The Inverse Function Theorem Completed

At order $k=0$, [F1] reduces the assertions to scalar component functions, and [L4] supplies closure under the stated operations; on an empty domain all these assertions are vacuous.

Fix $r\in \mathbb{N}$ and assume that all the closure assertions hold through order $r$.

Consider maps of class $C^{r+1}$. For a sum, scalar multiple, or componentwise product, [L2] on each coordinate line expresses every first partial derivative as a finite sum of products of $C^r$ functions. For a composite, [L3] supplies total differentiability, [L1] gives its total derivative, and [L5] identifies the first partials with the columns of that derivative; hence every first partial is a finite sum of products of first partials of the factors. The hypothesis in step 1.2 makes all these first partials $C^r$, so [F1] makes the resulting maps $C^{r+1}$. Together with the base case, this proves the result through every finite $k$.

Every cofactor and the determinant are polynomial expressions in the $C^r$ entries of $A$, so they are $C^r$ by [L2] and [L3]; [L1] identifies the only remaining factor needed for the inverse.

On $\mathbb{R}\setminus \{0\}$, repeated differentiation of $h(t)=t^{-1}$ gives $h^{(j)}(t)=(-1{)}^{j}j!t^{-j-1}$. This formula follows by induction from the reciprocal and product rules, and every derivative displayed is continuous on that domain.

Since $\det A$ never vanishes, its image lies in the domain of step 1.2. Thus $(\det A{)}^{-1}$ is $C^r$ by composition, and [L1] together with step 1.1 and [L3] makes every entry of $A^{-1}$ $C^r$.

By [L1], the local inverse exists, is $C^1$, and satisfies $Dg=(Df\circ g{)}^{-1}$ throughout its domain.

Suppose $1\le r<k$ and $g$ is $C^r$. Because $f$ is $C^k$, the entries of $Df$ are $C^{k-1}$, hence $C^r$ when $r<k$; closure under composition makes $Df\circ g$ $C^r$, and [L2] makes $Dg$ $C^r$. Therefore $g$ is $C^{r+1}$. Starting from step 1.1 and repeating this finite bootstrap reaches $C^k$.

Fix $a\in \mathrm{Reg}(f)$. By [L1], some $R>0$ has $B(a,R)\subseteq \mathrm{Reg}(f)$, so $a$ is an interior point of the regular locus.

Every point of the regular locus is interior by step 1.1; if the locus is empty, it is open by definition. Thus $\mathrm{Reg}(f)$ is open.

Let $y\in f[O]$ and choose $x\in O$ with $f(x)=y$. Apply [L1] at $x$ and replace its source neighbourhood by its intersection with $O$; the image of that smaller neighbourhood is an open neighbourhood of $y$ contained in $f[O]$.

Thus every point of $f[O]$ is interior. If $O=\mathrm{\varnothing }$, its image is the open empty set, so $f[O]$ is open in every case.

By [L1], $f[U]$ is open and the continuous bijection $f:U\to f[U]$ is open. Therefore its inverse $g:f[U]\to U$ is continuous.

Fix $y\in f[U]$. The unique global inverse agrees near $y$, by injectivity, with the local inverse from [L2]. Hence $g$ is $C^1$ near every image point and satisfies $Dg(y)=Df(g(y){)}^{-1}$ there. This proves the stated global $C^1$ diffeomorphism and derivative formula.

Since a $C^k$ map with $k\ge 1$ is $C^1$, [L1] supplies the open image and the unique global $C^1$ inverse $g:f[U]\to U$.

Around each $y\in f[U]$, the inverse $g$ agrees with the unique local inverse of $f$. By [L2] that restriction is $C^k$. Thus $g$ is locally, and hence globally, $C^k$, so the corestriction is a $C^k$ diffeomorphism.

Put $\tilde{U}=\{(x,\lambda ,y):(x,y,\lambda )\in U\}$ and $\tilde{F}(x,\lambda ,y)=F(x,y,\lambda )$. The coordinate permutation is linear, so $\tilde{U}$ is open and $\tilde{F}$ is $C^k$. Regard $u=(x,\lambda )$ as the first variable block and apply [L1] to $\tilde{F}$. It gives neighbourhoods $P,Q$, a unique $C^1$ map $\phi :P\to Q$, the equivalence $F(x,y,\lambda )=0$ if and only if $y=\phi (x,\lambda )$, and the combined-block derivative formula.

