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I've been confused about two sentence in Folland's real analysis text concerning $C^1$ diffeomorphisms and the inverse function theorem (the theorem I'm looking at is from Spivak's book Calculus on Manifolds, see e.g. here). Here's the paragraph from Folland's text which confuses me, in particular the last two sentences:

Let $G = (g_1, \ldots, g_n)$ be a map from an open set $\Omega \subset \mathbb{R}^n$ into $\mathbb{R}^n$ whose components $g_j$ are of class $C^1$, i.e., have continuous first-order partial derivatives. We denote by $D_x G$ the linear map defined by the matrix $\big((\partial g_i/\partial x_j)(x) \big)$ of partial derivatives at $x$. (Observe that if $G$ is linear, then $D_x G = G$ for all $x$.) $G$ is called a $C^1$ diffeomorphism if $G$ is injective and $D_x G$ is invertible for all $x \in \Omega$. In this case, the inverse function theorem guarantees that $G^{-1} : G(\Omega) \to \Omega$ is also a $C^1$ diffeomorphism and that $D_x (G^{-1}) = [D_{G^{-1}(x)} G]^{-1}$ for all $x \in G(\Omega)$.

Is the definition of $C^1$ diffeomorphism used by Folland a common one? I always thought it meant for the function and the inverse to be $C^1$. I suspect they are equivalent, but I feel stuck showing both directions (I think one direction needs the inverse function theorem, or is a consequence of it).

Consider $G:\Omega\to G(\Omega)$ and $G^{-1}:G(\Omega)\to \Omega$ being $C^1$, where $\Omega\subset\mathbb R^n$. Then $G:\Omega\to\mathbb R^n$ would be injective, but how can I justify that $\det D_x G\neq0$ for all $x\in \Omega$?

I'm also stuck with showing the other direction; $G,G^{-1}$ being $C^1$ maps is implied by $G$ being injective and having $\det D_x G\neq0$ for all $x\in \Omega$.

I understand the assumption of $G$ being injective, since local invertibility at each point alone does not imply global invertibility (take the complex exponential), still, I struggle with the two sentences in the paragraph. I simply can not make sense of the definition of a $C^1$ diffeomorphism as given by Folland. Any insights are appreciated.

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If $f\in C^1$ and has full rank derivative at all points, then by the inverse function theorem, $f$ sends open sets to open sets. If $f$ is in addition injective, then $f$ is a homeomorphism. Finally, one can show that $f^{-1}$ is differentiable at $b=f(a)$ if and only if $f^{-1}$ is continuous at $b$ and $\det f'(a)\neq 0$ (this can be proved essentially using only the definition of differentiability). The chain rule then facilitates computation of $(f^{-1})'$, which also shows that $f^{-1}\in C^1$. These facts combined show that the two definitions of $C^1$ diffeomorphism are consistent.

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If $G^{-1}$ is $C^1$, then using the chain rule, $$G\circ G^{-1}=Id\Rightarrow DG_x \circ (DG^{-1}_x)=D(Id)=Id,$$ which gives invertibility of the differential.

For the other direction, the fact that $G^{-1}$ is well-defined on $G(\Omega)$ follows from standard set theoretical arguments, but the proof that $G^{-1}$ is $C^{1}$ is really the challenging part here. The proofs I know use things like Banach's fixed point theorem and similar arguments, and are not very trivial (at least in my opinion). I believe that Folland should give a proof of this later in the chapter - I suggest you try to read it (since proving by yourself could be quite challenging).

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As Folland says, you need the inverse funcion theorem. Let $y\in G(\Omega)$ and set $y=G^{-1}(x)$, if $G$ is injective, of class $C^1$ and $D_xG$ is invertible, the inverse function theorem says that there exists and open neighborhood $U\subseteq\Omega$ of $x$ such that $G|_U : U\rightarrow G(U)$ is a diffeomorphism of class $C^1$ (it is of class $C^1$ and has inverse of class $C^1$), in special, $G(U)$ is an open neighborhood of $y$ and $G^{-1}$ restricted to $G(U)$ is of class $C^1$, from this, $G^{-1}$ is locally of class $C^1$, this implies that $G^{-1}$ is of class $C^1$.

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    $\begingroup$ You are using a "global inverse function theorem," so it would be nice to know where it comes from. Indeed, the OP acknowledges the inverse function theorem, but is not aware of a global version of one. $\endgroup$ Commented Oct 30, 2024 at 20:07
  • $\begingroup$ @Andrew Is the usual IFT, $U$ is a neighborhood of $x$. What allow us to obtain the global inverse $G^{-1}$ is the injectivity of $G$. $\endgroup$ Commented Oct 30, 2024 at 20:39
  • $\begingroup$ That's not true. You have to check that the local functions the inverse function theorem gives can be stitched together to give a $C^1$ function, but it's not clear how to do this. $\endgroup$ Commented Oct 30, 2024 at 21:32
  • $\begingroup$ I thought about it more, you are right. $\endgroup$ Commented Oct 31, 2024 at 0:10

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