I Is the Unit Closed Disk Minus the Origin Homeomorphic to the Unit Circle?

[davidge]

The unit closed disk minus the point $(0,0)$
$\mathbb{D}^1 \setminus (0,0): \bigg[(x,y) \in \mathbb{R}^2 | 0 < x^2 + y^2 \leq 1 \bigg]$
is homeomorphic to the unit circle
$\mathbb{S}^1: \bigg[(x,y) \in \mathbb{R}^2 | x^2 + y^2 = 1 \bigg]$
Since $\mathbb{D}^1 = \big(\mathbb{D}^1 \setminus (0,0) \big) \cup (0,0)$, is it correct to say that
$\mathbb{D}^1 \sim \mathbb{S}^1 \cup (0,0)$?

 

[davidge]

By the symbol $\sim$, I mean topological equivalence

 

[fresh_42]

The unit closed disk minus the point $(0,0)$
$\mathbb{D}^1 \setminus (0,0): \bigg[(x,y) \in \mathbb{R}^2 | 0 < x^2 + y^2 \leq 1 \bigg]$
is homeomorphic to the unit circle
$\mathbb{S}^1: \bigg[(x,y) \in \mathbb{R}^2 | x^2 + y^2 = 1 \bigg]$
Since $\mathbb{D}^1 = \big(\mathbb{D}^1 \setminus (0,0) \big) \cup (0,0)$, is it correct to say that
$\mathbb{D}^1 \sim \mathbb{S}^1 \cup (0,0)$?

I'm not sure about the homeomorphism, at least I cannot imagine one and I'm not sure what to do with the second dimension. I mean $\mathbb{D}^1-\{(0,0)\}$ is a cylinder with an open end and not a closed circle. But $\mathbb{D}^1 \sim \mathbb{S}^1 \cup (0,0)$ has to be wrong as the former is connected whereas the latter is not.

 

[davidge]

Please take a look at the image below

 

[fresh_42]

Please take a look at the image below
View attachment 204936

And what about $f \circ f^{-1}$? The embedding of $\mathbb{S}^1$ into $\mathbb{D}^1-\{(0,0)\}$ isn't the problem and surely can be inverted. But is $f^{-1}$ surjective or $f$ injective? What is $f(\frac{1}{\sqrt{8}},\frac{1}{\sqrt{8}})$ and $f(\frac{1}{\sqrt{32}},\frac{1}{\sqrt{32}})$ which are both on the disc?

 

[davidge]

is $f^{-1}$ surjective or $f$ injective? What is $f(\frac{1}{\sqrt{8}},\frac{1}{\sqrt{8}})$ and $f(\frac{1}{\sqrt{32}},\frac{1}{\sqrt{32}})$ which are both on the disc?

Do you mean $f^{-1}$? I don't understand, these functions seems to map different points on $\mathbb{S}^1$

EDIT: I got your point. Indeed those functions map two points of the disk into the same point of the circle. I didn't notice that when I was constructing the mappings.

 

[WWGD]

No, it is not correct; $\mathbb D^1$ is connected, pathconnected, but $\mathbb S^1 \cup (0,0)$ is not.

 

[davidge]

Thanks WWGD and fresh_42

So then what could we change on the disk so that the resultant object is homeomorphic to the circle?

 

[fresh_42]

Thanks WWGD and fresh_42

So then what could we change on the disk so that the resultant object is homeomorphic to the circle?

You can obviously take its boundary, but I assume that's not what you wanted to hear. I still don't think it's possible for dimensional considerations. But as it's topology I long have given up to say something for certain until I've seen a proof.

 

[ConfusedMonkey]

I agree with fresh_42. $D \setminus (0,0)$ is a 2-manifold, while the circle is a 1-manifold, so they can't be homeomorphic to each other. There's not much you can do about this.

 

[WWGD]

Actually, the disk minus the center is just homotopically equivalent but not homeomorphic to $\mathbb S^1$. Remove 2 points from $\mathbb D^1 -(0,0)$ does not disconnect the space, while removing any two points from $\mathbb S^1$ will disconnect.

