Local diffeomorphism is diffeomorphism onto image provided one-to-one. For the problem Guillemin & Pallock's Differential Topology 1.3.5, I am not confident with my proof.

Prove that a local diffeomorphism $f: X \rightarrow Y$ is actually a diffeomorphism of $X$ onto an open subset of $Y$, provided that $f$ is one-to-one.

Proof: First show that the local diffeomorphism $f: X \to$ image$f \subseteq Y$ is a bijection. We know it's surjective since we restricted to its image. Given $f$ is one-to-one, it is bijective.
We know local diffeomorphisms are open maps from the proof of 1.3.3: Let $N = f(X)$. By assumption we have a bijective local diffeomorphism $f: X \to N$.   To prove that $f$ is smooth let $x \in X$. There exists an open set $U \subseteq X$ around $x$ such that $f_U : U \to f(U)$ is a diffeomorphism.  Hence, there exists charts $(U_x, \phi)$ $(U_{f(x)}, \psi)$ of $x$ and $f(x)$ such that $f(U_x) \subseteq f(U_{f(x)})$ and 
$$\psi \circ f \circ \phi^{-1} : \phi(U_x) \to \psi(U_{f(x)})$$
is smooth as a map between Euclidean space.
According to the definition of smooth maps between smooth manifolds, if $f$ is a map from an $m$-manifold $M$ to an $n$-manifold $N$, then $f$ is smooth if, for every $p \in M$, there is a chart $(U, \phi)$ in $M$ containing $p$ and a chart $(V, \phi)$ in $N$ containing $f(p)$ with $f(U) \subset V$, such that is smooth from $\phi(U)$ to $\psi(V)$ as a function from $\mathbb{R}^m$ to $\mathbb{R}^n$.
Then we consider $f^{-1}: N \to X$. To prove that $f^{-1}$ is smooth let $y \in N$. There exists an open set $V \subseteq N$ around $y$ such that $f^{-1}_y : V \to f^{-1}(V)$ is a diffeomorphism.  Hence, there exists charts $(V_y, \phi^\prime)$ $(f^{-1}(V_y), \psi^\prime)$ of $y$ and $f^{-1}(y)$ such that
$$\psi^\prime \circ f^{-1} \circ \phi^{\prime-1} : \phi^\prime(V_y) \to \psi^\prime(f^{-1}(V_y))$$
is smooth as a map between Euclidean space.
Therefore, $f$ and $f^{-1}$ are smooth, $f$ is bijective, and hence $f$ is a diffeomorphism.
Thank you~
 A: Looks good. Some comments:

*

*I think you need to say like $N$ is a smooth (regular/embedded) submanifold because $N$ is an open submanifold because $N$ is open because $f$ is open because $f$ is a local diffeomorphism. This is the way $N$ is a smooth manifold by itself. I mean if you just have $N$ as immersed submanifold, then what would $f: X \to N$ even mean?


*The way that $N$ is a smooth manifold in (1) preceding is relevant for (3) as follows.


*Proving $f: X \to N$ I think is a wheel reinvention. Surely there should be some rule in the book that says restriction of range of a smooth map to any submanifold of range that contains image is also smooth (or at least that restriction of range to image is smooth).


*A shortcut to showing $f^{-1}: N \to X$ is smooth: This is true if and only if $f$ is an immersion. See here. (I think also true if and only if $f$ is a submersion. See here.) Finally, local diffeomorphism at $p \in X$ is equivalent to both-immersion at $p$-and-submersion at $p$

Actually, you can prove something stronger.
This says: Injective local diffeomorphism only if (smooth) embedding.
So what condition on embedding allows to have an if? Open. Actually:
Injective local diffeomorphism if and only if open embedding. I ask and answer here: Is open (topological) smooth embedding equivalent to injective local (homeomorphism) diffeomorphism?
You can also observe and then ask other things:

*

*Injective local diffeomorphism only if open immersion. So what condition on open immersion allows to have an if? Topological embedding.


*Open embedding only if local diffeomorphism. So what condition on local diffeomorphism allows us to have an if? Injective.
