How to show that $\Bbb R^m$ is not diffeomorphic to $\Bbb R^n$ when $n \neq m$

Question

Suppose that there is indeed an diffeomorphism $f: \Bbb R^m \to \Bbb R^n$, and $f(a) = 0$. Then $f \circ f^{-1} = identity$. In class my teacher asserts that $Df^{-1}_a \cdot Df_a = identity$ and $Df_a \cdot Df_a^{-1} = identity$. Why this is true and how does it imply that $f$ cannot exist?

Answer 1

This would imply $\mathbb{R}^n$ and $\mathbb{R}^m$ are isomorphic as vector spaces, which is of course a contradiction! The fact that $Df_a^{-1} \circ Df_a = Df_a \circ Df_a^{-1} = Id$ is just the chain rule.

Answer 2

The matrix $M=Df_a$ is of size $m\times n$ and $N=Df_a^{-1}$ of size $n\times m$. If $f$ is a diffeomorphism between the two spaces then the rule for taking derivative of compositions implies that $M N= 1_m$ (identity in ${\Bbb R}^m$) and $NM=1_n$. The rank, however, can not be bigger than the minimum value of $n$ and $m$ so $n=m$.