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  1. If there is a continuous mapping $f:\Bbb{R^2}\to\Bbb{R}$, will $f^{-1}$ also be continuous?
  2. If there is a differentiable mapping $g:\Bbb{R^2}\to\Bbb{R}$, will $g^{-1}$ be continuous/differentiable?

I don't know how to proceed here. To show that $f^{-1}$ is continuous, we will have to show that the inverse of every open set is an open set. How will we form open sets in $\Bbb{R^2}$ for $f^{-1}$? Do we only consider open sets in $\Bbb{R^2}$ whose elements are fibers of points in $\Bbb{R}$?

freebird
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    If $f^{-1}$ were continuous, $f$ would be a homeomorphism between $\mathbb{R}^2$ and $\mathbb{R}$. Does this seem possible? What sorts of properties of topological spaces are preserved under homeomorphism? – Eric Towers Jun 17 '14 at 04:59
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    If $g^{-1}$ exists, then $g$ is injective. Can you think of a differentiable injection from $\mathbb{R}^2$ to $\mathbb{R}$? – Eric Towers Jun 17 '14 at 05:00
  • For 2. If you want to apply the inverse function theorem, one more condition needed: the Jacobian matrix $Dg(x_0)$ is invertible $\forall x_0 \in \mathbb{R^2}$. Then $g^{-1}$ exists and is of class $C^1$. – Mark Jun 17 '14 at 05:29
  • @Mark- I think the inverse function theorem is applicable only for $\Bbb{R^n}\to\Bbb{R^n}$. The exponent $n$ has to be the same. – freebird Jun 17 '14 at 05:31
  • Yes, thank you @freebird. You are right. – Mark Jun 17 '14 at 05:47

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There can be not just no continuous invertible map (bijection) $f$ of $\mathbb R^2$ onto $\mathbb R$ , as Eric Towers pointed out, but not even a continuous map with continuous inverse into $\mathbb R$ (i.e. you can't even have a local bijection): Consider a compact subset $K$ of $\mathbb R^2$, then its image is ,as a subspace, Hausdorff, i.e.,$f|_K: \rightarrow f(K)$ is a continuous bijection between compact and Hausdorff, which is a homeomorphism . This is not possible by, e.g., Invariance of Domain. Since continuous local bijection is not possible, differentiable is not possible either.

user99680
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