Finding a Left Inverse for a Cylinder: Proving Injectivity of a Parametrization

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Homework Statement


Let ##S ## be a cylinder defined by ##x^2 + y^2 = 1##, and given a parametrization ##f(x,y) = \left( \frac{x}{ \sqrt{x^2 + y^2}}, \frac{y}{ \sqrt{x^2 + y^2} },\ln \left(x^2+y^2\right) \right)## , where ##f: U \subset \mathbb R^2 \rightarrow \mathbb R^3 ## and ## U = \mathbb R ^2 /{(0,0)}##

1. Find a left inverse of ##f##
2. Show that ##f## is injective.

Homework Equations

The Attempt at a Solution


I'm not even sure where to begin with this, My professor has done a very poor job of explaining how he wants us to go about these problems, and the book doesn't help much at all. The first thing that confuses me is that ##S## isn't a cylinder, it's a circle.

In class, we did inverses of simple functions which you could fairly easily solve, such as ##f(x,y) = (\sin xy, y)##, where we would just set it equal like this: ##y' = y## and ##\arcsin x' = xy##, then we would have ##f^{-1} (x',y') = (\frac{\arcsin (x'y')}{y'},y')##.

Also regarding the second question, I'm not aware of anyway to show that a function is injective unless its derivate is a square matrix, in which case you can take the determinant of the derivative and see where it's nonzero.
 
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Z90E532 said:

Homework Statement


Let ##S ## be a cylinder defined by ##x^2 + y^2 = 1##, and given a parametrization ##f(x,y) = \left( \frac{x}{ \sqrt{x^2 + y^2}}, \frac{y}{ \sqrt{x^2 + y^2} },\ln \left(x^2+y^2\right) \right)## , where ##f: U \subset \mathbb R^2 \rightarrow \mathbb R^3 ## and ## U = \mathbb R ^2 /{(0,0)}##

1. Find a left inverse of ##f##
2. Show that ##f## is injective.

Homework Equations

The Attempt at a Solution


I'm not even sure where to begin with this, My professor has done a very poor job of explaining how he wants us to go about these problems, and the book doesn't help much at all. The first thing that confuses me is that ##S## isn't a cylinder, it's a circle.
In R2, you are correct: the equation ##x^2 + y^2 = 1## represents a circle. In R3, though, the same equation represents a cylinder whose central axis lies along the z-axis. Your function f is a map from R2 to R3.
Z90E532 said:
In class, we did inverses of simple functions which you could fairly easily solve, such as ##f(x,y) = (\sin xy, y)##, where we would just set it equal like this: ##y' = y## and ##\arcsin x' = xy##, then we would have ##f^{-1} (x',y') = (\frac{\arcsin (x'y')}{y'},y')##.

Also regarding the second question, I'm not aware of anyway to show that a function is injective unless its derivative is a square matrix, in which case you can take the determinant of the derivative and see where it's nonzero.
What's the definition of a left inverse that you are using?
 
Mark44 said:
What's the definition of a left inverse that you are using?

The standard one: http://mathworld.wolfram.com/LeftInverse.html

I see what you're saying about ##S##, I didn't realize that at first.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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