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Let's say $f: \mathbb{R}^D\to\mathbb{R}^x$ is an invertible and differentiable Function (Jacobain Matrix is defined everywhere). Can we for sure say that $x=D$?

I mean, there are many functions that map vectors from $\mathbb{R}^D$ to $\mathbb{R}^D$ and they are also invertible and differentiable. For example, all linear transformations with independent columns. However, is there any invertible and differentiable transform for example from $\mathbb{R}^D$ to $\mathbb{R}^{D-1}$?

I guess the answer to these question is no, but I do not know how I can approach proving it.

Things I know and I have tried:

I can think of some invertible transforms between $\mathbb{R}^2$ to $\mathbb{R}^1$. You can just write the number in digits (the number $x$ should be $0.0<x<1.0$) and then take the odd digits after zero and make it the first number, and take the even digits after zero as the other number. I mean, $0.987654$ will become $0.975$ and $0.864$. However, this transformation is not differentiable.

Mrcrg
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milad aghajohari
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  • Here, you can see that there does not exist a continuous bijection between $\mathbb{R}$ and $\mathbb{R}^2$ : https://math.stackexchange.com/questions/47547/is-there-a-continuous-bijection-from-mathbbr-to-mathbbr2 The general case $\mathbb{R}^D \rightarrow \mathbb{R}^x$ should work the same. – TheSilverDoe Aug 24 '20 at 22:13

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