Local Diffeomorphism and diffeomorphism

Question

The following question was part of my quiz on smooth manifolds and I couldn't solve it. I tried again at home but in vain.

A smooth map $f: M\to N$ between manifolds is said to be a local diffeomorphism if, around each point $p \in M$ there exists a nbd $U_p$ of p such that $U_p$ is diffeomorphic to its image under f.

(a) Prove that any local diffeomorphism $f: \mathbb{R} \to \mathbb{R}$ is diffeomorphism onto its image.

(b) Give an example to show that this fact is not true if we consider a smooth map from $\mathbb{R}^2 $ to $\mathbb{R}^2$.

So, it is given that for every $p\in M$ there exists a nbd $U_p$ of p such that $U_p$ is diffe. to its image which implies: f is $C^1$ , $f(U_p)$ is open, $f^{-1} $ is $C^1$ but there might be for point p' a nbd $U_p'$ where a different diffeomorphism f' exists. But I have to prove the existence of a single function (Say F) and I am not getting any intuition.

So, can you please help?

Answer 1

As podiki commented, a local diffeomorphism $f : \mathbb R \to \mathbb R$ has the property $f'(x) \ne 0$ for all $x$. Since $f'$ is continuous for smooth $f$, we either have $f'(x) > 0$ for all $x$ or $f'(x) < 0$ for all $x$. Thus $f$ is either strictly increasing or strictly decreasing, hence $f$ maps $\mathbb R$ bijectively onto its image $J = f(\mathbb R)$ which is open in $\mathbb R$. By the inverse function theorem $f^{-1} : J \to \mathbb R$ is smooth.

Now consider the exponential map $\exp : \mathbb C \to \mathbb C, \exp(z) = e^z$, which is holomorphic. We can regard it as map $\mathbb R^2 \to \mathbb R^2$, and this map is smooth with Jacobian having determinant $\ne 0$ in all points. It is therefore a local diffeomorphism. However, it is not injective because $\exp(z) = \exp(z + 2\pi i)$ for all $z$.