# Does there exist a diffeomorphism on $\mathbb{R^2}$ that flattens out the boundary of a compact set at a point?

Given a compact subset of $\mathbb{R^2}$ with ${C^2}$ boundary $S$ and a point $x \in S$, can one find a diffeomorphism $f$ from $\mathbb{R^2}$ to $\mathbb{R^2}$ for which $f(x) = x$, the image $f(S)$ is a ${C^2}$ curve and such that, in a neighborhood of $f(x)$, the curve $f(S)$ coincides with the tangent line of $x$ in $S$?

I think this is possible but still have no idea how to construct such a function $f$.

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Isn't this a standard application of the implicit function theorem? –  Branimir Ćaćić Feb 19 '13 at 5:14
Thanks for editing and suggesting answers. But I don't see how the implicit function theorem can be applied directly. Here, I need to adjust the curve $S$ so that it's flattened at a point $x$. –  Songkiat Sumetkijakan Feb 19 '13 at 8:29

## 2 Answers

If by a diffeomorphism you mean a smooth map with smooth inverse the answer is no in general. If you can find such a diffeomorphism it would follow that $S$ is actually smooth in a neighborhood of $x$ not just $\mathcal C^2$ as you have constructed a smooth local submanifold chart of $S$ near $x$.

If otherwise you are satisfied with a $\mathcal C^2$ map having a $\mathcal C^2$ inverse the answer is yes: You may without loss of generality assume that $x = 0$ and choose some $\epsilon > 0$ such that $B_\epsilon(0) \cap S$ is a graph of some $\mathcal C^2$ function $f$. Choose a $\mathcal C^2$ function $g$ which coincides with $f$ on some intervall $]-\delta,\delta[$ and consider its graph $\{(g(y),y) \in \mathbb R^2 | y \in \mathbb R\}$. Then you may define your desired function via $\phi(x,y) = (g(y) + r,y) \mapsto (r,y)$. That this is indeed a $\mathcal C^2$ map with $\mathcal C^2$ inverse is clear since $h(x,y) := (g(y) + x,y)$ defines such a map and $\phi = h^{-1}$.

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By a diffeomorphism, I mean a function $f$ for which both $f$ and $f^{-1}$ are differentiable (can be approximated by a linear map).

I think your answer is going in a right direction. It's just that the function $\phi$ might affect the boundary on the other side(s). We will have to define $\phi$ to be identity map outside a region, possibly $B_\epsilon(0)$, and choose a function $r$ (instead of a constant $r$) so that $\phi$ is still $C^2$. Thank you.

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i do not understand; what do you mean by 'side', what do you want to keep fixed? –  wspin Feb 20 '13 at 16:19