What is the line going through points $(5, 5, 5), (2, 2, 2) \in \mathbb{R^3}$ when mapped it is mapped to a point in the real projective plane? So the real projective plane is homeomorphic under a function $f$ to $\mathbb{R^3} - (0, 0 ,0)$.
Hence lines in $\mathbb{R^3} - (0, 0 ,0)$ become points in the the real projective plane.
So what is the line going through points $(5, 5, 5), (2, 2, 2) \in \mathbb{R^3}$ when mapped it is mapped to a point in the real projective plane?
Is there numerous possiblites for the value of this point? Ie. are there numerous functions $f$ that can map the line to different points?
If that is the case, what would be considered the most 'natural' mapping $f$?
 A: The real projective plane can be defined as the set of line going through the origin in $\mathbb{R}^3$ and there is a map $f\colon\mathbb{R}^3\setminus\{0\}\to\mathbb{R}P^2$ which maps a point to the corresponding point in $\mathbb{R}P^2$ which is given by the unique line through the origin on which that point lies.
The line going through the points $(5,5,5)$ and $(2,2,2)$ happens to go through the origin (it is equal to $\{(\lambda,\lambda,\lambda)\mid \lambda\in\mathbb{R}\}$ which contains $0$) and so there is a point in $\mathbb{R}P^2$ which correspond to this line. Because they lie on the same line, you will also note that $f(5,5,5)=f(2,2,2)$. The map $f$ is continuous and is in fact also a quotient map. Because of this, it means we could define $\mathbb{R}P^2$ instead as a quotient of $\mathbb{R}^3\setminus \{0\}$ by an equivalence relation $\sim$ given by $(a,b,c)\sim(a',b',c')$ if and only if there exists a $\lambda\in\mathbb{R}\setminus 0$ such that $(a',b',c')=(\lambda a,\lambda b,\lambda c)$. Show that $x\sim y \iff f(x)=f(y)$.
If we then consider the map $\tilde{f}\colon \mathbb{R}^3\setminus\{0\}/{\sim}\to\mathbb{R}P^2$ given by $\tilde{f}([(a,b,c)]_{\sim})=f(a,b,c)$ then this map is a homeomorphism.
