Boundary and homeomorphism

[fresh_42]

Can you give an example of a map from the disc to $\mathbb{R}_{+}^{n}$?

Say $U$ is an open neighborhood of a point $(x,y)$ with $x^2+y^2=1$ on the closed disc $B$ as a manifold. We can chose a coordinate system, such that $(x,y)=(1,0)$ is the south pole of $B$ by an appropriate rotation and translation. Now the closed disc is completely within $\mathbb{R}_{+}^{n}$. Next we stretch $U$ in such a way, that all its boundary points come to rest on the boundary of $\mathbb{R}_{+}^{n}$, e.g. by a perspective mapping from the north pole.

The fraud here lies of course in the coordinate transformations $T$ where the work is done, but rotations as well as translations and the final stretching are continuous and bijective, i.e. a homoeomorphism.

This description is only to prevent me from doing any calculations, which probably won't show very much and the final stretching is like bending a curved piece of metal into a line, which is a bit difficult to write down. The imagination should do.

 

[lavinia]

You can see in many ways that $S^1×S^1$ is not homeomorphic to $S^2$. One way is to triangulate each and calculate the Euler characteristic from the triangulation. For the torus $S^1×S^1$ the answer is zero. For the sphere it is 2. This calculation is independent of the triangulation.

Also the sphere is simply connected while $S^1×S^1$ is not. This means that every closed loop on the sphere can be shrunk to a point - if one imagines that the loop is made of infinitely stretchable rubber - while the torus clearly has closed loops that can not be shrunk.

It is also true that every map of the sphere into $S^1×S^1$ is null homotopic - so it can not be a homeomorphism.

The sphere minus a point is contractible. That is: it can be continuously shrunk to a point. $S^1×S^1$ minus a point shrinks to a figure eight so it is not contractible.

 

[WWGD]

You can see in many ways that $S^1×S^1$ is no homeomorphic to $S^2$. One way is to triangulate each and calculate the Euler characteristic form the triangulation. For the torus $S^1×S^1$ the answer is zero. For the sphere it is 2. This calculation is independent of the triangulation.

Also the sphere is simply connected while $S^1×S^1$ is not. This means that every closed loop on the sphere can be shrunk to a point - if one imagines that it is made of infinitely stretchable rubber while the torus clearly has closed loops that can not be shrunk.

It is also true that every map of the sphere into $S^1×S^1$ is null homotopic - so it can not be a homeomorphism.

The sphere minus a point is contractible. That is: it can be continuously shrunk to a point. $S^1×S^1$ minus a point shrinks to a figure eight so it is not contractible.

I wonder if one can show that $S^2$ is also not a product space $Y \times Y$, nor even of the form $Y \times W$ EDIT: This may be false if we have $Y= S^2$ and Y={ $.pt$} , where {.pt} is a one-point space. but true otherwise.

. Is there any way to show $S^2$ is not the top space of a trivial bundle. I remember seeing an argument to the effect that $\mathbb R^{2n+1}$ is not a product space, i.e., it is not homeomorphic to the product $Y \times Y$ with the product topology..

 

[lavinia]

I wonder if one can show that $S^2$ is also not a product space $Y \times Y$, nor even of the form $Y \times W$ . Is there any way to show $S^2$ is not the top space of a trivial bundle. I remember seeing an argument to the effect that $\mathbb R^{2n+1}$ is not a product space, i.e., it is not homeomorphic to the product $Y \times Y$ with the product topology..

I am not sure. Let's think about it.

 

[zwierz]

another question: is the following implication true?: $\mathbb{R}^2=Y\times Y\Longrightarrow Y=\mathbb{R}$

 

[lavinia]

Thank you for the detailed reply.
Can you give an example of a map from the disc to $\mathbb{R}_{+}^{n}$?

For $n=2$ take a look at $z → (z + i)/(1 + iz)$ where $z$ is a point in the open unit radius disc centered at the origin.

 

[davidge]

For $n=2$ take a look at $z → (z + i)/(1 + iz)$ where $z$ is a point in the open unit radius disc centered at the origin.

