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 ??? ----- —Dan Gene Salamin wrote: ----- On Saturday, June 8, 2019, 11:12:30 PM PDT, Dan Asimov <dasimov@earthlink.net> wrote: ----- Obviously a *necessary* condition is that P be locally one-to-one Q x Q —> Q. Which is just saying that for all (x,y) in Q^2, we have dP/dx * dP/dY ≠ 0 (i.e., not both derivatives are zero). ------------------------------------------------------------ The relation dP/dx * dP/dY ≠ 0 does not imply that P is locally one-to-one.. Counterexample: P(x,y) = xy. ---------------------------------------------------------------- Puzzle (tricky!): Find an entire function f(z) — i.e., f(z) is analytic on all of C — that is locally one-to-one everywhere without being globally one-to-one ... or else prove that such cannot exist. —Dan -------------------------------------------------————— f(z) = exp(z). --------------------------------------------------------------------- -----