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Let $a > 0$, and set $B_a = \{x \in \mathbb{R}^n : |x|^2 < a \}$. Let $\phi : B_a \to \mathbb{R}^n$ be given by $\phi(x) = \frac{ax}{\sqrt{a^2 - |x|^2}}$. Prove that $\phi$ is a diffeomorphism of $B_a$ onto $\mathbb{R}^n$.

I am stucked in proofing that the above function is 1-1 and onto because in the definition of the function the scalar part depends on the norm of x as a vector, so if I equate 2 functions say one of them of y and the other of x there will be a difference in the scalar value on both sides so what shall I do?

related links: Show that the map is a diffeomorphism

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Hint: Show that the map $\psi : \Bbb R^n \to B_a$ defined by the equation $$\psi(y) = \frac{ay}{\sqrt{a^2 + \lvert y\rvert^2}}$$ is the inverse of $\phi$.

kobe
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  • Is that a prove for showing 1-1 & onto ..... how? –  Sep 10 '18 at 20:53
  • Yes. Since $\psi(\phi(x)) = x$ for all $x\in B_a$, $\phi$ is one-to-one; since $\phi(\psi(y)) = y$ for all $y\in \Bbb R^n$, $\phi$ is onto. – kobe Sep 10 '18 at 20:55