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~
Best Answer
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.