The image of $I$ is an open interval and maps $\mathbb{R}$ diffeomorphically onto $\mathbb{R}$? This is Problem 3 in Guillemin & Pallock's Differential Topology on Page 18. So that means I just started and am struggling with the beginning. So I would be expecting a less involved proof:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a local diffeomorphism. Prove that the image of $f$ is an open interval and maps $\mathbb{R}$ diffeomorphically onto this interval.

I am rather confused with this question. So just identity works as $f$ right?


*

*The derivative of identity is still identity, it is non-singular at any point. So it is a local diffeomorphism.

*$I$ maps $\mathbb{R} \rightarrow \mathbb{R}$, and the latter is open.

*$I$ is smooth and bijective, its inverse $I$ is also smooth. Hence it maps diffeomorphically.

Thanks for @Zev Chonoles's comment. Now I realized what I am asked to prove, though still at lost on how.


If $f$ is a local differeomorphism then the image must be connected, try to classify the connected subsets of $\mathbb{R}$ into four categories. Since $f$ is an open map, this gives you only one option left. I do not know if this is the proof the author has in mind. 
