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I want to show the following: Let $\pi:M\to N$ be a submersion. Then, every point of $M$ is in the image of a smooth local section of $\pi$.

Since $\pi$ is a submersion, it is also an immersion because for linear maps injectivity, surjectivity and bijectivity are equivalent. So $M$ and $N$ have the same dimension. Also, $\pi$ has constant rank $k$. Now take a $p\in M$. By the constant rank theorem there exist smooth coordinates $(x^1,...,x^k)$ centered at $p$ and $(v^1,...,v^k)$ centered at $\pi(p)$ in which $\pi$ has the coordinate representation $\pi(x^1,...,x^k)=(x^1,...,x^k)$, that is, $\pi$ is locally the identity map. So locally we can take its inverse as a local section which satisfies the proposition.

I'm not really convinced about the identity map part, is everything correct?

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  1. The theorem that says that injectiviy, surjectivity and bijectivity are equivalent is true only for endomorphisms : obviously you can have linear maps $u : E \rightarrow F$ between different spaces $E$ and $F$ which are injective but not surjective, or surjective but not injective. In particular there are (lots) of submersions which are not immersions.

  2. What you want is the submersion theorem : see http://en.wikipedia.org/wiki/Submersion_%28mathematics%29. The constant rank theorem is a generalization that is not needed here. What the submersion theorem says is, up to a suitable change of coordinates a submersion is locally like a linear projection $p : (x_1, \ldots, x_n, y_1, \ldots, y_m) \mapsto (x_1, \ldots, x_n)$ (which by the way is in general not injective). You should be able to construct the sections you want for such linear projections.

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