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This is part of the proof of the inverse function theorem:

Let $U,V\subset R^n$ and $f\colon U\to V$ a differentiable bijection. Also assume that $f$ is $C^1$ and $f^{-1}$ is differentiable. Show that $D(f^{-1})$ is continuous.

Using the chain rule, I have derived that $Df^{-1}(f(x))=Df(x)^{-1}$. Since $f$ is $C^1$, we know that $Df(x)$ is continuous. So what I basically have to show is that if a function $g\colon U\to V$ is continuous (with $U,V$ satisfying the above conditions), then its inverse $g^{-1}$ is continuous too. I have seen someone mention the invariance of domain theorem on this site, but I feel like that is too advanced for me now, in the sense that our teacher actually meant something easier (since this is a part of a small homework exercise). Unless I'm wrong, there should be a relatively easy to show this. Any hint?

Sha Vuklia
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Just remark that $Df^{-1}(x)=Df(f^{-1}(x))^{-1}$, $x\rightarrow f^{-1}(x), x\rightarrow Df(x)$ and $A\rightarrow A^{-1}$ defined on $Gl(\mathbb{R}^n)$ are continuous and $x\rightarrow Df^{-1}(x)$ is the composition of these functions.

Tsemo Aristide
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  • Why is $Df^{-1}(x)=Df(f^{-1}(x))^{-1}$? I would say that $Df^{-1}(x)=Df(f^{-1}(y))^{-1}$, where $f^{-1}(y)=x$. Correct me if I'm wrong. But yea, I do think that I can use $A\to A^{-1}$ on $GL(R^n)$ is continuous, so I think that solves my problem. Maybe what you wrote in the beginning was just some notation that I don't read correctly. – Sha Vuklia Nov 05 '17 at 02:53
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    $Df^{-1}(f(x))=Df(x)^{-1}$. Write $y=f(x), x=f^{-1}(y)$, you obtain $Df^{-1}(y)=Df(f^{-1}(y))^{-1}$. – Tsemo Aristide Nov 05 '17 at 03:01
  • Oh, of course! Thanks. I hope you don't mind if I ask one last question about this continuity of $A\to A^{-1}$. If I understand correctly; an invertible matrix $A$ that depends continuously on $x$ has a continuous inverse? If this a theorem or something? I would like to look up a proof, because I can't show that myself. – Sha Vuklia Nov 05 '17 at 03:03
  • $A\rightarrow A^{-1}$ is considered as a function defined on the set of invertible matrices $Gl(n,\mathbb{R})$, not on $U$. Note that $df:U\rightarrow Gl(n,\mathbb{R})$ since the differential of $f$ is invertible. – Tsemo Aristide Nov 05 '17 at 03:07
  • Okay, I understand that $A\to A^{-1}$ is defined on $GL(n,\mathbb R)$ and that $Df\colon U\to GL(n,\mathbb R)$, but how do we know that the function is continuous? Could we say that $A\mapsto A^{-1}$ is actually the function $A^{-1}A^{-1}$? I somehow have to get rid of the dependence of $x$, and consider a constant function that maps $A$ to $A^{-1}$, I think. – Sha Vuklia Nov 05 '17 at 03:18
  • just write its expression in terms of rational functions of the coefficients. – Tsemo Aristide Nov 05 '17 at 03:19