I consider the sphere $\mathbb S^n:=\{x\in\mathbb R^{n+1}: \|x\|=1 \}$ and the equivalence relation $x\sim y:\Leftrightarrow x=\pm y$.
How can it be shown that the inclusion $\mathbb S^n\rightarrow\mathbb R^{n+1}$ induces a well-defined bijection $\mathbb S^n/\sim\rightarrow\mathbb {RP}^n$ ?
If I could show that I would get the identification $\mathbb{ RP}^2=\mathbb S^2/\sim$, i.e the points of the projective plane $\mathbb {RP}^2$ can be identified with the antipodal points on $\mathbb S^2$.
My second question is, how can the projective lines in this image be described?
As requested. To prove surjectivity, suppose we have an arbitrary element $[x_0: \cdots : x_n]\in \mathbb{RP}^n$. We must find some element $y\in S^n/\sim$ that maps to it.
Since some $x_i\neq 0$, we have $\|(x_0, \ldots, x_n)\|\neq 0$ and hence $\displaystyle y=\frac{(x_0, \ldots, x_n)}{\|(x_0, \ldots, x_n)\|}\in S^n$.
The map in question is including this into $\mathbb{R}^{n+1}$ followed by the quotient map, i.e. $y\mapsto [y]$. But now $[y]=[x_0: \cdots : x_n]$ because $y$ is just a scalar multiple of $(x_0, \ldots , x_n)$.
There are several equivalent ways of describing/thinking about lines. I noticed Lines in projective space off to the right, and Georges answer is exactly how I'd describe it.