Is there an example of a smooth map between smooth manifolds which is surjective, but not a submersion?

I feel there can't be one, but don't know of a proof. Nor do I know of a counter-example. Kindly help!

up vote 8 down vote accepted

Yes! How about $f:\mathbb{R}\to\mathbb{R}$, $f(x) = x^3$? At $x=0$, $df = f'(x) = 3x^2$ is not surjective. (In general, you can perturb a submersion to have points where the differential does not have full rank.)

Edited in response to your question below: Sard's theorem guarantees that critical values of $f$ will be of measure zero. However, critical points of $f$ need not have measure zero. For example, the smooth map $$f(x) = \begin{cases} (x-2)e^{-(x-2)^{-2}}, & 2\leq x \\ 0, & -2<x<2 \\ -(x+2)e^{-(x+2)^{-2}}, & x\leq -2 \end{cases}$$ has a critical value at $0$, but its critical point set is $[-2,2]$.

As another (stupid) example, take a disconnected manifold and project it onto one of its components by sending the other component to a point. For example, $S^1\sqcup S^1\to S^1$ by $id\sqcup *$.

  • Thanks. Will the set of such points be finite, or of measure zero? – whydoiask Apr 18 '13 at 11:56
  • I want to say Sard guarantees measure zero. – Neal Apr 18 '13 at 12:10
  • 1
    @SurajKrishna Sard's theorem (en.wikipedia.org/wiki/Sard's_theorem) guarantees that the critical values of $f$ will be of measure zero. However, the critical points need not have measure zero. For example, "stretch" the example I gave so that the preimage of $0$ is a whole interval. – Neal Apr 18 '13 at 14:31

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