Real Projective Plane in 3 different way While reading the book Basic Topology by M.A. Armstrong , there are three interpretation of Real Projective Plane as follows:
(a) Take the unit sphere $S^n$ in $E^{n+1}$ and partition it into subsets which contain exactly two points, the points being antipodal (at opposite ends of a diameter). $P^n$ is the resulting identification space. We could abbreviate our description by saying that $P^n$ is formed from $S^n$ by identifying antipodal points.
(b) Begin with $E^{n+1}-\{0\}$ and identify two points if and only if they lie on the same straight line through the origin. (Note that antipodal points of $S^n$ have this property.)
(c) Begin with the unit ball $B^n$ and identify antipodal points of its boundary sphere.
But I can not visualize why (a) and (c) are same. Please some one help me to get an intuition of it.
 A: Let's take the definition $P^n=S^n/[z\sim -z]$ as in $(a)$ and work towards $(c)$. We'll use homogeneous coordinates to display the points of $P^n$, so $[z_1,\dots,z_n]\in P^n$ represents the equivalence class of $z=(z_1,\dots,z_n)\in S^n$.
To start we'll use radial coordinates to view $B^n$ as the cone $(S^{n-1}\times I)/(S^{n-1}\times\{0\})$. The boundary sphere is then the subspace $S^{n-1}\times\{0\}$. With this understood we define
$$\widetilde\varphi:B^n\rightarrow P^n$$
to be the map $\widetilde\varphi(z,t)=[t\cdot z,\sqrt{1-t^2}]$, where the positive square root is taken in the last coordinate.
This map is well-defined and moreover satisfies $\widetilde\varphi(z,1)=\widetilde\varphi(-z,1)$ for all $z\in S^{n-1}$. Thus if we let $Q^n$ be the quotient space of $B^n$ obtained by identifying antipodal points of $S^{n-1}$, then there is an induced map
$$\varphi:Q^n\rightarrow P^n.$$
The last thing to check is that this map is a homeomorphism, which I'll leave up to you. The point is that given $[z_1,\dots,z_n]\in P^n$, either $z_n=0$, or $z_n\neq 0$ and there is a unique representative of $[z_1,\dots,z_n]$ with $z_n>0$. This was the reason for specifying above that the positive square root be taken. With this it's not difficult to see that $\varphi$ is a continuous bijective. Writing down a continuous inverse is not difficult. Alternatively, $Q^n$ is compact, so if you are happy that $P^n$ is Hausdorff, then having a continuous bijection is sufficient.
