# Is this a manifold?

I am trying to get started with differential geometry, and am having a difficult time wrapping my head around the concept of a manifold. One thing that would make it easier to understand would be if authors would give more examples of things that aren't manifolds. Anyway, here's one potential example that I came up with that I think will help me along quite a bit if I can understand it.

So, let's say we're in $\Bbb R^3$ and we have the unit sphere $x^2+y^2+z^2=1$. Even I can tell that this is a manifold. But now let's take the disc $z=0, x^2+y^2<2$, cut a hole out of it at $x^2+y^2<1$ and "attach" it to our sphere. Now, is the resulting object a manifold? Why or why not? What happens if we take our disc again and cut out the hole $x^2+y^2 \leq 1$?

I can't see how this object violates any explicit part of the definition of a manifold, but it just doesn't seem right.

Thanks for helping me get some sleep again.

• So you're looking at the sphere with like a ring around it thats touching it? This won't be locally homeomorphic to $\mathbb R^2$. – Eric O. Korman Aug 14 '14 at 0:16
• Yep, that's the idea. – bob.sacamento Aug 14 '14 at 0:16

Let's reduce the dimension by $1$, and consider the unit circle $x^2+y^2=1$, together with a whisker, $y=0$, $x\ge 1$. Does the point $(1,0)$ have a neighborhood that is homeomorphic (diffeomorphic) to an interval $I=\{|t|<\epsilon\}\subset\Bbb R$? Suppose you had such a homemorphism, and suppose it sends $(1,0)$ to $0\in I$. Can you see an elementary reason that this is impossible?
• @Ted Shifrin Thanks for the answer, and the positive feedback. Now, to your question, my messed up thinking is that there wouldn't be a homeomorphism between $I$ and our "circle+whisker" because the point where the two objects connect would have to map to two different coordinates on $I$. But then I think: we're supposed to be able to cover the object with overlapping "coordinate patches", and that seems to me -- incorrectly, no doubt -- to offer a way out. – bob.sacamento Aug 14 '14 at 17:04
• No, if we had a parametrization there'd just be a single point (which I assumed was $0$) mapping to the junction point. Of course, realistically, this cannot happen, as the elementary proof shows. (Pull out $0$ from the interval and you have two pieces. Pull out the junction point from a neighborhood in the "circle+whisker" and you'll have three pieces.) – Ted Shifrin Aug 14 '14 at 17:58