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.

 A: 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. 
A: Hint: Since $f$ is a local diffeomorphism, $f$ is a local homeomorphism.  Use this fact to prove that $f$ is an open map.  By continuity, the image of $f$ is connected.  What is the only type of open, connected subset of $\mathbb{R}$?
Here's the proof that $f$ is an open map.  Let $U \subseteq \mathbb{R}$ be an open set.   Given $y \in f(U)$, there exists $x \in U$ such that $f(x) = y$. Take an open set $U_x$ around $x$ for which $f_{U_x}$ is a homeomorphism onto its image.  Since $U \cap U_x$ is an open subset of $U_x$, $f(U \cap U_x)$ is open in $f(U_x)$.  However, $f(U_x) \subseteq \mathbb{R}$ is open, and therefore, $f(U \cap U_x) \subseteq \mathbb{R}$ is open.  We have
$$y \in f(U \cap U_x) \subseteq f(U)$$
and so $f(U)$ is open.
For the second part, it suffices to show that the local diffeomorphism $f: \mathbb{R} \to (a, b)$ is a bijection.  We know it's surjective, so assume it's not $1-1$.  Then by Rolle's theorem (or MVT), there is some point $x \in \mathbb{R}$ such that $f'(x) = 0$.   However, the pushforward $df_x$ at $x$ must be a linear isomorphism.  This is a contradiction.
