Let φ:ℝ→ℝ, φ(x)=x^3. Which conclusion is correct?

It is a homeomorphism but not a diffeomorphism, because φ^-1(x)=√[3]x is not differentiable at 0

It is a diffeomorphism, because its Jacobian 3x^2 is always positive

It is a diffeomorphism, because it is bijective and smooth

It is an isometry, because it preserves distances

Exercise · ZW-SMOOTH-DIFFEOMORPHIC-DEFORMATION-011 · ★★★ Advanced

Let \(\varphi:\mathbb{R}\to\mathbb{R}\), \(\varphi(x)=x^3\). Which conclusion is correct?

Choose one option.

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C. It is a homeomorphism but not a diffeomorphism, because \(\varphi^{-1}(x)=\sqrt[3]{x}\) is not differentiable at 0

The inverse \(\sqrt[3]{x}\) is not differentiable at 0, and \(D\varphi(0)=0\) makes the Jacobian vanish; so it is a homeomorphism but not a diffeomorphism.

A smooth diffeomorphic deformation \(\varphi\) is a bijective, smooth transformation whose inverse is also smooth.

\[\varphi^{-1}\text{ is smooth}\]