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 Answer 1


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)

  • $\begingroup$ 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... $\endgroup$
    – ggeolier
    Oct 19, 2020 at 17:22
  • $\begingroup$ 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. $\endgroup$
    – ggeolier
    Oct 19, 2020 at 17:24
  • $\begingroup$ 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 $\endgroup$ Oct 19, 2020 at 17:25
  • $\begingroup$ You are right of course about the other half of the disk. I still don't see why I get two connected components.. $\endgroup$
    – ggeolier
    Oct 19, 2020 at 17:28
  • $\begingroup$ 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 $\endgroup$ Oct 19, 2020 at 17:37

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