2

Here's the question : A smooth bijective map of manifolds need not be a diffeomorphism. In fact, show that $$f:\mathbb{R^1}\rightarrow {R^1}$$ $$x\rightarrow f(x)=x^3,$$ is an example.

I would like to do this problem, but I'm really not sure I understand it. The question does mean that there are smooth bijective map of manifold without as the map is a diffeomorphism. Could someone explain to me the meaning of this question?

  • 3
    Do you understand the words used? Do you have trouble proving that $f$ is a bijection, that it is smooth, that it is not a diffeomorphism? – Najib Idrissi Apr 27 '15 at 19:43
  • Yes, all are well known to me. It is only the meaning of the question that I do not understand. –  Apr 27 '15 at 19:47
  • 4
    You need to prove that $f$ is smooth, $f$ is bijective, but $f$ is not a diffeomorphism. – Najib Idrissi Apr 27 '15 at 19:48
  • I am a Spanish. That may be why I do not understand the meaning of the question. –  Apr 27 '15 at 19:48
  • @J.G Guillemin & Pollack define a diffeomorphism from $X$ to $Y$ as a map $f: X \rightarrow Y$ which is a bijection, is smooth, and whose inverse is smooth. This is more than just being a smooth bijection, as it also includes the bit with "whose inverse is smooth." (To clear up the language issue: smooth bijection just means a map which is smooth and is a bijection; it doesn't mean it's smooth "as a bijection" in some way that requires the inverse to be smooth.) – aes Apr 27 '15 at 23:02

1 Answers1

-1

The important fact here is that you need to take a closer look at the inverse $f^{-1}$.

The inverse of $x^3$ is $\sqrt[3]{x}$ the differential of the inverse $\frac{d}{dx}f^{-1}=\frac{1}{3 \sqrt[3]{x^2}}$ outside of $0$ and nonexistent at $0$ therefore $f$ is no diffeomorphism since the inverse is not differentiable at $0$.

ThorbenK
  • 1,465