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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~

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    $\begingroup$ Looks good to me. $\endgroup$ – Dan Rust Jul 2 '13 at 9:14

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