Why is the set of all lines through the origin homeomorphic to the circle S1?

[jacobrhcp]

Homework Statement

Prove P1 is homeomorphic to S1

The Attempt at a Solution

P1 is the set of all lines through the origin. I have shown for myself that this is homeomorphic to S1/R, where R is given by: (x,y) in R iff x=y or x=-y.
This also shows there is a continuous surjective map f from S1 to S1/R. To prove S1 and S1/R are homeomorphic I need to show that f is injective, and that it has continuous inverse.

However, I think there is something fundamentally wrong with my understanding of the problem, because I cannot imagine the map f being injective. To me it feels as though S1/R is the same as the 'northern hemisphere' of S1, because it sends -x and x to the same line through the origin (where x is an element of the real plane on the circle S1). So wouldn't that imply that f(x)=f(-x) while x is not equal to -x, so f is not injective, so P1 is not homeomorphic to S1?

Anyone who can tell me why I'm fundamentally off here, and give me a push in the right direction, thank you.

Jacob

 

[futurebird]

Think of the origin as a point at the top (north pole?) of the circle. As the lines pass through the origin they project on to the circle like a stereographic projection.

 

[jacobrhcp]

Ah I was taking the centre as origin... it confused me deeply.

But then again, isn't S1={x in the real plane | ||x||=1}... so why do we get to say the origin is on the top of the circle instead of in the middle?

This new view will probably help me take a better shot at the problem later this evening.