The set $B$ is open in $\mathbb{R}^n$ and the inverse function $g : B \to A $ is of class $C^r$

130 Views Asked by At

Theorem: Let $A$ be open in $ \mathbb{R}^n $; let $ f : A \to \mathbb{R}^n $ be of class $ C^r $; let $B=f(A)$. If $f$ is one to one on $A$ and if $Df(x)$ is non singular for $ x \in A $, then the set $B$ is open in $\mathbb{R}^n$ and the inverse function $ g:B \to A $ is of class $ C^r $.

Proof: first prove that if $\phi:A \to \mathbb{R}$ is differentiable and if $\phi $ has a local minimum at $ x_0 \in A $, then $ D\phi(x_0)=0 $. Then goes on to show that the set $B$ is open in $ \mathbb{R}^n $. Given $ b \in B $, we show B contains some open ball $ B(b,\delta) $ about b. Then we show the function $g$ is continuous. Next, given $ b \in B $, we show $g$ is differentiable at $b$. In the end, show that the inverse function $g$ is of class $ C^r$.

This is some general manner to prove for this theorem, and can someone explain the big idea in this proof? Especially that I am not sure why we have go demonstrate that $\phi $ has a local minimum at $ x_0 \in A $. I don't get from the beginning.