I'm looking to confirm some properties of the total derivative of a diffeomorphism. Suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is a diffeomorphism. Then its total derivative at a point $p\in\mathbb{R}^n$, denoted $Df(p): \mathbb{R}^n \rightarrow \mathbb{R}^m$, can be seen as multiplication by some $m\times n$ matrix.
I am wondering whether it is true that $Df(p)$ is also bijective.
$f$ being a diffeomorphism means that it is smooth and has a smooth inverse $g : \mathbb{R}^m \to \mathbb{R}^n$ (i.e. $f \circ g = id$ and $g \circ f = id$). The chain rule shows $df(g(q)) \circ dg(q) = did(q) = id$ and $dg(f(p)) \circ df(p) = did(p) = id$ for all $p, q$. Hence $df(p)$ is an isomorphism of vector spaces and in particular is a bijection. Moreover, we see that this is possible only in case $m = n$.