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I have a question about diffeomorphism between $\mathbb{R}^m$ and $\mathbb{R}^n$.

From this page of the internet we have the following definition:

Let $U\subseteq\mathbb{R}^m$ and $V\subseteq\mathbb{R}^n$. A function $F:U\to V$ is called a Diffeomorphism from $U$ to $V$ if $F$ has the following properties:

a) $F:U\to V$ is bijective.

b) $F:U\to V$ is smooth.

c) $F^{−1}:V\to U$ is smooth.

But in this post, it is proven that there is no diffeomorphism between $\mathbb{R}^2$ and $\mathbb{R}^3$. In fact, the spaces $\mathbb{R}^m$ and $\mathbb{R}^n$ are not diffeomorphic when $m \neq n$. Therefore, there cannot be a diffeomorphism between $\mathbb{R}^m$ and $\mathbb{R}^n$. But by this definition, as the symbol $\subseteq$ is used, it implies that the open sets $U$ and $V$ can be $\mathbb{R}^m$ and $\mathbb{R}^n$. So, the definition is "wrong", in the sense that there is no diffeomorphism between $\mathbb{R}^m$ and $\mathbb{R}^n$?

Would the definition be correct if the symbol $\subset$ was used? That is, is it possible to construct diffeomorphism between open sets of $\mathbb{R}^m$ and $\mathbb{R}^n$?

Mrcrg
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    The definition is fine, but the set of diffeomorphisms between U and V might be empty, in which case the dets are not diffeomorphic. – LBE Jun 16 '20 at 21:17
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    Also, in your definition it says nothing about openness ( even though for practical reasons it should be assumed) but this is very relevant for instance if you consider non open subobjects, where the Notion of smoothness is well defined. E.g. it has a manifold structure. – LBE Jun 16 '20 at 21:24
  • I think there is a bit of confusion regarding my question. This definition gives the impression that there may be a diffeomorphism between open subsets of $\mathbb{R}^m$ and $\mathbb{R}^n$. I know it can't exist between $\mathbb{R}^m$ and $\mathbb{R}^n$, But can it exist between open subsets? – Mrcrg Jun 16 '20 at 21:29
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    @Mrcrg no, it cannot exist between open subsets when $m\neq n$. It only makes sense when $U$ and $V$ are differentiable manifolds – Masacroso Jun 16 '20 at 21:33
  • Thaks @Masacroso, do you provide an answer, with a demonstration that there can be no such diffeomorphism? – Mrcrg Jun 16 '20 at 22:05

1 Answers1

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Suppose that $m\neq n$ and $U\subset \mathbb{R}^m,\, V\subset \mathbb{R}^n$ are open, then $U$ and $V$ are not diffeomorphic.

Proof: in first place note that if $U$ and $V$ are diffeomorphic then they are necessarily locally diffeomorphic, that is, if $f:U\to V$ is a diffeomorphism then the restriction of $f$ to any open ball of $U$ is an embedding (this means that it is diffeomorphic into it image). Say we pick $g:=f|_{\mathbb B (0,1)}$.

Also note that diffeomorphism is an equivalence relation because composition of diffeomorphisms is again a diffeomorphism, what follows from the chain rule. Also there exists trivial diffeomorphisms between any open ball an the entire space, that is, $\mathbb B (0,1)\subset \mathbb{R}^m$ and $\mathbb{R}^m$ are diffeomorphic, therefore the question reduces to show that $\mathbb{R}^m$ and $Y:=\operatorname{img}(g)$ are not diffeomorphic.

Then suppose that $h: \mathbb{R}^m\to Y$ is a diffeomorphism, then as the trivial embedding $i:Y \hookrightarrow \mathbb{R}^n$ is smooth we will had that $h\circ i:\mathbb{R}^m \to \mathbb{R}^n$ is also a differentiable embedding, but now it follows from the matrix representation of the Fréchet derivative at a point $x\in \mathbb{R}^m$ of any differentiable map $d:\mathbb{R}^m\to \mathbb{R}^n$, that if $m\neq n$ then $\partial d(x)$ is not invertible, therefore $h\circ i$ cannot be locally invertible at any point, however $i$ is locally invertible at any point, therefore from the chain rule we find that $h$ is not locally invertible at any point so $h$ cannot be a diffeomorphism and so our original function $f$ neither.$\Box$

Masacroso
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