I Boundary and homeomorphism

[fresh_42]

@fresh_42 and @WWGD
I think I see what you are saying.

Suppose a parametrization $g: (-r,r) \rightarrow \mathbb{S}^1 \\ t \mapsto x$ where $(-r,r)$ is an open interval of $\mathbb{R}$.
Now define $f_1: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, \sqrt{r²-x²} \bigg{)}$
This function does not cover the entire circle $\in \mathbb{R}^2$, because the points $(r,0)$ and $(-r,0)$ $\in \mathbb{R}^2$ are excluded. Also, we would need to define another function $f_2: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, - \sqrt{r²-x²} \bigg{)}$

Is my above understanding correct?

Your notation is a bit confusing as I assume $f_1=g$ and $x=t$. A better parameterization is simply done be the angle. You can "run" through the circle by $p\, : \,[0,2\pi) \longrightarrow \mathbb{R}^2$ defined as $p(\varphi)=(\cos \varphi, \sin \varphi)$. This is a continuous embedding. But all points apart from the circle are not hit, so it's only injective, not surjective. A homeomorphism $h$ requires both, such that it can be inverted. In addition both mappings ($h$ and $h^{-1}$) have to be continuous.

$\mathbb{S}^1$ is a circle, $\mathbb{R}^2$ a plane. How could there be a bijection? So let's see what the stereographic projection does:

You can map all points smoothly from the circle to the line ($x-$axis): the southern half of the circle will get you the points between $-1$ and $+1$ and the northern half the points outside. Except for the northpole itself, which is infinitely far away. So you have to extend the real line by an infinitely far away point, which is called projective line (see link above). So now if the circle $\mathbb{S}^1$ and the projective line $\mathbb{P}(1,\mathbb{R})$ are topologically the same, it therefore can neither be the real line $\mathbb{R}^1$ nor the real plane $\mathbb{R}^2$.

 

[davidge]

you can also use some topological invariants to show the non-existence of a homeomorphism

Like the theorem that says that $A \times B$ is homeomorphic to $C$ iff both $A$ and $B$ are homeomorphic to $C$, or like the compactness property

Thanks @fresh_42 for the explanation. It is interesting to see in details how the stereographic projection works.

Now the problem is that if the circle is'nt homeomorphic to $\mathbb{R}^2$, then the definition of a boundary on a manifold doesn't seem to work in this case, because that definition says a manifold don't have a boundary if the neighborhood of each of its points is homeomorphic to $\mathbb{R}^n$. In the case of the circle, we know it does not have a boundary, but yet it's not homeomorphic to $\mathbb{R}^2$.

Or should we just use the fact that a manifold is required to be locally Euclidean and conclude that the circle is locally homeomorphic to $\mathbb{R}^2$? But if we proceeded this way, then we would find that there're no manifolds with boundaries at all and the whole concept of boundary would become meaningless.

 

[WWGD]

Actually circle is locally homeomorphic to $\mathbb R^1$ , not to $\mathbb R^2$. It is more like a twisted line than a solid object, informally.

 

[fresh_42]

Like the theorem that says that $A \times B$ is homeomorphic to $C$ iff both $A$ and $B$ are homeomorphic to $C$, or like the compactness property

What do you mean? If $A=B=C=\mathbb{S}^1$ is the circle, then $A \times B = \mathbb{S}^1 \times \mathbb{S}^1$ is a torus, which is not homeomorph to a circle $\mathbb{S}^1$ (wrong dimensions). Don't confuse homeomorphisms with homotopy mappings.

Now the problem is that if the circle is'nt homeomorphic to $\mathbb{R}^2$, then the definition of a boundary on a manifold doesn't seem to work in this case, because that definition says a manifold don't have a boundary if the neighborhood of each of its points is homeomorphic to $\mathbb{R}^n$. In the case of the circle, we know it does not have a boundary, but yet it's not homeomorphic to $\mathbb{R}^2$.

It gets automatically confusing here, as terms are used a bit ambiguously when it comes to manifolds compared to ordinary topological spaces. As manifolds are also topological spaces, this is a very unlucky situation, which in my language is solved by an extra name.
Let's see whether I can get it right.

