# ’Stereographic projection’

This is problem 1.12 in Armstrong's Basic Topology:

’Stereographic projection’ $\pi$ from the sphere minus the north pole to the plane.

Work out a formula for $\pi$ and check that $\pi$ is a homeomorpism. Notice that $\pi$ provides us with a homeomorphism from the sphere with the north and south poles removed to the plane minus the origin.

I am struggling to even get started on this one.

Can anyone give me a hint about the first step i should take to find this projection?

• Could you please explain where you're stuck? (It would also be appropriate in the future to cite where your statements come from, particularly if they're taken verbatim from a published source.) – Andrew D. Hwang Jan 24 '16 at 21:16
• Yes, i will do that, and i clearly explained where i was stuck. I wanted some help to get started, which i got. I have solved it now btw, in the comments below the answer. – JKnecht Jan 24 '16 at 22:57
• I can't speak for the down-voter or those who voted to close, but 1. The question reads "Work out formulas...", then states those formulas. 2. Those formulas clearly define a continuous mapping from the sphere minus two poles to the punctured plane and back, and 3. are easily shown to be inverse mappings. Overall, it wasn't clear to me where you needed help, or even what remained to be done. Anyway, glad to hear you got things worked out. – Andrew D. Hwang Jan 25 '16 at 0:27
• @Andrew D. Hwang The question is taken word for word from Armstrong's Basic Topologoy. An here we have five people saying its missing context or other details. I wanted help to get started and 1. added the solution + 2. added the sentence about 2 dimension which i found online. The reasons i added these two things was to help the people who might give an answer. – JKnecht Jan 25 '16 at 9:27

Given a point $x$ on $S^n$, use linear algebra to work out the line between the north pole $(1,0,\dots,0)$ and $x$, and then find the projection of that line onto a given plane $\{(x_1, \dots, x_n): x_1 = y\}$, like in the image below:
• Yes, thx i solved it. Following the notation they use suppose we project to $\bf{x}$$=(x,y,z). Then a line from the north pole (0,0,1) through \bf{x} is l(t)=(tx, ty, 1+t(z-1)). And this line hits the plane when 1+t(z-1) = 0 or t = 1/(z-1). The projection then becomes \pi (X,Y,Z)=\frac{1}{1-z} *(x, y) – JKnecht Jan 24 '16 at 22:52 • This is clearly continuous when z \neq 1, and has a continuous inverse. To calculate the inverse \pi^{-1} i use \|\bf{x}\|^2$$=1$ and $x = (1-z)X$, $y=(1-z)Y$ and using pythagoras theorem and a little algebra i get $$\pi^{-1}(x, y, z) = (\frac{2X}{1 + X^2 + Y^2}, \frac{2Y}{1 + X^2 + Y^2}, \frac{−1 + X^2 + Y^2}{1 + X^2 + Y^2})$$ – JKnecht Jan 24 '16 at 22:52