Let $$\mathbb{P}^n = \{l \ \mid \ l \text{ is a line through the origin in } \mathbb{R}^{n+1} \}$$ denote the $n$-dimensional real projective space. What is an explicit quotient map from $D^n$ to $\mathbb{P}^n$, if we define the equivalence relation $\sim$ on $D^n$ as follows: $x \sim y$ if and only if $x = y$ or $x,y \in S^{n-1}$ and $x = \pm y$ $-$ so we identify the antipodal points on $S^{n-1}$ (here $D^n$ is the closed unit disk in $\mathbb{R}^n$ and $S^{n-1}$ the unit sphere in $\mathbb{R}^{n}$).
So we need to find a surjection $\pi: D^n \to \mathbb{P}^n$ such that for all $x,y \in D^n$, $x \sim y$ if and only if $\pi(x) = \pi(y)$.
For $x \in \mathbb{R}^{n+1}$, by letting $l_x$ be the line through $x$ and the origin in $\mathbb{R}^{n+1}$, I though about the following function: for $x = (x_1, \cdots, x_n) \in D^n,$ $$\pi: D^n \to \mathbb{P}^n, \pi(x_1, x_2, \cdots, x_n) = \left\{ \begin{array}{ll} l_{(0,0,0, \cdots, 0, 1)}, \text{ if } x = (0,0, \cdots, 0), \\
l_{p(x)}, \text{ otherwise } \end{array} \right., $$ where $$p(x) = \left( \frac{x_1}{||x||_{\mathbb{R}^n}} \cdot \sin(||x||_{\mathbb{R}^n} \cdot \pi), \cdots, \frac{x_n}{||x||_{\mathbb{R}^n}} \cdot \sin(||x||_{\mathbb{R}^n} \cdot \pi), \cos(||x||_{\mathbb{R}^n} \cdot \pi) \right).$$
If $x = y$, then of course $\pi(x) = \pi(y)$. If $x = \pm y$ and $x,y \in S^{n-1}$, then $||x||_{\mathbb{R}^n} = ||y||_{\mathbb{R}^n} = 1,$ so $p(x) = p(y) = (0,0, \cdots, 0, 1)$, hence $\pi(x) = \pi(y)$.
If $\pi(x) = \pi(y)$, then, in particular, we have that $$ \cos(||x||_{\mathbb{R}^n} \cdot \pi) = \cos(||y||_{\mathbb{R}^n} \cdot \pi), $$ hence $$ ||x||_{\mathbb{R}^n} \cdot \pi = \pm ||y||_{\mathbb{R}^n} \cdot \pi + 2k\pi, k \in \mathbb{Z}, \text{ so } ||x||_{\mathbb{R}^n} = \pm ||y||_{\mathbb{R}^n} + 2k, k \in \mathbb{Z}. $$
However, since $x,y \in D^n$, we have that $||x||_{\mathbb{R}^n}, ||y||_{\mathbb{R}^n} \leq 1$, so we get that $k = 0$, hence $||x||_{\mathbb{R}^n} = ||y||_{\mathbb{R}^n}$. Thus, from the other point-wise equalities, if $||x||_{\mathbb{R}^n} \neq 1$, we get that $x = y$, hence $x \sim y$. However, if $||x||_{\mathbb{R}^n} = 1$, we do not necessarily get that $x \sim y$, so this function is not a quotient map. How should I choose a different function?
Best Answer
Let $U\subseteq S^n\subseteq\Bbb R^{n+1}$ be the upper closed sphere. Then the projection on the last component defines an homeomorphism $\varphi:U\to D^n$. On the other hand we have a map \begin{align} &U\to\Bbb P^n& &u\mapsto\{\lambda u:\lambda\in\Bbb R\} \end{align} which induces an homeomorphism $U/{\sim}\to\Bbb P^n$.