$f:A\rightarrow B$ and $f^{-1}:B\rightarrow A, \quad A\subset \mathbb{R}^{\alpha} , B\subset \mathbb{R}^{\beta}$ are bijective differentiable, and I aim to show that $\alpha = \beta$.
To do this i've computed the jacobian of $$(f \circ f^{-1})'(f(x)) = id_{\beta \times \beta} $$
and $$(f^{-1} \circ f)'(x) = id_{\alpha \times \alpha} $$
Is it true that the rank of the matrices $f'(x)$ and $f^{-1}(f(x))$ must be equal (I'm guessing this is a result of the inverse function theorem?) and if so does that show that $\alpha = \beta$
Let $x_0\in A$ and $y_0=f(x_0)$, we have $Df(x_0):\mathbb{R}^{\alpha}\to\mathbb{R}^{\beta}$ and $Df^{-1}(y_0):\mathbb{R}^{\beta}\to\mathbb{R}^{\alpha}$ linear operators.
By chain rule,
$Id=D(f^{-1}\circ f)(x_0)=Df^{-1}(y_0)Df(x_0)$
and
$Id=D(f\circ f^{-1})(y_0)=Df(x_0)Df^{-1}(y_0)$,
then $Df(x_0)$ is a linear isomorphism. Thus $\ker Df(x_0)=0$ and $\mbox{im} Df(x_0)=\mathbb{R}^{\beta}$ , by Rank–nullity theorem
$\alpha=\dim(\mathbb{R}^{\alpha})=\dim(\ker Df(x_0))+\dim(\mbox{im}Df(x_0))=\dim (0)+\dim(\mathbb{R}^{\beta})=\beta.$
You can find this theorem in Hoffman, Kenneth, and Ray Kunze. "Linear Algebra. 1971." Englewood Cliffs, New Jersey, page 71 theorem 2.