I Boundary and homeomorphism

[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