Homeomorphism from $S^1\setminus(0,1)$ to $\mathbb{R}$

Question

I am trying to derive a bijection between $S^1\setminus\{(0,1)\}$ and the real line, but I am stuck on using the most obvious way

Let the top point of the circle be $(0,1)$, and the blue line hits some point $x = \begin{bmatrix} w \\h \end{bmatrix}$ on the circle and touches the real line at $f(x)$

Find a bijection between $S^1\setminus\{(0,1)\}$ and $\mathbb{R}$

Let $c$ be the length of the blue line, then by Pythagorean

$1^2+f(x)^2 = c^2$

The hypothenus of the small triangle $I$ is given by

$(f(x)-w)^2 + h^2 = c_1^2$

And the hypothenus of the small triangle $II$ is given by

$(1-h)^2 + w^2 = c_2^2$

$c = c_1 + c_2$

It seems proceeding this way I will have

$1+f^2(x) = (\sqrt{(f(x)-w)^2 + h^2} + \sqrt{(1-h)^2 + w^2})^2$

But I need to some how extract the $f(x)$ from the RHS...

Does anyone see how to proceed?

Answer 1

This is actually the right idea! It also generalizes nicely. Typically, this is called the Stereographic projection, and you can find proofs for the fact that it is a homeomorphism from any of the following sites:

http://www.math.ku.dk/~moller/e02/3gt/opg/S29.pdf (Lemma 2, which is a proof of a fact given by munkres ch. 29)

http://www.math.colostate.edu/~renzo/teaching/Topology10/Notes.pdf (example 1.8.5)

Theorem 59.3 in Munkres also proves that the stereographic projection is a homeomorphism.

It might help to define your map differently:

Let $f: S^1 \to \mathbb{R}$ be defined by $$f(x,y)=\frac{1}{1-y}(x)$$

Notice that this is really just an easier characterization of a "straight line" through some $(x_1,y_1) \in S^1-\{0,1\}$. For example ,we know the slope of the line, it's obvious: $\frac{1-y_1}{-x_1}$, and the $y-intercept$ is $1$, so we have a line $y=\frac{1-y_1}{-x_1}x+1$, but we want it to intersect the $x-axis$, so we set $y=0$ and then obtain that $x=\frac{x_1}{1-y_1}$, which is exactly our map.

More generally: $$f(x_1,...,x_{n+1})=\frac{1}{1-x_{n+1}}(x_1,...x_n)$$ gives a stereographic projection.

This should be a much easier function to prove is a homeomorphism, but the idea is exactly the same.

Answer 2

I'm sorry to say I don't have any great ides for solving your equation for $f(x)$. You could try getting the radical that involves $f(x)$ by itself and squaring both sides but it'll be pretty awful, I would assume. Thankfully there is a significantly easier approach utilizing more geometry, less algebra.

Small triangle II, with vertices $(0,1),\, (w, h)$, and $(0,h)$, is similar to the "overall" triangle with vertices $(0, 1),\, (f(x), 0)$, and $(0, 0)$.

Thus the ratios of heights and widths are the same; $$\frac{h}{1} = \frac{w}{f(x)},$$

and you can easily solve for $f(x)$.