TITLE: Show that if a function $f : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable with differentiable inverse then $m = n$
QUESTION [5 upvotes]: So far I have:
$\boldsymbol{f^{-1}} \circ \boldsymbol{f}(\boldsymbol{a}) = \boldsymbol{a}
\implies [\boldsymbol{D}(\boldsymbol{f^{-1}}(\boldsymbol{a}) \circ \boldsymbol{f}(\boldsymbol{a}))] = I_n
\implies [\boldsymbol{D}\boldsymbol{f^{-1}}(\boldsymbol{f}(\boldsymbol{a}))][\boldsymbol{D}\boldsymbol{f}(\boldsymbol{a})] = I_n $ by the chain rule.
Interpreting $[\boldsymbol{D}\boldsymbol{f^{-1}}(\boldsymbol{f}(\boldsymbol{a}))]$ as the matrix composed of row-reduction operations, $[\boldsymbol{D}\boldsymbol{f}(\boldsymbol{a})]$ row-reduces to $I_n$. Now $[\boldsymbol{D}\boldsymbol{f}(\boldsymbol{a})]$ is a $m \times n$ matrix, therefore we have $m \le n$: Is this convincing? How do I make it rigorous?
This looks like a similar question to Existence of an inverse to differentiable function

REPLY [2 votes]: The simplest way to show is the following: Assume that $m>n$, and $I_m :\mathbb R^m\to\mathbb R^m$ the identity map. ($I_m$ is also the identity $m\times m$ matrix). Then for $a\in\mathbb R^μ$
$$
I_m=f\circ f^{-1}\quad\Longrightarrow\quad I_m=DI_m =Df\big(f^{-1}(a)\big) Df^{-1}(a).
$$
But $Df\big(f^{-1}(a)\big)$ is an $m\times n$ matrix, and its rank is at most $n$, while $Df^{-1}(a)$ is an $n\times m$ matrix, and its rank is also at most $n$. Hence their product has also rank at most $n$, which  contradict the fact that their product is the $m\times m$ identity matrix, with rank $m>n$.