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I saw here (A surjective map which is not a submersion) that a smooth differentiable map $f:M\to N$ between two manifolds $M$ and $N$ is not necessarily a submersion. A counterexample is $f:\mathbb{R}\to\mathbb{R}$, $x\to x^3$, then at $x=0$ its differential is not surjective. Is it possible to make general extra assumptions on the map $f$ such that it is a submersion, and if so, what are they?

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    $\begingroup$ What could they possibly be? Certainly you can't make further topological conditions; a homeomorphism is as good as a map can be. You could say "the map admits local smooth sections", but that's straightforwardly equivalent to being a submersion. $\endgroup$ – user98602 May 20 '16 at 14:37
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If $f$ is a surjective map of constant rank, then it is a submersion. Let's prove this. The main engine is the rank theorem.

$F:M \to N$ is a smooth map with constant rank $r$. $dim(M) = m$ and $dim(N) = n$. Then, given $p \in M$, we can choose charts around $p$ and $F(p)$ in which $F$ has the coordinate representation $(x_1, ..., x_r, x_{r+1}, ..., x_m) \to (x_1, ..,x_r, 0,..,0)$ if $m \leq n$.

The proof I know uses contradiction, but I would like to see a different proof if anyone knows one. Anyway, here goes the one I know. Suppose $F: M \to N$ as above, and is surjective, and it is has constant rank $r < n$.

Then, given a point $p \in M$, we can put coordinate charts $U$ and $V$ around $p$ and $F(p)$ , respectively, such that we have the form $(x_1,..., x_n) \to (x_1, ..., x_r, 0,..,0)$. AS a manifold, $M$ has a basis of pre compact coordinate balls. So, we can assume $U$ is some pre compact coordinate ball, since it certainly contains such a ball containing $p$. We can also assume that $F(\overline{U}) \subseteq V$. Now, $F(\overline{U})$ is identified with a subset of $\{(y_1,...,y_r,0,...,0)| y_1,...,y_r \in \mathbb{R} \}$ in $\mathbb{R}^n$, and so $F(\overline{U})$ clearly contains no open subset. If it did, then the image of this set in $\mathbb{R}^n$ would have to be open, and so could not be a subset of that set of points. Thus, $F(\overline{U})$ is closed, as the image of a compact set, and has empty interior, since it contains no open set. Thus, is nowhere dense.

$M$ has a countable basis of such coordinate balls, and so we can obtain a countable cover of $M$, $\{ \overline{U_{a_n}} \}$ with $F(\overline{U_{a_n}})$ nowhere dense.

So, then $M = \bigcup \overline{U_{a_n}}$ gives $F(M) = F(\bigcup \overline{U_{a_n}}) = \bigcup F(\overline{U_{a_n}})$. This is the union of countably many nowhere dense sets, and as such it has empty interior, and so, in particular, is nowhere dense. So, $F$ was not surjective in the first place.

I think one could probably actually give a shorter proof using Sard's theorem. If $F$ had constant rank and is not a submersion, then every point in $F$ is a critical point. So, every value of $F$ is a critical value. So, $F(M)$ has measure $0$ in $N$, and consequently $F$ is not surjective.

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  • $\begingroup$ This is lovely. It appears as Theorem 4.14 in the second edition of Lee's smooth manifolds book. $\endgroup$ – Arrow Apr 27 at 9:42

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