How does $\mathbb{P^1}$ identify with $S^1$? Let $\mathbb{P^1}$ be the set of all lines $\mathbb{R^2}$  passing through the origin.
And by the definition of the unit circle $S^1 =  \left[0,1\right]/\sim$ , $\left[0,1\right]$ are identified.
What does it mean that  $\mathbb{P^1}$ identifies with $S^1$?
What I supposed is that by definition $ S^1 \subset \mathbb{P^1}$.
Am I missing the point? What does it mean by "identifies with"?
 A: Consider $C=\{(x,y):x^2+y^2=1, x\geq 0, \}$, it is the half-circle it is an interval. every line in $\mathbb{R}^2$ which is different of $\{y=0\}$ meets this half-circle in exactly one point. The line $\{y,0\}$ meets $C$ in $(-1,0)$ and $(1,0)$. This defines a bijection between the projective plane and the quotient of $C$ with $(1,0)$ and $(-1,0)$ identified which is the circle.
A: You already know by definition, $\Bbb RP^1$ is homeomorphic to $S^1/x\sim -x$ where $x\in S^1$.
Hint: 1. We have a commutative diagram
$$\require{AMScd} \begin{CD} D^1 @>i>> S^1\\ @Vf VV @VVgV\\ D^1/\sim @>k>> S^1/\sim \end{CD}$$
where $i$ is a natural inclusion map and $k$ is an induced map that makes the diagram commutes and $f,g$ are quotient maps. Here, $D^1/\sim$ is just an identification of two endpoints. Using the definition of quotient topology, show that $k$ is a continuous map.


*Recall the theorem: a continuous bijective map $f:X\to Y$ where $X$ is compact and $Y$ is Hausdorff is a homeomorphism.


*Conclude $D^1/\sim\ \cong\ S^1/\sim$.
Remark. Using the same argument, we can show that $D^n/\sim\ \cong\ S^n/\sim$
A: Take the line y=1  in $\mathbb{R^2}$. Then, clearly you can define a bijection between this line and the set of lines through the origin without x=0 (every such line cuts y=1). So, the whole set of lines through the origin is in bijection with $\mathbb{R}$ plus a point, which itself is one-to-one with the circle.
Now you can also show that this is not only a bijection but also an homeomorphism if you define a certain natural topology on the set of lines through the origin.
