I often heard that it is possible to show by using the inverse function theorem that if a function is smooth (i.e. arbitrarily often differentiable), a bijection between open sets, and has a non-singular jacobian, then it is a smooth diffeomorphism. but somehow the inverse function theorem that I know and that wikipedia seems to know, only states that if it is a continuously differentiable bijection with nonzero jacobian, then its inverse function is also continuously differentiable. But how do you get from this, to the statement that I proposed above? I don't see the implication.
[Math] Diffeomorphism from Inverse function theorem
inversemultivariable-calculusreal-analysis
Best Answer
Let $f(p)=q$, $J_f(p)\ne 0$. Then $f$ maps a suitable neighborhood $U$ of $p$ diffeomorphically onto a neighborhhod $V$ of $q$. Let $g:\ V\to U$ be the inverse of $f\restriction U$. The statement of the inverse function then contains the formula $$dg(y)=\left(df\bigl(g(y)\bigr)\right)^{-1}\qquad(y\in V)\ .$$ This can be interpreted as $$dg=\iota\circ df\circ g\ ,$$ where $\iota:\ L\mapsto L^{-1}$ denotes inversion in $GL({\mathbb R}^n)$.
Assume that $g$ is $r$ times continuously differentiable for some $r\geq1$. As both $f$ and $\iota$ are infinitely differentiable it follows by the chain rule that $dg$ is $r$ times continuously differentiable as well, and this implies that $g$ is in fact $r+1$ times continuouslay differentiable.