If a mapping is bijective and regular, then the mapping is a diffeomorphism?

Let me ask a question which appears in the book 'Elementary Differential Geometry' written by O'Neil.

The questions is: prove that if a one-to-one and onto mapping $f:\Bbb R ^n \to \Bbb R ^n$ is regular, then it is diffeomorphism.

In the book, "$f$ is regular" means that the tangent map of $f$ is one to one. I think certainly I need to use the inverse function thorem. $f$ is a mapping so $f$ is in the class $\mathcal C ^1$. Since $f$ is regular and in the class $\mathcal C^1$, its Jacobian is invertible, so the derivative of $f$ is invertible.

Therefore, I might apply the inverse function theorem to $f$: there exists an open set in $\Bbb R ^n$ in which the inverse of $f$ exists and the inverse belongs to the class $\mathcal C ^1$ so $f$ is a diffeomorphism. But to prove $f$ is a diffeomorphism, shouldn't I show the inverse of $f$ is in the class $\mathcal C ^1$ for every point in $\Bbb R ^n$?

I would really appreciate any help. Thank you for reading.

• When applying the inverse mapping theorem, you only get a local diffeomorphism, not a global one! – Alex M. Jun 1 '15 at 12:45

Because $f$ is bijective, it has an inverse. Your inverse function theorem argument shows $f^{-1}$ is "locally $C^{1}$" at an arbitrary point (though you might want to supply a bit more detail to show the "arbitrary point" part). But being $C^{1}$ is a local condition, so you've actually shown $f^{-1}$ is $C^{1}$.

On a tangent, the fact that $f$ maps $\mathbf{R}^{n}$ to $\mathbf{R}^{n}$ (a space of the same dimension) is crucial: It's not generally true that a regular injection is a diffeomorphism onto its image. The plane curve $c:(-\pi, \pi) \to \mathbf{R}^{2}$ defined by $c(t) = (\sin 2t, \sin t)$ is regular and injective, but its image is a figure-8, which is not diffeomorphic to an interval.

• Thanks, so the problem is just find the local diffeomorphism...got it.. Let me just ask one thing. Is injectivity and surjecticity necessary? Doesn't the inverse ftn thm guarantee the existence of the local inverse without bijectivity? – Mathcho Jun 1 '15 at 14:24
• In fact, i just showed the mapping is a local diffeomorphism...isn't it different from the diffeomorphism? – Mathcho Jun 1 '15 at 14:30
• The inverse function theorem does guarantee local bijectivity, but (as Alex M's comment hints) there exist local diffeomorphisms, such as the complex exponential map, that aren't injective. In your situation, however, the regular map $f$ has a global inverse by hypothesis, and $f^{-1}$ is regular by your inverse function theorem argument; this means $f$ is a diffeomorphism. :) – Andrew D. Hwang Jun 1 '15 at 14:35
• Aha! I think i got the whole picture now. At any point of R^n the local inverses are the same(what you call the global inverse)since if f is a bijection, the inverse is unique....so this fact and my argument prove the mapping is actually a global diffeomorphism...thanks a lot! I truly appreciate your help :) – Mathcho Jun 1 '15 at 14:50