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.
Best Answer
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)