1. ## Projection

Proposition: Let $\displaystyle S$ be a circle on $\displaystyle \displaystyle \Sigma$ and let $\displaystyle T$ be its projection on $\displaystyle \mathbb{C}$. Then (a) if $\displaystyle S$ contains $\displaystyle (0,0,1)$, $\displaystyle T$ is a line; (b) if $\displaystyle S$ does not contain $\displaystyle (0,0,1)$, $\displaystyle T$ is a circle.

If $\displaystyle S$ contains $\displaystyle (0,0,1)$ then the circle in not entirely contained on the sphere. Is this an intuitive reason why the projection is a line? Whereas a circle that does not contain $\displaystyle (0,0,1)$ is a circle completely contained on the sphere. Thus its projection is a circle.

2. Originally Posted by manjohn12
Proposition: Let $\displaystyle S$ be a circle on $\displaystyle \displaystyle \Sigma$ and let $\displaystyle T$ be its projection on $\displaystyle \mathbb{C}$. Then (a) if $\displaystyle S$ contains $\displaystyle (0,0,1)$, $\displaystyle T$ is a line; (b) if $\displaystyle S$ does not contain $\displaystyle (0,0,1)$, $\displaystyle T$ is a circle.

If $\displaystyle S$ contains $\displaystyle (0,0,1)$ then the circle in not entirely contained on the sphere. Is this an intuitive reason why the projection is a line? Whereas a circle that does not contain $\displaystyle (0,0,1)$ is a circle completely contained on the sphere. Thus its projection is a circle.
I cannot think of a great intuitive explanation why lines and circles go to lines and circles, I think you need to do some algebraic calculations. However, this is how I remember it. If a circle passes through (0,0,1) on $\displaystyle \Sigma$ then it passes through the point at infinity. Thus, when you map it onto $\displaystyle \mathbb{C}$ (i.e. project) it means the image becomes unbounded. Since circle are bounded it cannot be a circle. It has to be a line then. And if a circle is on $\displaystyle \Sigma$ and not through (0,0,1) it means it stays away from the point at infinity, therefore it stays bounded. Thus, such a circle gets projected back onto a circle.

For intuition you might want to think in 2d. Imagine a circle sitting on top of the real number line. Can you visualize this?

3. Yes. E.g. the circle is homeomorphic to $\displaystyle [0,1]$. I was thinking that since $\displaystyle (0,0,1)$ is the point at infinity, a circle that contains it also contains points not on the sphere. The part that is contained on the sphere can be "stretched" out into a line on $\displaystyle \mathbb{C}$.

Or maybe think of lines as infinitely long circles.