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 $\mathbb{R}^n$?
Best Answer
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.$\Box$