But this is complex, not real.

 

[lavinia]

But this is complex, not real.

The complex plane is homeomorphic to $R^2$

 

[davidge]

@fresh_42 I have something to add to our discussion regarding your post #34. I was reading some books on topology that are available at the university library and according to Differential Topology by Victor Guillemin and Allan Pollack: "The Whitney theorem states that every k-dimensional manifold embeds in $\mathbb{R}^{2k}$." Does that theorem implies that a homeomorphism (or as mentioned in that book, a diffeomorphism) can be defined from the k-dimensional manifold to $\mathbb{R}^{2k}$? If so, why did you say earlier that the circle (a 1-dimensional manifold) cannot be embedded in $\mathbb{R}^2$?

 

[WWGD]

another question: is the following implication true?: $\mathbb{R}^2=Y\times Y\Longrightarrow Y=\mathbb{R}$

Up to homeomorphism?

 

[WWGD]

@fresh_42 I have something to add to our discussion regarding your post #34. I was reading some books on topology that are available at the university library and according to Differential Topology by Victor Guillemin and Allan Pollack: "The Whitney theorem states that every k-dimensional manifold embeds in $\mathbb{R}^{2k}$." Does that theorem implies that a homeomorphism (or as mentioned in that book, a diffeomorphism) can be defined from the k-dimensional manifold to $\mathbb{R}^{2k}$? If so, why did you say earlier that the circle (a 1-dimensional manifold) cannot be embedded in $\mathbb{R}^2$?

But a circle can be embedded in $\mathbb R^2$. The standard { $(x,y) \in \mathbb R^2 : x^2+y^2 =1$} is an embedding of the standard unit circle.

RE my previous post, Borsuk-Ulam theorem https://en.wikipedia.org/wiki/Borsuk–Ulam_theorem states that every continuous map f: $\mathbb S^n \rightarrow \mathbb R^n$ has a point x with f(x)=f(-x), meaning there can be no embedding of the n-circle into $\mathbb R^n$.

 

[fresh_42]

@fresh_42 I have something to add to our discussion regarding your post #34. I was reading some books on topology that are available at the university library and according to Differential Topology by Victor Guillemin and Allan Pollack: "The Whitney theorem states that every k-dimensional manifold embeds in $\mathbb{R}^{2k}$." Does that theorem implies that a homeomorphism (or as mentioned in that book, a diffeomorphism) can be defined from the k-dimensional manifold to $\mathbb{R}^{2k}$? If so, why did you say earlier that the circle (a 1-dimensional manifold) cannot be embedded in $\mathbb{R}^2$?

I haven't said it can't be embedded. I said it's of a different dimension and it can't be onto. There is no homeomorphism because there is no bijection. Every time you take a compass and draw a circle, you probably embed it in $\mathbb{R}^2$. And as a manifold, the total topological space is already merely the circle and there is no $\mathbb{R}^2$ where it lives in.

 

[davidge]

I said it's of a different dimension and it can't be onto

There is no homeomorphism because there is no bijection.

Ah, ok. Thank you

 

[davidge]

But a circle can be embedded in $\mathbb R^2$. The standard { $(x,y) \in \mathbb R^2 : x^2+y^2 =1$} is an embedding of the standard unit circle.

yes, I see. And can it also be embedded in $\mathbb{R}^p$, $p > 2$?

 

[zwierz]

Up to homeomorphism?

sure

 

[WWGD]

yes, I see. And can it also be embedded in $\mathbb{R}^p$, $p > 2$?

Yes, since $\mathbb R^p$ can be embedded in $\mathbb R^{p+n} ; n\geq 1$ using $(x_1,x_2,...,x_p) \rightarrow (x_1, x_2,..,x_p , 0,0,..,0)$ ; n zeros.
Then compose the embeddings to get a new one.

