, V\cap S$ and the range is not $\Bbb R^3$ so the map is effectively an open map, but only between those two sets. (So $\phi(U)= V\cap S $ is open in $V\cap S$ but that is obvious in this case , where $\phi$ is the homeomorphism)
Jan
24
asked is this set a regular surface?
Jan
18
asked proving that this ideal is radical or the generator is irreducible