Daniel
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 Sep24 awarded Notable Question Jul2 awarded Curious Jul2 awarded Inquisitive Jun20 awarded Yearling Jun23 awarded Popular Question Jun20 awarded Yearling Feb19 asked preimage of a connected under a covering map has unique representation into slices Feb15 comment a proof concerning fundamental group and lifting of paths you are right xd thanks Feb15 accepted a proof concerning fundamental group and lifting of paths Feb15 asked a proof concerning fundamental group and lifting of paths Feb11 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 Feb8 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) Feb3 accepted prove that this formula defines a measure on $(X,\mathfrak{A})$ Feb2 asked prove that this formula defines a measure on $(X,\mathfrak{A})$ Jan28 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. Jan28 asked the differential of a regular map between varieties Jan26 comment is this set a regular surface? that's what I mean, so I can't see the contradiction Jan25 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. Jan25 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) Jan24 asked is this set a regular surface?