Can a function from R^n to R^n fail to be one-to-one, and also lack critical points? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-26T00:07:32Z http://mathoverflow.net/feeds/question/97190 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97190/can-a-function-from-rn-to-rn-fail-to-be-one-to-one-and-also-lack-critical-poin Can a function from R^n to R^n fail to be one-to-one, and also lack critical points? Lawrence D'Anna 2012-05-17T05:12:32Z 2012-05-17T05:35:32Z <p>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?</p> http://mathoverflow.net/questions/97190/can-a-function-from-rn-to-rn-fail-to-be-one-to-one-and-also-lack-critical-poin/97192#97192 Answer by Nik Weaver for Can a function from R^n to R^n fail to be one-to-one, and also lack critical points? Nik Weaver 2012-05-17T05:35:32Z 2012-05-17T05:35:32Z <p>$f(x,y) = (e^x \cos y, e^x \sin y)$?</p>