# Can a function from R^n to R^n fail to be one-to-one, and also lack critical points? [closed]

I feel like there should either be a straightforward example of such a thing, or a homological reason why it can't exist. The reason why I ask is that if the Hessian matrix of a Lagrangian is non-singular then its dynamics are well defined, and (locally) you can pass to a Hamiltonian via the Legendre transformation, but it seems possible that two velocities (tangent vectors) could have the same momentum (cotangent vector, associated via the Legendre transform). In this case you'd have to formulate the Hamiltonian version on a cover of the cotangent bundle, instead of the cotangent bundle itself. Does this situation occur?

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$\arctan(x)$, for example. –  Ryan Budney May 17 '12 at 5:40
Or $e^x$. Homology and Morse theory hardly seem relevant. –  Angelo May 17 '12 at 7:24
A proper generic smooth map $\Bbb R^n\to\Bbb R^n$ without critical points must be one-to-one (injective) and onto (surjective). –  Sergey Melikhov May 17 '12 at 8:33
$f(x,y) = (e^x \cos y, e^x \sin y)$?