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Let $(x,y)\in\mathbb R^2$ and $M$ a manifold defined by $M=\left\{ (x,y)\in\mathbb R^2\, |\, y^2+x=0 \right\}$. Let $\pi$ be a projection $\pi(x,y)=(x)$. Let $\phi:\mathbb R\to\mathbb R$ be a diffeomorphism.

Let me write $\tilde\pi=\pi_M$, then $\tilde\pi(x,y)=-y^2$. Note that $\tilde\pi$ is degenerate at $0$.

My question is: let $\Phi=\tilde\pi^{-1}\circ\phi\circ\tilde\pi$. It is clear that $\Phi|_{M\backslash\left\{ 0 \right\}}$ is a diffeomorphism. Does a smooth (or continuous) extension $\Phi'$ of $\Phi$ exists? Does this necessarily implies that $\Phi'(0,0)=(0,0)$?

I thought I could use the "smooth extension theorem" but $M\backslash \left\{0 \right\}$ is not closed. Any reference to useful reading is appreciated.

PepeToro
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  • How did you define $\tilde{\pi}^{-1}$? – DiegoMath Jul 29 '14 at 14:35
  • I guess $\tilde\pi^{−1}(x)=(x,\pm\sqrt{-x})=(-y^2,\pm y)$ is the correct way of taking $\tilde\pi^{-1}$, but since I know which $y$ we are taking since the beginning, we could set $\tilde\pi^{-1}(y)=(-y^2,y)$. – PepeToro Jul 29 '14 at 16:43

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