# Some examples and non-examples of topological manifolds (w boundary or not)

I'm trying to check if these spaces are topological manifolds (i.e. locally euclidean and $$T_2$$) with or without boundary.

I would like to know if I made any mistakes, both in the answers or in the reasoning leading to them (i.e. if I give the correct answer for the wrong reasons).

1. $$D^2$$ the closed disk in $$\mathbb{R}^2,$$ quotiented by identifying all points on $$S^1.$$

I think this is not a topological manifold: indeed, I can identify $$D^2$$ with the half sphere $$S^2_{\geq 0}$$, and under this homeomorphism (which takes $$(x,y)$$ to $$(x,y,1-x^2-y^2$$)) the points on $$S^1$$ are left fixed.

Hence, contracting $$S^1$$ to the point the half sphere becomes something like a baloon.

The baloon is not a manifold since a neighborhood of the point $$P$$ corresponding to $$S^1$$ will become contractible after removing $$P,$$ while something homeomorphic to a disk would retract to $$S^1$$ after removing a point.

On the other hand, I think it is a manifold with boundary, where the only boundary point is $$P$$. This is because a neighborhood of $$P$$ will be homeomorphic to the positive ($$x\geq 0, y\geq0$$) portion of a disk centered in $$0$$ by a homeomorphism sending $$P \mapsto 0.$$

2. The closed disk $$D^2,$$ quotiented by identifying the diameter given by all $$(x,0)$$ with $$-1 \leq x \leq 1.$$

This is not a top. manifold because a point on $$S^1$$ will have a neighborhood that will be contractible after removing a point.

I think it is not a manifold with boundary. Indeed if I picture this space as a disk with the diameter pinched to the center $$0$$, then taking a neighborhood of $$0$$ and removing $$0$$ from it I get two connected components, while a half disk of $$\mathbb{R}^2$$ remains connected after removing any point.

3. The closed disk $$D^2$$ where you identify $$(-1,0)\sim (1,0)$$

Certainly this is not a topological manifold for the same reason as above. I think this is a manifold with boundary; in this case the boundary is given by all points on $$S^1$$ except for $$(1,0) \sim (-1,0),$$ since these points have a neighborhood homeomorphic to a disk.

1- A balloon is exactly the same thing as $$S^2$$, so it is a manifold without boundary.

Your argument does not work: why would a neighbourhood of $$P$$ become contractible after removing $$P$$ ?

2- Your justification is correct, although you only really need the second part (indeed, half disks and disks remain connected when you remove a point)

3- It is not a manifold with boundary: the points that are identified have no neighbourhood that is a disk or a half disk (for the same reason as above)

• for 1: Yes you are right I was confusing myself. For 3: I don't see how I can use the same reasoning as 2. I was thinking that if I have a neighborhood of $(-1,0)$, which is a half disk, then after identifying $(0,1)\sim (-1.0)$ I would get the other half of the disk. In any case I don't see why those points can't have a neighborhood homeomorphic to a disk or half disk... Oct 19, 2020 at 17:22
• For 2: I wanted to say that even without looking at the identification on the diameter we still can't have a top. manifold without boundary beacause points on $S^1$ will have ngbhds homeomorphic to half disks and not disks. Oct 19, 2020 at 17:24
• If you remove that point, the neighbourhood has two connected components. You don't get the other half of the disk because the boundary points other than $(-1,0)$ and $(1,0)$ are not identified Oct 19, 2020 at 17:25
• You are right of course about the other half of the disk. I still don't see why I get two connected components.. Oct 19, 2020 at 17:28
• A neighbourhood of this point will be two half disks connected by a single point on their boundary. If you remove that point, you get two half disks, i.e. two connected components Oct 19, 2020 at 17:37