TITLE: A surjective map which is not a submersion
QUESTION [13 upvotes]: 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!

REPLY [14 votes]: 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 $id\sqcup *$.