Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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?

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.