In our syllabus, we proved the local inverse function theorem: Let $f: E\to\mathbf{R}^n$ with $E\subseteq\mathbf{R}^n$ open be $C^1$. If $\vec{a}\in E$ with $\det f'(\vec{a})\neq 0$, then there exist open sets $U\ni\vec{a}$ and $V\ni f(\vec{a})$ such that $f:U\to V$ is a $C^1$-diffeomorphism.
I have two questions which involve this theorem.
First, I am asked to prove the following 'global' version of the inverse function theorem, which states: Let $f:E\to\mathbf{R}^n$ with $E\subseteq\mathbf{R}^n$ open be injective and $C^1$ such that $\det f'(\vec{x})\neq0 $ for all $\vec{x}\in E$. Then $f$ is a $C^1$-diffeomorphism $E\to f(E)$. Of course, by restricting the codomain to $f(E)$, we automatically get the surjectivity and thus bijectivity. However, I don't see how to proceed any further.
Could someone provide some insight into this?
I proof the global version of the inverse function theorem using the following local version:
I will proof:
Proof: Note that we need to proof three things to conclude $f$ is a $C^1$ diffeomorphism $E\to f(E)$: $(1)$ $f$ is bijective, $(2)$ $f$ is a $C^1$ map, and $(3)$ $f^{-1}$ is a $C^1$ map. The second criterion is true by assumption.
Because $f:E\to\mathbb{R}^n$ is injective, $f:E\to f(E)$ is bijective since it is trivially surjective, and thus $f$ is invertible. An inverse map is, whenever it exists, uniquely determined. Furthermore, because for any $a\in E$, $\det(f(a))\neq0$, the local version of the inverse function theorem gives that there exists open balls $U\ni a$ and $V\ni f(a)$ such that $f:U\to V$ is a $C^1$ diffeomorphism. Because of uniqueness of inverse functions, we see that $f^{-1}:V\to U$ is $C^1$. Since this holds for any $a\in E$, we see that $f^{-1}$ is continuously differentiable for every $f(a)\in f(E)$. Therefore (remember that $f:E\to f(E)$ is bijective) $f^{-1}$ is a $C^1$ map; it follows that $f$ is a $C^1$ diffeomorphism $E\to f(E)$. $\blacksquare$