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 $r^2+s^2=1$.

  • 1
    $\begingroup$ Create an inverse function instead of proving injectivity and surjectivity. It should be geometrically clear what the inverse should be. Also, you can think about Pythagorean triples, Weierstrass trigonometric substitutions, etc. $\endgroup$ – stressed out Mar 1 at 4:05


$$t\overset{\varphi}\mapsto{\displaystyle \left({\frac {2t}{1+t^{2}}},{\frac {t^2-1}{1+t^{2}}}\right)}$$

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 $\mathbb{R}$.


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


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.

  • $\begingroup$ So I just need to show that $\psi \circ f=I$? Where $f$ is your function and $I$ is the identity. $\endgroup$ – AColoredReptile Mar 1 at 4:15
  • $\begingroup$ @AcoloredReptile Exactly. Because a function is bijective if and only if it's invertible. Also, $\varphi$ is obviously continuous. If you prove injectivity and surjectivity separately, you will still need to show that the inverse of your function is continuous too. Otherwise, you haven't proved that it's a homeomorphism. And to be more precisely, you have to show that $\varphi\circ\psi = 1$ and $\psi\circ\varphi = 1$. $\endgroup$ – stressed out Mar 1 at 4:18

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