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One-to-One in R2

  1. Aug 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Let f = (f1,f2) be the mapping of R2 into R2 given by f1=excos(y), f2=exsin(y).
    (1) What is the range of f?
    (2) Show that every point of R2 has a neighborhood in which f is one-to-one.
    (3) Show that f is not one-to-one on R2.

    3. The attempt at a solution
    (1) Since ex is nonzero and cos and sin cannot be 0 at the same time, then the origin is not in the range, so the range is R2\{0,0}.

    (2) Calculating the Jacobian we get:
    J = e2xcos2y + e2xsin2y= e2x, which is nonzero, so every point has a neighborhood in which f is one-to-one.

    (3) This is where I'm really stuck. I assume we need to find a point that is mapped by two different ordered pairs but I don't know how to go about finding those points.

    Any help on #3 or comments on the previous parts is appreciated. Thanks.
     
  2. jcsd
  3. Aug 13, 2008 #2

    HallsofIvy

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    Think about [itex]y_2= y_1+ 2\pi[/itex].
     
  4. Aug 13, 2008 #3
    Oh, duh. I guess it didn't occur to me to hold x constant, so the points (0,pi) and (0,3*pi) both map to (-1,0), hence f is not one-to-one. Thanks.

    Does the rest look okay?
     
  5. Aug 13, 2008 #4
    (1) and (2) looks fine to me.
     
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