Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I often heard that it is possible to show by using the inverse function theorem that if a function is smooth(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.

share|cite|improve this question
up vote 6 down vote accepted

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.

share|cite|improve this answer

Let $LI^m$ and $L^m$ be the set of invertible linear transformations of $\mathbb{R}^m$ in itself and the set of linear transformations of $\mathbb{R}^m$ in itself respectively. Define $\operatorname{Inv}: LI^m\to L^m$ by $\operatorname{Inv}(A)$ is the inverse matrix of $A$. First you can note that if $f$ is a function satisyfing your conditions, then $$(f^{-1})'(y)=[f'(g(y))]^{-1}$$

This implies that $(f^{-1})'=(\operatorname{Inv})\circ f'\circ f^{-1}$. Because $\operatorname{Inv},f\in C^\infty$, you can conclude from the last formula (by using chain rule) that $f^{-1}\in C^\infty$.

Update: I do not understood OP's comment in the bounty, nevertheless I give here more explanations:

Since $f'\neq 0$ we have that $\operatorname{Inv}(f'(g(y)))$ is well defined for all $y$, so the derivative of $(f^{-1})$ dependes on functions that are differentiable again. By a induction argument OP's can finish the proof. Also the proof that $\operatorname{Inv}$ is $C^\infty$ was given in the comments by @ronno. As I have wrote to @Lipschits in the comments, I know another proof of the fact that $\operatorname{Inv}$ is $C^\infty$, but it is really big and @ronnos's proof is more straightforward.

share|cite|improve this answer
mhmm, I don't know what $I(A)$ is, did you mean $Inv(A)$? and what do we know about $g$? – user66906 May 17 '13 at 22:05
Yes, you are right, let me fix it. – Tomás May 17 '13 at 22:29
thank you. i think i have the idea now. the only thing that i do not know is but why $(Inv) \in C^{\infty}$? – user66906 May 17 '13 at 22:43
@Lipschitz, to prove that $\operatorname{Inv\in C^\infty}$ is a big calculation. I will try to resume it and then I post it here. – Tomás May 18 '13 at 0:29
$Inv$ is $C^\infty$ since on $GL_n(\mathbb{R})$, it is a quotient of polynomials, with the denominator non-vanishing. You of course now that given a basis, taking the entries of the matrix from a linear operator is $C^\infty$ – ronno May 20 '13 at 9:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.