Wait. I don't see why an onto and locally one-to-one continuous function f : C —> C needs to be a covering map (even assuming it's analytic). In fact it doesn't have to be. An example by Lawrence Zalcman defined by f(z) = Integral exp(z^2) dz (integrating over any curve from 0 to z), is such an example. —Dan ————— * Proof: f'(z) = exp(z^2) is never 0, so f is locally one-to-one. Also f(0) = 0 and f(-z) = f(z), so if any nonzero value c were omitted, then -c must also be omitted — contradiction since an entire function cannot omit two values (Picard's theorem). Hence f is onto. But it's not a covering map C —> C since the analytic covering maps C —> C are only the functions of form az + b as Andy mentions. Andy Latto wrote: ----- On Sun, Jun 9, 2019, 14:08 Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote: The only analytic bijections of the Riemann sphere are the homographic functions w=(az+b)/(cz+d). If the function is analytic on the finite complex plane, it cannot have a pole, so c=0. Thus the only such functions are linear, w=az+b. This shows impossibility under the added assumption that the function can be extended from the complex plane to the Riemann sphere, but that additional assumption is not needed. F is a covering map on the plane. Since the plane is simply connected, it can't have more than one sheet. Andy ----Gene On Sunday, June 9, 2019, 10:52:44 AM PDT, Dan Asimov <dasimov@earthlink.net> wrote: Now assume further that the entire function f : C —> C is onto. (Unlike exp(z), which omits 0.) Fun question: ----- Does there exist an analytic function f : C —> C that is a) locally one-to-one (equivalently, f'(z) is nowhere zero), and b) onto ??? ----- -----