Show that if a function $f : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable with differentiable inverse then $m = n$ 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
 A: Such a function $f$, which is differentiable, bijective and has a differentiable inverse, is called a diffeomorphism.
Claim: If $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a diffeomorphism, then $Df(x)$ is bijective for all $x\in\mathbb{R}^n$. 
Proof:
Let us consider $f^{-1}\cdot f(x)=x$ and $f\cdot f^{-1}(x)=x$. 
Deriving the first equation gives you 
$$Df^{-1}(f(x))Df(x)=I_n$$
and the second equation gives 
$$Df(f^{-1}(y))Df^{-1}(y)=I_m$$
Setting $y=f(x)$, we see that the second equation becomes
$$Df(x)Df^{-1}(f(x))=I_m$$
But this just means that $Df(x)$ is an invertible matrix with inverse $Df(x)^{-1}=Df^{-1}(f(x))$.
$\square$
Can a $m\times n$ matrix be invertible if $m\not=n$?
No, therefore $f$ being a diffeomorphism implies $m=n$.
Note: This also applies to diffeomorphisms $f:U\rightarrow V$ where $U\subset\mathbb{R}^n$, $V\subset\mathbb{R}^m$ open (with the same proof, being differentiable is a local property). The result is called invariance of dimension/invariance of domain.
For merely continuous maps $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ which are bijective and have a continuous inverse (called homeomorphisms), this result also holds true, but requires more machinery (using singular homology it is trivial, but one can also prove it in an "elementary" way using Brouwer's fixed point theorem).
A: 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 $m>n$. 
