# Submersion implies every point is in the image of a local section

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?

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.