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?
Best Answer
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.