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My textbook gives the following definition of a diffeomorphism:

Let $A \subseteq \mathbb{R}^k$ and $B \subseteq \mathbb{R}^n$. We say that a function $\psi: A \to B$ is a $C^p$-diffeomorphism $(p \geq 1)$ if $\psi$ is a homeomorphism and if $\psi$ and $\psi^{-1}$ are functions of class $C^p$, in the sense that there exist open sets $\tilde{A} \subseteq \mathbb{R}^k$ and $\tilde{B} \subseteq \mathbb{R}^n$ such that $A \subseteq \tilde{A}$ and $B \subseteq \tilde{B}$ and such that $\psi$ and $\psi^{-1}$ can each be extended to $C^p$-functions over $\tilde{A}$ and $\tilde{B}$ respectively.

Note that later this definition is used in a context where $k < n$.

I was under the impression that it is impossible for a diffeomorphism to be between sets of different dimensions, see e.g. here. It seems like the definition tries to get around this by introducing $\tilde{A}$ and $\tilde{B}$, but I think that just makes $\phi$ a diffeomorphism between $\tilde{A}$ and $\tilde{B}$, which are again of different dimensions. Note that this is the second time the word diffeomorphism was defined, the first definition earlier on in the text was

Let $A, B \subset \mathbb{R}^n$ be open. A function $f: A \to B$ is a $C^p$-diffeomorphism if it is a bijection in $C^p$ and its inverse is also in $C^p$.

This definition made sense to me and the new one seems to be a slight generalization, but I don't see how it could work.

zaq
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    It don't say the extension to $\tilde A$ and $\tilde B$ was a diffeomorphism, or even a bijection. You are correct, there no diffeomorphisms from spaces of different dimensions. But there are subsets of $\mathbb R^n$ which are not of dimension $n.$ For example, a circle in a plane. – Thomas Andrews Aug 17 '24 at 03:25
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    In the case of two open sets in $\mathbb R^n$ the definition means the same, because $\tilde A=A$ and $\tilde B=B$ works. You'll encounter at least one more even more general definition of diffeomorphism. – Thomas Andrews Aug 17 '24 at 03:28
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    So, in short, the only thing the extension of $\psi$ needs to be is $C^p.$ And the only thing the extension of $\psi^{-1}$ needs to be is $C^p.$ The two extensions dont need to be inverses of each other, and they can't be, if $k\neq n.$ – Thomas Andrews Aug 17 '24 at 03:34
  • Oh, okay, that makes sense! Thanks! – zaq Aug 17 '24 at 03:40
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    For example, if $A={(x,y)\in \mathbb R^2\mid x^2+y^2=1},$ and $B={(x,y,z)\mid x^2+4y^2=1, z=0},$ and $\psi:(x,y)\mapsto (x,y/2,0)$ we get the simple extension of $\psi$ $(x,y)\mapsto (x,y/2,0)$ defined for $\tilde A=\mathbb R^2$ and the extension of $\psi^{-1}$ as inverse $(x,y,z)\mapsto (x,2y)$ defined on all of $\mathbb R^3.$ – Thomas Andrews Aug 17 '24 at 03:41

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