 

[WWGD]

Ah, ok. Thank you

Interesting that we have obstructions both on the injection side, through Borsuk-Ulam, and on the EDIT 2 surjection side. I think there is a result that, under some conditions, the image of a differentiable curve into higher dimension is meager in the higher-dimensional space. But this obviously excludes space-filling curves (can any of these be differentiable?). This last result is a theorem, but I can't remember its name now. I will edit when I remember it. EDIT: It is Sard's theorem, ( embedded in here ; ) : https://en.wikipedia.org/wiki/Sard's_theorem )but we need to figure out how to use it here to show there is no EDIT2 Surjection or at least no differentiable Surjection. I remember that for a map $f: \mathbb R^n \rightarrow \mathbb R^{n+k}$ the set of non-critical points has measure zero and I think most of the points in the domain are critical points , using the Jacobian , when the matrix is not square. I will try to tighten this up. This means the image will have Lebesgue measure zero in the image space $\mathbb R^{n+k}$ .

EDIT: This all assumes the use of Lebesgue Measure.

 

[davidge]

Interesting that we have obstructions both on the injection side, through Borsuk-Ulam, and on the EDIT 2 surjection side. I think there is a result that, under some conditions, the image of a differentiable curve into higher dimension is meager in the higher-dimensional space. But this obviously excludes space-filling curves (can any of these be differentiable?). This last result is a theorem, but I can't remember its name now. I will edit when I remember it. EDIT: It is Sard's theorem, ( embedded in here ; ) : https://en.wikipedia.org/wiki/Sard's_theorem )but we need to figure out how to use it here to show there is no EDIT2 Surjection or at least no differentiable Surjection. I remember that for a map $f: \mathbb R^n \rightarrow \mathbb R^{n+k}$ the set of non-critical points has measure zero and I think most of the points in the domain are critical points , using the Jacobian , when the matrix is not square. I will try to tighten this up. This means the image will have Lebesgue measure zero in the image space $\mathbb R^{n+k}$ .

EDIT: This all assumes the use of Lebesgue Measure.

It seems interesting to see things from that point of view. Unfortunately, I don't have enough knowledge on these topics, though.

I was thinking of another way of seeing whether a mapping is a diffeomorphism or a homeomorphism is by computing the determinant of the Jacobian matrix. If it vanishes in some point, then there could not be an inverse mapping and this violates condition for the mapping to be a diffeomorphism or a homeomorphism. But then I notice that if the coordinate functions of the domain of the mapping are independent, the determinant will vanish. So the conclusion would be that there could not be homeomorphisms nor diffeomorphisms (even locally) if the coordinate functions of the domain are independent. Is this conclusion correct?

 

[WWGD]

It seems interesting to see things from that point of view. Unfortunately, I don't have enough knowledge on these topics, though.

I was thinking of another way of seeing whether a mapping is a diffeomorphism or a homeomorphism is by computing the determinant of the Jacobian matrix. If it vanishes in some point, then there could not be an inverse mapping and this violates condition for the mapping to be a diffeomorphism or a homeomorphism. But then I notice that if the coordinate functions of the domain of the mapping are independent, the determinant will vanish. So the conclusion would be that there could not be homeomorphisms nor diffeomorphisms (even locally) if the coordinate functions of the domain are independent. Is this conclusion correct?

Hi, will get to it when I can, sorry caught up now with Sql Server. Remember, though, that for invertibility, your matrix must be a square matrix, i.e., must be$n \times n$ , meaning that if $n \neq p$ then $\mathbb R^n , \mathbb R^p$ cannot be diffeomorphic ( nor homeomorphic, but that is harder to show; we use invarinace of domain theorem )

 

[davidge]

Hi, will get to it when I can, sorry caught up now with Sql Server. Remember, though, that for invertibility, your matrix must be a square matrix, i.e., must be$n \times n$ , meaning that if $n \neq p$ then $\mathbb R^n , \mathbb R^p$ cannot be diffeomorphic ( nor homeomorphic, but that is harder to show; we use invarinace of domain theorem )

No problem.
I was assuming a square matrix, i.e. a transform from a n-dim space to another n-dim space