# Local Diffeomorphism and diffeomorphism

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.

• A bijective local diffeomorphism is a diffeomorphism. So you need to show that if the domain and codomain are $\mathbb{R}$ injectivity is implied but an example in dimension 2 that isn't injective. Nov 12, 2021 at 18:42
• @podiki I got it that I have to prove that local diffeomorphism is injective. Do you have any ideas on how it can be proved?
– user775699
Jan 10, 2022 at 6:50
• If the function $f:\mathbb{R}\to\mathbb{R}$ is a local diffeomorphism, it means that the slope at each point is nonzero. Then the function is increasing (if $f'>0$) or decreasing (if $f'<0$). In either case it is injective. Jan 10, 2022 at 15:18
• @podiki Can you please also help with example in dimension 2 that isn't injective. I am unable to think about it.
– user775699
Jan 16, 2022 at 9:34

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 $$z$$.