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I'm trying to understand the proof of the theorem (given in the link) that states "Stereographic projection maps circles of the unit sphere, which do not contain the north pole, to circles in the complex plane."

Link to the proof

In the proof it states "In order to obtain an equation for the projection points (x, y) ∈ C of the circle c under stereographic projection, we substitute (1) into Equation (2), which yields"

Why does plugging in the pre image of points from the image plane into an arbitrary plane give me an equation of the points under stereographic projection? What is the significance of using the pre image?

This is an idea for your use:

enter image description here

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Remember, all transformations have two interpretations, a passive and and active one.

He's looking to express the equation for the plane, which is given in coordinates of $\Sigma$ (that is, $x,\,y,\,h$), in terms of coordinates of the complex $2d$ plane (that is, $x,y$).

Right before he states that $R(x,y)$ is precisely the function that does such a thing, namely, expressing a point from $\Sigma$ in terms of $x,y$.

In doing so, the equation of the plane, which involves $x,y,h$ becomes an equation for just the two coordinates of the plane $x,y$. This allows him to check what kind of geometric object we have within $\mathbb{C}$.

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  • $\begingroup$ Overall I have a better understanding but I'm trying to make sense of your first sentence. Is the passive transformation moving the plane to the sphere? Also he eventually gets the equation of a circle but how do I know this equation describes the equation of the projected points and not some random circle on the image plane? $\endgroup$ – Kevin Nov 20 '15 at 17:14
  • $\begingroup$ $R(x,y)$ could be understood as function that maps a point in the plane to a point in the sphere. But at the same time, it is expressing a point of the sphere with only coordinates of a point in the plane. That's what I meant here. As to your second question: it's the reasoning that tells you this. Let's see: You have a circle on the sphere, which is defined by the slicing plane intersecting the latter. You express the plane in terms of coordinates of the plane $x,y$ and finally you see these last coordinates do define a circle. $\endgroup$ – MASL Nov 20 '15 at 18:47
  • $\begingroup$ More graphically I guess you could say he looks for points in the plane that map the circle on the sphere, and only then he realizes those points describe as well a circle on the plane. $\endgroup$ – MASL Nov 20 '15 at 18:47

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