Full question is: Define a map $\psi:S^1\setminus\{(0,1)\}\to\mathbb{R}$ as for each point $p\in S^1$ take the line through $(0,1)$ and $p$ and define $\psi(p)$ as the intersection of this line and the $x$-axis.
So I believe I want $\psi$ to be the map $(x,y)\to \frac{x}{1-y}$, I took the line $y=\frac{y-1}{x} +1$, and found the $x$ value when $y=0$.
So now I need to show that $\psi$ is a homeomorphism. I believe after showing it is bijective, showing it is continuous is simply a matter of showing $S^1\setminus\{(0,1)\}$ is open in $S^1$. Then using that the pre image of open sets is open, thus continuous. However I'm having some problems showing this function is injective and surjective.
Injective:
Suppose $f(x,y)=f(r,s)$ then $\frac{x}{1-y}=\frac{r}{1-s}$
then $x(1-s)-r(1-y)=0$. Somehow I want to show that either $x\not=r$ or $y\not=s$ but I'm not seeing how to do this. I know that $x^2+y^2=1$ and $r^2+s^2=1$.
Best Answer
Hint:
$$t\overset{\varphi}\mapsto{\displaystyle \left({\frac {2t}{1+t^{2}}},{\frac {t^2-1}{1+t^{2}}}\right)}$$
is the inverse of your map from $\mathbb{R} \to S^1\setminus\{(0,1)\}$. This will prove that your initial map is both injective and surjective. Note that at $x=\infty$, we get $\varphi(\infty)=(0,1)$ but $\infty\not\in\mathbb{R}$ of course.
If you want to see how these two maps are obtained, assuming you don't know it already, just fix a line at $(0,1)$ and change the slope to sweep $\mathbb{R}$. This will give a (bi-continuous) one-to-one correspondence between $S^1\setminus\{(0,1)\}$ and $\mathbb{R}$ as long as the slope of the line doesn't become parallel to the $x$-axis. And it never becomes parallel to the $x$-axis unless the line is tangent to the circle at $(0,1)$.
Meanwhile, note that just proving your function $\psi$ is continuous is not enough. You should also show that its inverse is continuous. In this case, $\varphi$ is given by two polynomials whose denominators never become zero. So, $\varphi$ is continuous on $\mathbb{R}$.
Addendum
More precisely, if you wonder how $\varphi$ is found, first write the equation of the line passing through $(0,1)$ with the varying slope $t$
$$y-1=t(x-0) \implies y=tx+1$$
You want to see how $t$ determines a point on the circle. So, you have to find its intersection with the unit circle. So, your intersection point must satisfy $x^2+y^2=1$. This gives
$$x^2+(tx+1)^2=1$$
Now you've found an equation which is quadratic in $t$. So, you'll find two solutions. One solution is $t=0$ which corresponds to the point $(0,1)$ where our line has been fixed to the circle, and the other solution gives us the point that we're looking for.