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In order to show that $\mathbb{R}P^n$ has the cellular structure $e^0 \cup \ldots \cup e^n$ we have the characteristic map, we take $u \in D^n$ and define $$ \Phi(u) = [u_1,\ldots,u_n,\sqrt{1-\| u \| ^2}]$$

What I really want to see is that $\Phi|D^n-S^{n-1}$ maps homeomorphically onto $\mathbb{R}P^n - \mathbb{R}P^{n-1}$. Unfortunately my geometric intuition of what $\mathbb{R}P^n - \mathbb{R}P^{n-1}$ is, is falling down.

So if we restrict the function $\Phi$ to $D^n - S^{n-1}$ we are restricting it to the interior of the $n$-disk and hence $\| u \| < 1$, then under our map we have an element in $\mathbb{R}P^{n}$ with its $n+1$ co-ordinate non-zero. (Actually I think I see how this works now).

If we consider $\mathbb{R}^n$ to be the set of all real sequences $(x_1,x_2,\ldots,x_n)$ such that $x_i=0$ for $i>n$ then $\mathbb{R}^n \subset \mathbb{R}^{n-1}$. As a result $S^{n-1} \subset S^{n}$; in fact $S^{n-1}$ is the intersection of $S^{n}$ with the plane $x_{n+1}=0$. Now under the equivalence relation $x \sim -x$ is the same in $S^{n-1}$ as it is in $S^n$, therefore $\mathbb{R} P^{n-1} \subset \mathbb{R} P^n$. Thus when we have an element of $\mathbb{R}P^n$ with the $n+1$ coordinate non-zero we have that its complement is the embedded copy of $\mathbb{R}P^{n+1}$, and thus the map $\Phi$ maps $D^n - S^{n-1}$ homeomorphically onto $\mathbb{R}P^n - \mathbb{R}P^{n-1}$

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You did not exactly ask a question! – Mariano Suárez-Alvarez Mar 23 '11 at 1:40
@Mariano - while I was typing it kind of made more sense - I was then just hoping my ramblings were (somewhat) correct! :) – Juan S Mar 23 '11 at 1:45
up vote 1 down vote accepted

$P^n-P^{n-1}$ can be seen as the set of lines in $\mathbb R^{n+1}$ which are not parallel to the coordinate hyperplane $\pi_0=\{x\in\mathbb R^{n+1}:x_{n+1}=0\}$. Such a line intersects the hyperplane $\pi_1=\{x\in\mathbb R^{n+1}:x_{n+1}=1\}$ in exactly one point, so there is a map $f:P^n-P^{n-1}\to\pi_1$. It is easy to see that this is an homeomorphism.

Similarly, each of the lines in $P^n-P^{n-1}$ intersects the northern open hemisphere $S_+^n=\{x\in\mathbb R^{n+1}:\lVert x\rVert=1, x_{n+1}>0\}$ in exactly one point, and this determines a map $g:P^n-P^{n-1}\to S_+^n$ which, again, is easily seen to be an homeo. This is your map.

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