# Projective geometry well defined bijection

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?

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$\mathbb R\mathbb P^n$ is $\mathbb R^{n+1}$ modulo scalar multiplication and $\pm1$ are among the scalars. –  Hagen von Eitzen Jan 6 '13 at 21:53
Thanks I just noticed that $[\lambda x_0:...:\lambda x_n]$ lies in the unit sphere iff $|\lambda|=1$ but I still do not see the bijection. What is meant with the projective lines in this image? –  Montaigne Jan 6 '13 at 21:57
Is the problem with "well-defined" or with "bijection"? All parts follow quickly from the definition of injective/surjective and just picking elements. If two points on the sphere map to the same class, then they differ by a scalar multiple, which you've pointed out must be $\pm 1$ and hence they are the same class in $S^n/\sim$. Surjective: pick a class, then by rescaling any representative you can make it on the sphere (of unit length). –  Matt Jan 7 '13 at 1:04
Thanks, may you could explain the surjective part a little bit more in detail, I know exactly what you mean but do not write do write it down formally. Do you also have an idea what is meant with projective lines in this image? –  Montaigne Jan 7 '13 at 13:45
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 $(x_0, \ldots , x_n)$.