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Let $\Omega \subseteq \mathbb{R}^n$ and $p \in \Omega$. Let $f:U \to V$ be a bijection of open sets $p \in U \subseteq \Omega$ amd $f(p) \in V \subseteq \mathbb{R}^n$. If $f^{-1}: V \to U$ is differentiable at $p$, then $df_p: \mathbb{R}^n \to \mathbb{R}^n$ is invertible.

Suppose $f$ is a bijection. Since $f^{-1}$ is differentiable at $p$, it is continuous at $p$. Let $f^{-1}(f(x))=x$. Then
\begin{align*} df^{-1}(f(x))&=1\\ df_p^{-1}df_p&=1\\ \end{align*}

Following Dan Shved's suggestion I applied the chain rule, but I'm not sure that $f$ is differentiable - thus I'm not sure if $df_p$ exists. If $f^{-1}$ were continuously differentiable this would be easier because I could invoke the Inverse Function theorem.

emka
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  • I'm confused. $\Omega$ doesn't seem to play any role in the problem. Also, the thing that you said about $f(p)$ is as strange to me as it is to you. – Dan Shved Apr 16 '13 at 09:01
  • I fixed that. $U$ should be a neighborhood of $p$ in $\Omega$. – emka Apr 16 '13 at 09:02
  • This makes more sense, but still we can just say that $U$ is an open subset of $\mathbb{R}^n$. Doesn't change anything at all. Also, I'm fairly sure that you're right and $f^{-1}$ should be differentiable at $f(p)$. After all, $f^{-1}$ doesn't even need to be defined at $p$. – Dan Shved Apr 16 '13 at 09:04
  • I'm just wondering if I'm on the right track. Is this an Inverse Function Theorem question? – emka Apr 16 '13 at 09:08
  • I'd say it's easier. Looks like the chain rule should be enough. But I can't be completely sure when the problem statement is so foggy. – Dan Shved Apr 16 '13 at 10:26
  • I edited the problem and fixed my solution a bit. Though I'm stuck. – emka Apr 21 '13 at 04:07
  • $f^{-1}:V\to U$ differentiable at $p$ makes no sense, since $p\in U$, not $p\in V$. Should that be "$f^{-1}$ differentiable at $f(p)$"? – Marc van Leeuwen Jun 03 '13 at 07:39

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The problem statement is either incorrect, incomplete, or both. Certainly, in order to say anything about $df_p$, the assumption on $f^{-1}$ should be made at $f(p)$. But the mere fact that $f^{-1}$ is differentiable at $f(p)$ is not enough. For example, $f(x)=x^{1/3}$ is a bijection of $(-1,1)\subset \mathbb R$ onto itself. Its inverse $f^{-1}(x)=x^3$ is differentiable at $0=f(0)$, but $f$ is not differentiable at $0$.

The following amended statement is correct:

Let $f:U \to V$ be a bijection of open sets $p \in U \subseteq \Omega$ amd $f(p) \in V \subseteq \mathbb{R}^n$. If $f^{-1}: V \to U$ is differentiable at $\mathbf{f(p)}$, then $df_p: \mathbb{R}^n \to \mathbb{R}^n$ is invertible provided it exists.

Indeed, the chain rule applies to $\mathrm{id}=f^{-1}\circ f$ at $p$ and yields $\mathrm{id}=df^{-1}_{f(p)} \circ df_p$. Hence $df_p$ is invertible.

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