On $\tilde{U}$, the block map $H(x,\lambda ,y)=(x,\lambda ,F(x,y,\lambda ))$ is $C^k$. Its derivative at the base point is block triangular with identity on the first block and invertible block $D_{y}F$ on the second, so [L3] supplies a local inverse $G$. By [L2], $G$ is $C^k$. The identity $H(G(u,z))=(u,z)$ forces $G(u,z)=(u,\psi (u,z))$; hence $u\mapsto \psi (u,0)$ is a $C^k$ solution of the equation. After intersecting the neighbourhoods, uniqueness in step 1.1 identifies this solution with $\phi$.

Step 1.1 already supplies the displayed derivative formula for the unique $C^1$ solution, while step 2.1 upgrades that same solution to $C^k$. Thus all regularity, equivalence, uniqueness, and derivative claims hold, including $p=0$.

The map $f$ is closed. Indeed, let $A\subseteq U$ be closed and let $y$ lie in the closure of $f[A]$. Choose a compact closed target ball $K$ about $y$ contained in $V$. By [L1], $f^{-1}[K]$ is compact; $A\cap f^{-1}[K]$ is compact by [L4], and its image is compact by [L3], hence closed in $V$. Every sufficiently small neighbourhood of $y$ meets that image, so $y\in f[A]$.

By [L2], $f[U]$ is open, and by step 1.1 it is closed. It is nonempty, so connectedness of $V$ gives $f[U]=V$. For $y\in V$, [L1] makes $f^{-1}(y)$ compact. Local injectivity makes this fibre discrete, and its cover by neighbourhoods meeting the fibre in one point has a finite subcover; hence the fibre is nonempty and finite.

Write $f^{-1}(y)=\{x_1,\dots ,x_s\}$. Choose pairwise disjoint open local-inverse neighbourhoods $O_i$ of the $x_i$, with open images $W_i$. The closed set $U\setminus {\bigcup }_i{O}_i$ has closed image by step 1.1 and that image omits $y$. Therefore $W:=\left({\bigcap }_i{W}_i\right)\setminus f[U\setminus {\bigcup }_i{O}_i]$ is an open neighbourhood of $y$. Its preimage is the disjoint union of $O_i\cap f^{-1}[W]$, and each restriction is a $C^1$ diffeomorphism onto $W$.

Over a neighbourhood supplied by [L1], every fibre meets each sheet exactly once. Sending a fibre point to its unique sheet gives a bijection from every fibre there to the same nonempty finite sheet index set. Thus fibre cardinality is locally constant.

For each positive natural $d$, let $V_d$ be the set of target points with fibre cardinality $d$. Step 1.1 makes every $V_d$ open, and the $V_d$ form a disjoint cover of $V$. If two were nonempty, one and the union of all the others would disconnect $V$. Hence exactly one $V_d$ is nonempty, and it is all of $V$.

The entries of $Df$ are continuous, and [L1] expresses $x\mapsto \det Df(x)$ as a polynomial in them. Hence the Jacobian determinant is continuous on $U$.

By [L2], its image is a connected subset of $\mathbb{R}$. Regularity excludes zero. If the image contained both a negative and a positive value, order-convexity would force it to contain zero, a contradiction. Nonemptiness therefore leaves exactly one sign throughout $U$.

For the implication from constancy to zero derivative, a constant map has increment zero for every displacement. The zero linear map therefore leaves an identically zero remainder, so $Df(a)=0$ at every $a\in U$.

For the implication from zero derivative to constancy, fix $x,y\in U$. By [L1], a finite polygonal path in $U$ joins them. On each affine segment, [L2] makes every scalar component of the composite differentiable with derivative zero; [L3] makes that composite constant on the segment. The endpoint values agree successively along the finite path, so $f(x)=f(y)$. Since $x,y$ were arbitrary, $f$ is constant.