Differential Topology: 1-dimensional manifold

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Homework Statement


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


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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.
 

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  • #2
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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 [tex]\phi: A \to B[/tex] is onto (surjective), you need only show that for every [tex]b\in B[/tex], there is [tex]a \in A[/tex] such that [tex]\phi(a) = b[/tex]. So, for each point [tex](x, y)[/tex] on the top half of [tex]S^1[/tex] (be careful about your definition of "top half"), just exhibit a point [tex]\xi \in (-1, 1)[/tex] with [tex]f_1(\xi) = (x, y)[/tex]. 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 [tex]\phi: A \to B[/tex]; the existence of an inverse map [tex]\phi^{-1}: B \to A[/tex] presupposes that [tex]\phi[/tex] is bijective. Remember that a function is always defined over its entire domain; the existence of [tex]\phi^{-1}(b)[/tex] means that there is some [tex]a[/tex] such that [tex]\phi(a) = b[/tex].

You should expect the harder part of this to be showing that [tex]f_1[/tex] and its inverse are smooth, not showing that [tex]f_1[/tex] is bijective.
 
  • #3
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To show that a map [tex]\phi: A \to B[/tex] is onto (surjective), you need only show that for every [tex]b\in B[/tex], there is [tex]a \in A[/tex] such that [tex]\phi(a) = b[/tex]. So, for each point [tex](x, y)[/tex] on the top half of [tex]S^1[/tex] (be careful about your definition of "top half"), just exhibit a point [tex]\xi \in (-1, 1)[/tex] with [tex]f_1(\xi) = (x, y)[/tex]. Your definition makes this easy.

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

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

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

What I dont 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 [tex]\sqrt{1-x^2}[/tex] is infinitely differentiable on [tex](-1,1)[/tex].
 
  • #4
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The composition of smooth functions is smooth; that should simplify your job.
 
  • #5
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So, suppose I decompose [tex]\phi (x)=\sqrt{1-x^2}[/tex] into two functions, namely, [tex]g(x)=1-x^2[/tex] which is definitely smooth everywhere and [tex]h(x)=\sqrt{x}[/tex] which is not smooth at the origin. But, I notice that it is difficult to decompose [tex]\phi (x)[/tex] into two smooth functions. Can I conclude that [tex]\phi (x)[/tex] is not smooth if it cannot be written as a decomposition of smooth functions.
 
  • #6
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Look more closely; for your purposes, does it matter that the square root function is not smooth at the origin?
 
  • #7
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I see what you mean. The range of [tex]1-x^2[/tex] is [tex](0,1)[/tex] on the domain [tex](-1,1)[/tex]. Thank you.
 
  • #8
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Actually [tex](0, 1][/tex], but you got the idea.
 

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