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Let $M$ and $N$ be smooth manifolds and $\pi:M\rightarrow N$ be a smooth map. A local section of $\pi$ is a a smooth map $\sigma:U\rightarrow M$ defined on some open subset $U\subseteq N$ such that $\pi\circ\sigma=id_U$.

Local section theorem says the following :

Suppose $M$ and $N$ are smooth manifolds and $\pi:M\rightarrow N$ is a smooth map. Then, $\pi$ is a smooth submersion if and only if every point of $M$ is in the image of a smooth local section of $\pi$.

I was confused by the fact that if any point is in some image of smooth local section of $\pi$ then $\pi$ is a smooth submersion. I think this fact implies that a locally diffeomorphic smooth map $F:M\to N$ is automatically a smooth submersion because we can always for each point $p$ with its neighborhood $U$ s.t. $F(U)$ open in $N$ and $F|_U:U\to F(U)$ is a diffeomorphism, and then we may define $\sigma=(F|_U)^{-1}:F(U)\to U\subset M$, which is the smooth local section desired. Then by the local section theorem 4.26, we should have $F$ be a smooth submersion.

What's wrong with my argument?

Anthony
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    It is indeed the case that a (surjective) local diffeomorphism is a submersion, and this is indeed a proof – Exit path Mar 20 '23 at 05:16
  • Thanks for the comment. I think you're right. I ask the above question because proposition 4.8 which is emphasized to be limited in the case where the domain and codomain manifold are both without boundary. However, after reviewing the proof, I see that the limitation results from a use of 4.5 in only one of the direction. So, smooth local diffeo indeed directly gives smooth submersion and immersion. – Anthony Mar 20 '23 at 06:08

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