0
$\begingroup$

${\large let \ U\subset \mathbb{R^n,}V\subset \mathbb{R^m}}$ be open subsets. map $f:{\large U }\to{\large V }$ is called a diffeomorphism if is bijective and both $f$ and $f^{-1}$ are continuously differentiable.

once there is diffeomorphism $f:{\large U }\to{\large V }$ then $Df:\mathbb{R^n}\to\mathbb{R^m}$ will be ${\large linear \ isomorphism }$ hence $n=m$. how is this a linear isomorphism??. i know $Df(a)$ is a linear map , but why this is one one and on to ??

  • 0
    $Df$ is not a map from $\mathbb R^n$ to $\mathbb R^m$. You mean $Df(a)$ for some $a \in U$. – 2017-02-10
  • 0
    Please avoid `\large` in titles. (You also use some for no reason in the body of the question but I left them.) – 2017-02-10
  • 0
    @OpenBall yes i mean $Df(a)$, for some $a$ in open set $U$ – 2017-02-10

3 Answers 3

0

Given $a \in U$, $f \circ f^{-1} = I_{\mathbb R^m}$, so as $f^{-1}$ is differentiable at $f(a)$, $Df(f^{-1}(f(a))) \circ Df^{-1} (f(a)) = I_{\mathbb R^m}$, i.e. $Df(a) \circ Df^{-1} (b) = I_{\mathbb R^m}$, where $b = f(a)$. Similarly, $Df^{-1}(b) \circ Df(a) = I_{\mathbb R^n}$, so $Df(a)$ is invertible and $[Df(a)]^{-1} = Df^{-1}(b)$; it is an isomorphism since it is linear and continuous with a continuous inverse by the assumptions in the question.

4

Hint: What do you get when you apply the chain rule to $$f \circ f^{-1} = \text{Id}?$$

4

$f\circ f^{-1}=Id, f^{-1}\circ f=Id$ implies $Df\circ Df^{-1}=Id, Df^{-1}\circ Df=Id$. It implies $Df$ is an isomorphism.