Smooth Mapping Between Unit Circle and Curve in R^2?

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In R^2, the unit circle defined by x^2 + y^2 = 1 can be smoothly mapped to the curve x^4 + y^2 = 1. One proposed mapping involves using the functions (sqrt(x), y) for non-negative x and (sqrt(-x), y) for negative x, but concerns arise about smoothness at x=0. An alternative approach suggests projecting along a direction that avoids tangency to the circle, which may help maintain smoothness. The discussion also references Riemann's mapping theorem, hinting at the possibility of an analytic map. Overall, the smoothness of the mapping remains a key point of inquiry, particularly at the origin.
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Hi, I have been told that in R^2 the unit circle {(x,y) | x^2 + y^2 = 1} is smoothly mappable to the curve {(x,y) | x^4 + y^2 = 1}.

Can someone please tell me what this smooth map is between them? I can only think of using the map (x,y) --> (sqrt(x), y) if x is non-negative and (sqrt(-x), y) if x is negative. Thanks for any help.
 
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Both x^2+ y^2= 1 and x^4+ y^2= 1 loop around the origin. Draw the line from the origin through a point on the circle. Where that ray crosses the second graph is s(x,y).
 
is there a smoothness problem at x=0? (in answer #1)
 
The advantage of Halls' answer seems to be that he is projecting along a direction that never becomes tangent to the circle. I.e. #1 projects horizontally, and #2 projects radially. Still it is not so trivial for me to prove #2 is actually smooth, as the equation I am getting for r is still undefined at x=0, although it seems to extend.

An abstract approach is Riemann's mapping theorem, with extension to the boundary, that apparently gives an analytic map.
 
HallsofIvy: Thanks, that's a nice bijection. I clearly need to think more geometrically for this type of problem.

mathwonk: I don't think there is a problem at x=0 (for Halls' map) because you can just define r to be 1 for x=0 and then it is smooth on S1.
 
well you have prove it is smooth.
 

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