# One-to-One in R2

1. Aug 13, 2008

### linearfish

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. Aug 13, 2008

### HallsofIvy

Staff Emeritus
Think about $y_2= y_1+ 2\pi$.

3. Aug 13, 2008

### linearfish

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?

4. Aug 13, 2008

### konthelion

(1) and (2) looks fine to me.