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Consider the Inverse Function Theorem in $\mathbb{R}$:

Let $O ∈\mathbb{R}$ be open for $f:O → \mathbb{R}$. If $f$ is continuously differentiable, and for a $x_0 ∈O$, $f'\left(x_0\right) \ne 0$. Then there exist an open interval $\boldsymbol{I}$ about $x_0$ and open interval $\boldsymbol{J}$ about the image of $f\left(x_0\right)$ such that $ f: \boldsymbol{I} → \boldsymbol{J}$ is one to one and onto. Futhermore, $f^{-1}: \boldsymbol{J} → \boldsymbol{I}$ is continuously differentiable, and if $y ∈\boldsymbol{J}$, $x ∈\boldsymbol{I}$ such that $f\left(x\right) = y$, then \begin{split} (f^{-1})'(y) = \frac{1}{f'(x)} \end{split}

Compared with the Inverse Function Theorem in $\mathbb{R^n}$:

Let $O ∈\mathbb{R^n}$ be open for $\boldsymbol{F}:O → \mathbb{R^n}$. If $\boldsymbol{F}$ is continuously differentiable, and for a $\boldsymbol{x_*} ∈O$, $\boldsymbol{DF}\left(\boldsymbol{x_*}\right)$ is invertible. Then there exist a neighborhood $\boldsymbol{I}$ about $\boldsymbol{x_*}$ and an neighborhood $\boldsymbol{J}$ about the image of $\boldsymbol{F}\left(\boldsymbol{x_*}\right)$ such that $ \boldsymbol{F}: \boldsymbol{I} → \boldsymbol{J}$ is one to one and onto. Futhermore, $\boldsymbol{F}^{-1}: \boldsymbol{J} → \boldsymbol{I}$ is continuously differentiable, and if $\boldsymbol{y} ∈\boldsymbol{J}$, $\boldsymbol{x} ∈\boldsymbol{I}$ such that $\boldsymbol{F}\left(\boldsymbol{x}\right) = \boldsymbol{y}$, then \begin{split} \boldsymbol{DF}^{-1}(\boldsymbol{y}) = [\boldsymbol{DF}(\boldsymbol{x})]^{-1} \end{split}

I noticed that for the Inverse Function Theorem in $\mathbb{R^n}$, it is possible for all $\boldsymbol{x} ∈O$ $(O ∈\mathbb{R^n})$ to satisfy the requirements of the theorem, yet still have $\boldsymbol{F}: O → \mathbb{R^n}$ be not one to one. For example: \begin{equation} \boldsymbol{F}(x,y) = \left(e^x\cos \left(y\right),e^x\sin \left(y\right)\right) \end{equation} $\boldsymbol{F}$ is continuously differentiable, and for any $\boldsymbol{x} ∈\mathbb{R^n}$, the determinant of $\boldsymbol{DF}\left(\boldsymbol{x}\right)$ does not equal to $0$. Yet, $\boldsymbol{F}$ is not one to one (e.g. the points $\left(0,0\right)$ and $\left(0,2π\right)$ produce the same result). This, I presume, is due to the fact that the Inverse Function Theorem is only true for the functon $\boldsymbol{F}$ on its restrictions (from $\boldsymbol{I}$ to $\boldsymbol{J}$).

Question:

1) Could this also happen for the Inverse Function Theorem in $\mathbb{R}$? That is, could $f: O → \mathbb{R}$ be continuously differentiable, and $f'\left(x\right) \ne 0$ for all $x ∈\mathbb{R}$, yet still have $f: O → \mathbb{R}$ be not one to one? If no, why are they different?

2) Could this also happen to the onto property of the function?

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To be explicit, I am taking continuous differentiability here to mean that the the function is continuous, and has a continuous first derivative.

If $f^\prime (x) \neq 0$ for all $x\in O$ by continuity it must be the case that $f^\prime (x)>0$ for all $x\in O$ or $f^\prime (x) <0$ for all $x\in O$.

In either case, $f$ is strictly monotonic and hence must be one-to-one.

This function may not be onto. Consider any affine function defined on an open interval that is not the entire real line.

I don't have a good explanation for why the difference, but my suspicion is that, in multiple dimensions, you can have saddle/inflection points in some dimensions but still have the Jacobian be invertible. However, this is decidedly not the case in one variable.

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  • $\begingroup$ That doesn't satisfy the requirement that $f'(x) \ne 0$ for all $x$ $\endgroup$ – bli00 Dec 12 '16 at 3:25
  • $\begingroup$ Yes, of course. Was being sloppy. Apologies. $\endgroup$ – Theoretical Economist Dec 12 '16 at 3:26
  • $\begingroup$ @thestateofmay Answer edited. $\endgroup$ – Theoretical Economist Dec 12 '16 at 3:38

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