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

• Z90E532
In summary: I think I might have been misunderstanding what he meant when he said "circle." In any case, it doesn't really matter, because we can solve the equation for ##x## regardless.The equation for a circle in R2 is just ##x^2 + y^2 = 1##, so we can solve for ##x## just like we would any other equation. However, because ##S## is a cylinder, its equation in R3 is a bit more complicated. We need to take the equation of the circle and multiply it by the height of the cylinder, which is just ##x^3 + y^3 = 1##. So, our equation for ##x## in R3 would be something like this:
Z90E532

## 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.

## 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.

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.

## 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.

## 1. What is a left inverse?

A left inverse is a value or set of values that can be multiplied with a given value or set of values to produce the identity element, or 1. In other words, it undoes the operation of multiplication by returning the original value or set of values.

## 2. Why is finding a left inverse important?

Finding a left inverse is important in mathematics and science because it allows for the solving of equations and the manipulation of values in a way that preserves the original information. It also allows for the simplification of complex expressions.

## 3. How do you find a left inverse?

To find a left inverse, you need to first identify the operation that is being used (such as addition, multiplication, etc.), and then use inverse operations to "undo" the original operation. For example, if the operation is addition, the left inverse would be the subtraction of the original value.

## 4. Can there be more than one left inverse?

Yes, there can be more than one left inverse for a given value or set of values. This is because different operations can produce the same result when multiplied with the original value. For example, in addition, both 3 and -3 are left inverses of 0.

## 5. What are the applications of finding a left inverse?

The concept of a left inverse has many applications in various fields of science and mathematics. It is commonly used in linear algebra, cryptography, and signal processing, among others. It also plays a crucial role in proving the existence of inverse functions in calculus.

• Calculus and Beyond Homework Help
Replies
5
Views
486
• Calculus and Beyond Homework Help
Replies
4
Views
883
• Calculus and Beyond Homework Help
Replies
6
Views
803
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
400
• Calculus and Beyond Homework Help
Replies
3
Views
946
• Calculus and Beyond Homework Help
Replies
3
Views
716
• Calculus and Beyond Homework Help
Replies
5
Views
841
• Calculus and Beyond Homework Help
Replies
2
Views
720
• Calculus and Beyond Homework Help
Replies
5
Views
894