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I'm reading "Differential Geometry of Curves and Surfaces of Manfredo Docarmo" I'm doing the exercises of the chapter 2. Here is the definition of regular surface that we are following: enter image description here

I have problems with this exercise: enter image description here

Is with the first part, I think that it's not true that it's a regular surface (the second it is). I don't know how to prove that something is not a regular surface.

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What happens if you take the point $p$ on the boundary of the unitary disc? How would you construct the homeomorphism? –  Sigur Jan 24 '13 at 23:50

2 Answers 2

You're correct that the first one is not a surface. Here's how to check that condition $2$ fails. Take a point $(x,y,0)$ in the boundary, that is where $x^2 + y^2 = 1$. Take any open neighborhood $V$ of this, and intersect $V \cap S$. This set $V \cap S$ needs to be homeomorphic to an open set $U$. This would imply that $V \cap S$ was itself open. Can you see why this cannot be the case?

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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) –  Daniel Jan 25 '13 at 21:08
    
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. –  Daniel Jan 25 '13 at 21:36
    
$V \cap S$ is open in its own topology -- the subset topology. We can forget about $\mathbb{R}^3$ here. –  Isaac Solomon Jan 26 '13 at 0:29
    
that's what I mean, so I can't see the contradiction –  Daniel Jan 26 '13 at 1:07
    
If you check, you'll see that $V \cap S$ cannot actually be open in its own topology. –  Isaac Solomon Jan 27 '13 at 2:16

It is because in any neighborhood of a boundary point none of the orthogonal projections to the coordinate plane give a parametrization. This is a proposition in that section,and is often used to show that something is not a manifold.

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