The boundary $\partial S$ of a set $S$ in a topological space $X$ can be defined as $\partial S = \overline{S} \cap \overline{X - S}$.
As topological space, a manifold is automatically closed as well as open, which makes its boundary automatically empty. This is a direct result of the fact, that the manifold is the entire available space. No embeddings into some outer space!

As an embedded structure, like a ball in usual space, it has the the topology of this outer space and a boundary there: The open ball $\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 < 1\}$ has the boundary $\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 = 1\}$, the sphere, although it doesn't belong to the open ball anymore, or in case of $\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 \leq 1\}$ it has also the sphere as boundary which this time is part of the set, the closed ball. The space itself, here $\mathbb{R}^2$, has no boundary.

But as we don't want to have embeddings, when it comes to manifolds, we have to say goodbye to outer spaces.

Instead we have an atlas of our manifold $M$, which gives us the family of regular charts on open sets. Now to speak of a boundary, we consider the prototype of something with a boundary, namely the set $\mathbb{R}^n_+ = \{x \in \mathbb{R}^n\,\vert \,x_n \geq 0\}$. Next we allow besides our regular charts, some additional boundary charts for our sets $U \subseteq M$, that is a chart that is a 1-to-1 map $\phi$ such that $\phi(U) \subseteq \mathbb{R}^n_+$ is open. Regular points are those covered by regular charts, and boundary points are points which are not regular, that means covered by a boundary chart plus the condition $\phi(p)_n =0$. This way the boundary property of $\mathbb{R}^n_+$ is transported into terms of the manifold via its atlas.

So as for the circle, all remains as it is, i.e. no boundary at all since it has no interior.

However, if you mean the disc $M=\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 \leq 1\}$ instead, then the circle of this disc is its boundary in the sense that it contains the boundary points (charts which map to $\mathbb{R}^2_+=\{(x,y)\in \mathbb{R}^2\,\vert \,y\geq 0\}$).

If you take away these points, i.e. $M=\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 < 1\}$, then it has no boundary points, as all points are covered by regular charts.

Or should we just use the fact that a manifold is required to be locally Euclidean and conclude that the circle is locally homeomorphic to $\mathbb{R}^2$?

The circle is a one dimensional curve and as such locally homeomorph to an open interval of the straight line $\mathbb{R}^1$.

But if we proceeded this way, then we would find that there're no manifolds with boundaries at all and the whole concept of boundary would become meaningless.

No, not really meaningless, just "handle with care". (See above, if I got it right. Pretty late here now.)
One last remark: The charts in an atlas don't have to be global, that is there is usually not a single chart that maps $M$ to $\mathbb{R}^n$, so there isn't a global homeomorphism needed. We only need local charts like in a real world atlas: pages of open sets which map to open sets in Euclidean space.

 

[davidge]

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

So as for the circle, all remains as it is, i.e. no boundary at all since it has no interior.

Would another way to put it be to give a counter example e.g. the cylinder $[a,b] \times \mathbb{S}^1$ is not homeomorphic to $\mathbb{R}^2$ because $[a,b]$ and $\mathbb{S}^1$ are both compact, while $\mathbb{R}^2$ is not. So the cylinder is not locally (nor globally) homeomorphic to $\mathbb{R}^2$, which means it has a boundary.

No, not really meaningless, just "handle with care". (See above, if I got it right. Pretty late here now.)
One last remark: The charts in an atlas don't have to be global, that is there is usually not a single chart that maps $\mathbb{R}^n$, so there isn't a global homeomorphism needed. We only need local charts like in a real world atlas: pages of open sets which map to open sets in Euclidean space.

Yes, I see now

What do you mean? If $A=B=C=\mathbb{S}^1$ is the circle, then $A \times B = \mathbb{S}^1 \times \mathbb{S}^1$ is a torus, which is not homeomorph to a circle $\mathbb{S}^1$ (wrong dimensions). Don't confuse homeomorphisms with homotopy mappings.

Oh, I thought at first that there could be homeomorphisms between spaces of different dimensions.

 

[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