Differential Topology: 1-dimensional manifold

[lmedin02]

Homework Statement

Given S¹={(x,y) in R²: x²+y²=1}. Show that S¹ is a 1-dimensional manifold.

Homework Equations

The Attempt at a Solution

Let f₁:(-1,1)->S¹ s.t. f₁(x)=(x,(1-x²)^1/2).

This mapping is a diffeomorphism from (-1,1) onto the top half of the circle S¹.
I was trying to write a prove that f₁ is indeed a diffeomorphism, but I am having trouble showing the onto part.

I argued that f₁ is onto by defining the inverse map and showing that the inverse map is 1 to 1 and hence f₁ is onto.

 

[ystael]

This mapping is a diffeomorphism from (-1,1) onto the top half of the circle S¹.
I was trying to write a prove that f₁ is indeed a diffeomorphism, but I am having trouble showing the onto part.

To show that a map \phi: A \to B is onto (surjective), you need only show that for every b\in B, there is a \in A such that \phi(a) = b. So, for each point (x, y) on the top half of S^1 (be careful about your definition of "top half"), just exhibit a point \xi \in (-1, 1) with f_1(\xi) = (x, y). Your definition makes this easy.

I argued that f₁ is onto by defining the inverse map and showing that the inverse map is 1 to 1 and hence f₁ is onto.

This is confused. Again take a map \phi: A \to B; the existence of an inverse map \phi^{-1}: B \to A presupposes that \phi is bijective. Remember that a function is always defined over its entire domain; the existence of \phi^{-1}(b) means that there is some a such that \phi(a) = b.

You should expect the harder part of this to be showing that f_1 and its inverse are smooth, not showing that f_1 is bijective.

 

[lmedin02]

To show that a map \phi: A \to B is onto (surjective), you need only show that for every b\in B, there is a \in A such that \phi(a) = b. So, for each point (x, y) on the top half of S^1 (be careful about your definition of "top half"), just exhibit a point \xi \in (-1, 1) with f_1(\xi) = (x, y). Your definition makes this easy.

Ok, let (x,y) \in S^1 choose \xi \in(-1,1) s.t. f_1(\xi) = (x, y). This implies \xi =x.

By the top half of the circle I am excluding the end points at -1 and 1.

To show that f_1 is smooth I need to show that it has continuous partial derivatives of all orders. So I computed the Jacobian matrix of f_1 and observe that it is continuous in the domain (-1,1).

What I don't know how to prove is why all the partial derivatives exists and are continuous? I observed that all I need to prove is that the function \sqrt{1-x^2} is infinitely differentiable on (-1,1).

 

[ystael]

The composition of smooth functions is smooth; that should simplify your job.

 

[lmedin02]

So, suppose I decompose \phi (x)=\sqrt{1-x^2} into two functions, namely, g(x)=1-x^2 which is definitely smooth everywhere and h(x)=\sqrt{x} which is not smooth at the origin. But, I notice that it is difficult to decompose \phi (x) into two smooth functions. Can I conclude that \phi (x) is not smooth if it cannot be written as a decomposition of smooth functions.

 

[ystael]

Look more closely; for your purposes, does it matter that the square root function is not smooth at the origin?

 

[lmedin02]

I see what you mean. The range of 1-x^2 is (0,1) on the domain (-1,1). Thank you.

 

[ystael]

Actually (0, 1], but you got the idea.