 

[WWGD]

Do you mean $f^{-1}$? I don't understand, these functions seems to map different points on $\mathbb{S}^1$

EDIT: I got your point. Indeed those functions map two points of the disk into the same point of the circle. I didn't notice that when I was constructing the mappings.

I think what fresh is trying to say is that, e.g., $(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}) = (\frac {2x}{\sqrt{(2x)^2+(2y)^2}},\frac{2y}{\sqrt{(2x)^2+(2y)^2}} )$ , since the 4's cancel each other out. The two spaces are homotopically-equivallent. Basically, your map crushes whole lines about the origin to a point, i.e., $f(x)=f(kx)$

 

[WWGD]

Informally, what monkey and fresh said may be right, informally. When you go from a disk -(0,0) to a circle, you are " lowering the dimension" in the sense that you are crushing a whole (half-open) annulus into a point, the point on the boundary circle. The map $x \rightarrow kx$ describes that. EDIT: Re dimensionality, notice that the disk -{pt} contains open 2-balls, while $\mathbb S^1$ does not.

 

[davidge]

When you go from a disk -(0,0) to a circle, you are " lowering the dimension" in the sense that you are crushing a whole (hal-open) annulus into a point, the point on the boundary circle

So would this be a good point to make precise the definition of boundary? In our case, the boundary (the circle) is where the original manifold (the disk) fails to be a manifold ?

You can obviously take its boundary, but I assume that's not what you wanted to hear

yea

 

[fresh_42]

So would this be a good point to make precise the definition of boundary? In our case, the boundary (the circle) is where the original manifold (the disk) fails to be a manifold ?

I think that's way too complicated (and too special) to define a boundary.

Formally and in general a boundary is the closure minus the inner points. Closure means with all accumulation points included, and inner points mean points with an open set around, that is still in it. I'm not quite sure whether accumulation point is the correct term here, I mean all points with the property that each open neighborhood of them contains a point of the set.

For manifolds there is a special case: manifold with boundary. It has to be especially treated as manifolds are usually not considered embedded anywhere and thus as topological spaces themselves, they don't have boundaries in the topological sense. So in the case of manifolds, their boundaries are defined via their atlases, i.e. the boundary points of their charts.

 

[WWGD]

This may be a rigorous , though somewhat clunky argument, using invariance of domain https://en.wikipedia.org/wiki/Invariance_of_domain , if there was a continuous injection ( which there would be , if there was a homeomorphism) , then take an open set ( open ball) in the Disk-{pt} as an open set in $\mathbb R^2$ . Its image ( if there was a homeomorphism, which would preserve dimension) , would also be open in $\mathbb R^2$ . But the circle does not contain open balls. Maybe overkill.

 

[fresh_42]

This may be a rigorous , though somewhat clunky argument, using invariance of domain https://en.wikipedia.org/wiki/Invariance_of_domain , if there was a continuous injection ( which there would be , if there was a homeomorphism) , then take an open set ( open ball) in the Disk-{pt} as an open set in $\mathbb R^2$ . Its image ( if there was a homeomorphism, which would preserve dimension) , would also be open in $\mathbb R^2$ . But the circle does not contain open balls. Maybe overkill.

Perhaps not an overkill but dependent on the surrounding space $\mathbb{R}^2$. I found your argument by removing points somehow better. I also thought about a mapping that peels the disc like an apple, but this only disguises the dimension problem: you cannot peal thin enough (I think).

 

[WWGD]

Perhaps not an overkill but dependent on the surrounding space $\mathbb{R}^2$. I found your argument by removing points somehow better. I also thought about a mapping that peals the disc like an apple, but this only disguises the dimension problem: you cannot peal thin enough (I think).

Yes, I guess you're right, embedding data is unnecessary baggage. EDIT: Re the peeling, the peeling becomes too thin when the cutset itself -- a loop -- changes dimension to a pair of points.