Reformulations of Inverse Function Theorem

Question

Inverse Function Theorem: Let $U\subset\mathbb{R^n}$ be open, $f:U\longrightarrow\mathbb{R^n}$ be $C^k$ such that for $a\in U,\quad d_a f:\mathbb{R^n}\longrightarrow\mathbb{R^n}$ is invertible. Then there exists a neigbourhood $V\subset U$ with $a\in V$ such that $f|_V$ is a $C^k$ diffeomorphism, and $W=f(V)\subset \mathbb{R^n}$ is open. Furthermore, $f^{-1}:W\longrightarrow V$ is $C^k$ and $$(d_a f)^{-1}=d_{f(a)}f^{-1}.$$

I have problem in proving following two corollaries:

Corollary 1: Let $f\in C^1$ be from an open set $U\subset \mathbb{R^m}$ to $\mathbb{R^n}.$ Let $0\in U,f(0)=0,$ and that $d_0 f$ is injective. Then there exists an open neighbourhood $V\subset\mathbb{R^n}$ of $0$, an open set $U'\subset U$ with $0\in U'$ such that $f(U')\subset V,$ and a diffeomorphism $\phi:V\longrightarrow V$ such that $$\phi(f(x_1,...,x_m))=(x_1,...,x_m,0,...,0).$$

Corollary 2: Let $f\in C^1$ be from an open set $U\subset \mathbb{R^m}$ to $\mathbb{R^n}.$ Let $0\in U,f(0)=0,$ and that $d_0 f$ is surjective. Then there exists an open set $W\subset\mathbb{R^m}$ of $0$ and a diffeomorphism $\psi:W\longrightarrow \psi(W)\subset U$ such that $$f(\psi(x_1,...,x_m))=(x_1,...,x_n).$$

I appreciate any hint.

Answer 1

For Corollary 1:

The linear map $d_0f$ is injective, so its image is an $m$-dimensional subspace of $\mathbb{R}^n$. To make things easier, let $\Lambda\colon \mathbb{R}^n\to \mathbb{R}^n$ be an invertible linear map such that the image $\Lambda\circ d_0f$ is the subspace $\{(x_1,\ldots, x_m, 0,\ldots, 0)\}\subset \mathbb{R}^n$.

Now, define a map $F\colon U\times \mathbb{R}^{n-m}\to \mathbb{R}^n$ by $F(x,y) = \Lambda f(x) + (0,y)$. You should check that $d_0F$ is an invertible linear map, and hence by the inverse function theorem it has a local inverse $F^{-1}$.

By the definition of $F$, we have $F(x,0) = \Lambda f(x)$, so that $(x,0) = F^{-1}\Lambda f(x)$. Thus you need only define $\phi = F^{-1}\Lambda$.

For Corollary 2:

The linear map $d_0f$ is surjective, so there is some $n$-dimensional subspace of $\mathbb{R}^m$ that is mapped isomorphically onto $\mathbb{R}^n$ by $d_0f$. To make things easier, let $\Lambda\colon \mathbb{R}^m\to \mathbb{R}^m$ be an isomorphism so that $d_0f\circ \Lambda$ maps the subspace $\{(x_1,\ldots, x_n,0,\ldots,0)\}\subset \mathbb{R}^m$ isomorphically onto $\mathbb{R}^n$. Let $W = \Lambda^{-1}(U)$.

Now, define a map $F\colon W\to \mathbb{R}^n\times\mathbb{R}^{m-n}$ by $F(x) = (f\Lambda(x), x_{n+1},\ldots, x_{m})$. You should check that $d_0F$ is an invertible linear map, and hence by the inverse function theorem has a local inverse $F^{-1}$.

Let $\pi\colon \mathbb{R}^n\times \mathbb{R}^{m-n}\to \mathbb{R}^n$ be the projection map $\pi(x,y) = x$. By the definition of $F$, we have $\pi F = f\Lambda$, and thus $\pi = f\Lambda F^{-1}$. You now need only define $\phi = \Lambda F^{-1}$.