# Differential Topology: 1-dimensional manifold

## Homework Statement

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

## 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.

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.

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 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 $$\sqrt{1-x^2}$$ is infinitely differentiable on $$(-1,1)$$.

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

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.

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

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

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