[Math] Prove that function F is diffeomorphism

Question

I have a question in the book 'Elementary differential geometry'
Prove that if a one-to-one and onto mapping $$F:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$$ is regular, then it is diffeomorphism.
I need your help.Thank you for reading

Answer 1

$F$ needs to be bijecive (one-to-one and onto) and its inverse differentiable. $F$ is clearly a bijection (by the statement already given!). Since $F$ is regular, $F'(x) \neq 0$ for all $x \in \mathbb{R}$. By the inverse function theorem, for $b=F(a)$, $$(F^{-1})'(b) = \frac{1}{F'(a)}.$$ This is clearly well-defined, since $F$ is regular. Thus $F$ is a diffeomorphism.

Edit: This only holds for functions of one variable, i.e. $n=1$. But the generalization shouldn't be a problem, see for example http://en.wikipedia.org/wiki/Inverse_function_theorem.