On proving that Constant Rank Theorem $\implies$ Inverse Function Theorem

Question

I am referring to the discussion on presented here: Rank theorem implies inverse function theorem. We known that the composition of diffeomorphisms is a diffeomorphism and that we have our diffeomorphisms $\alpha \circ f \circ \beta$ and $\alpha, \beta$. Thus $\alpha^{-1} \circ \alpha \circ f \circ \beta \circ \beta^{-1} = f$ is a diffeomorphism, and thus $f$ invertible at some neighborhood of a point $p \in U$. But how do we draw the equivalence between "$f$ is invertible in some neighborhood of a point $p$" and the Jacobian determinant of $f$ at $p$ is non-zero? We do know that $f$ has rank $n$, and thus its Jacobian has $n$ linearly independent columns. But as $f$ is defined on $n + m$ dimensional space, then its Jacobian determinant could be zero, while the determinant of the $n\times n$ submatrix is non-zero, right? Do we use projection mapping somehow to argue about the Jacobian, or how is the proof finished?

Answer 1

The rank theorem in the linked question says that if $U \subset \mathbb{R}^{n+m}$ is open and $f : U \to \mathbb{R}^{n+k}$ is a $C^r(U)$ function with constant rank $n$, then for each $z = (x,y) \in U$ there are diffeomorphisms $\alpha$ defined on some open neighborhood of $f(z)$ to an open subset of $\mathbb{R}^{n+k}$ containing $(x,0)$ and $\beta$ defined on some open subset of $\mathbb{R}^{n+m}$ to some open subset $U' \subset \mathbb{R}^{n+m}$ with $z \in U' \subset U$ such that $(\alpha \circ f\circ \beta)(x,y) = (x,0)$. This means that by diffeomorphisms around $z$ and $f(z)$ we can transform $f$ into the projection map $(x,y) \mapsto (x,0)$.

In the inverse function theorem we are in the special case $m = k = 0$. This means that $\alpha \circ f\circ \beta = id$ around $\beta^{-1}(p)$.

The Jacobian of $f$ at $p$ is an $(n \times n)$-matrix $Jf(p)$. Clearly its determinant is $\ne 0$ iff $Jf(p)$ has rank $n$. Therefore we get

1. If $\det Jf(p) \ne 0$, then via the constant rank theorem we get $\alpha \circ f\circ \beta = id$ around $\beta^{-1}(p)$. Hence $f = \alpha^{-1} \circ \beta^{-1}$ around $p$ which shows that $f$ is invertible in some open neighborhood of $p$. Its inverse has the form $\beta \circ \alpha$ in this neighborhood.

2. If $f$ is invertible in some neighborhood of $p$, we have a smooth map $g$ defined on an open neigborhood of $q = f(p)$ which is a local inverse for $f$. This shows that $Jf(p)$ and $Jg(q)$ are inverse matrices, therefore $\det Jf(p) \ne 0$.