Equivalent definition of the real projective space This is the definition of the real projective space in John Lee's book. However what I know is that the real projective space is defined by the quotient space of $S^{n+1}$ by identifying the antipodes. How is the two definitions equivalent? I tried to use the definition of quotient space but cannot find a way to prove the equivalence rigorously... Could anyone please help me?

Consider the inclusion map $S^n \hookrightarrow \mathbb{R}^{n+1}$. This is of course continuous, and thus the composition $$\begin{array}{ccccccccc} S^n & \xrightarrow{i} & \mathbb{R}^{n+1} \backslash \{0\} &\\ & \searrow{f} & \downarrow{\pi} \\ & & \mathbb{R}P^n \end{array}$$ is continuous. Note that $f(x)=f(-x)$. Thus, $f$ induces a map $F$ in the quotient by $$\begin{array}{ccccccccc} S^n & \xrightarrow{i} & \mathbb{R}^{n+1}\backslash \{0\} &\\ \downarrow{\pi'} & \searrow{f} & \downarrow{\pi} \\ S^n/\sim & \xrightarrow{F} & \mathbb{R}P^n .\end{array}$$ We have that $F$ is continuous. Note that it is also a bijection. Since $S^n/\sim$ is compact (it is the image by $\pi'$ of $S^n$, which is compact) and $\mathbb{R}P^n$ is Hausdorff (the book may have a proof of that), it is a homeomorphism.
The quotient $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{RP}^n$ can be broken up into two quotients $\mathbb{R}^{n+1} \setminus \{0\} \to S^{n} \to \mathbb{RP}^n$. The first sends $x$ to $\frac{x}{|x|}$ (considering $S^{n}$ as the unit circle embedded in $\mathbb{R}^{n+1}$); the second identifies the antipodes. The composition of these quotients identifies each linear subspace with one point and is precisely $\pi$ as described in that excerpt.
• Why is the map $x/norm(x)$ a quotient map? – Keith Jan 9 '18 at 14:50
• @Keith It identifies all $x$ with equal $\frac{x}{|x|}$ to a point; i.e., it identifies each half-line emanating from the origin to a point. – BallBoy Jan 9 '18 at 14:52
• Actually this is the part I am stuck at. I think that $x/norm(x)$ is an open surjective map, therefore a quotient map. But I cannot rigorously prove why the map is open. Could you please help me? – Keith Jan 9 '18 at 14:52
• @Keith Consider an open ball centered at $x$. Projecting on $\mathbb{R}^2$ formed by the vector $x$ and any orthogonal dimension, we have an open circle, whose image under the map will be open curve contained between the two tangents from the origin to the circle. Considering all orthogonal dimensions, we get an open ball in $S^n$. – BallBoy Jan 9 '18 at 15:00