# Why is $D^n/\sim$ homeomorphic to $\mathbb{RP}^n$?

Let $\sim$ be the equivalence relation on $D^n$ (the $n$-dimensional unit disc) which identifies antipodal points on the boundary $\partial D^n = S^{n-1}$. Show that $D^n/\sim$ is homeomorphic to $\mathbb{RP}^n$, the real projective space of dimension $n$.

Edit: We define $\mathbb{RP}^n$ as the set of real lines through the origin in $\mathbb{R}^{n+1}$, with the origin deleted, with the quotient topology determined by the map which sends a nonzero vector in $\mathbb{R}^{n+1}$ to the line containing it.

• but you didn't specify the topology. – Ittay Weiss Mar 15 '15 at 21:13
• can you think of a function from the set of lines you describe to the set of pairs of points in $D^n$? (hint: intersection). Are these points always antipodal (hint: yes). What does that tell you? – Ittay Weiss Mar 15 '15 at 21:14
• The function which sends a line (in, say, $\mathbb{R}^n \times \{0\}$) to its intersection with $\partial D^n$, which is a pair of antipodal points in $D^n$. I'm still thinking about what this tells you. – Randy Randerson Mar 15 '15 at 21:18
• I'm not sure what it tells you. Can you say what it tells you about? – Randy Randerson Mar 15 '15 at 21:26
• it tells you that you should review what quotient spaces are. – Ittay Weiss Mar 15 '15 at 22:03

Take these local coordinates, to proof ${D^{n + 1}} \simeq P{\mathbb{R}^n}$ \begin{gathered} {U_i} = \{ \left. {[{x_0}, \cdots ,{x_n}]} \right|{x_i} \ne 0\} \subset P({\mathbb{R}^{n + 1}}) \hfill \\ {\varphi _i}:{U_i} \to {\mathbb{R}^n},[{x_0}, \cdots ,{x_n}] \to (\frac{{{x_0}}}{{{x_i}}}, \cdots ,\frac{{{x_{i - 1}}}}{{{x_i}}},\frac{{{x_{i + 1}}}}{{{x_i}}}, \cdots ,\frac{{{x_n}}}{{{x_i}}}) \hfill \\ {\psi _i}:{\mathbb{R}^n} \to {U_i},({a_1}, \cdots ,{a_n}) \to [{a_1}, \cdots ,{a_{i - 1}},1,{a_i}, \cdots ,{a_n}] \hfill \\ \end{gathered}