# Given a smooth map which is open, is it a submersion?

A submersion between smooth manifolds is an open map. Is the converse true? That is, is a smooth open map $f:M\to N$ between smooth manifolds a submersion? We can additionally assume that it is surjective, if necessary, because that is the only case I am interested in.

I can see how it is almost true, taking a chart $U$ about $m\in M$ and considering a chart $V$ about $f(m)\in N$ contained in the image of $U$ (which is open). Using the isomorphism between the charts and the tangent spaces, I'm fairly sure this gives us a submersion, but I feel there is a slight gap in my argument. So either, does my argument work, or is it true but my argument is incomplete, or is it false?

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What about $x \mapsto x^3, \mathbb{R} \to \mathbb{R}$? –  Akhil Mathew Mar 2 '11 at 22:47
As Akhil mentions, the answer is no. And it's "almost always no" in the sense that one can homotope any submersion into an open map which is not a submersion -- the model is the straight-line homotopy between $x \longmapsto x$ and Akhil's map. –  Ryan Budney Mar 2 '11 at 22:52
@Akhil, @Ryan: Why not post an answer? –  Jonas Meyer Mar 2 '11 at 23:06
@Ryan - but this just says that there are non-open maps in every path component of the mapping space that contains a submersion, not the density or otherwise of submersions. (it's a bit cheeky, I know) –  David Roberts Mar 2 '11 at 23:30

A nonconstant holomorphic function is an open map on $\mathbb{C}\cong\mathbb{R^2}$, but is a submersion only if its derivative is never zero. So for example, $z\mapsto z^2$, a.k.a. $(x,y)\mapsto (x^2-y^2,2xy)$ is a counterexample.