Homeomorphism between $\mathbb{P}^n$ and $D^n / R$

Question

I've to find a homeomorphism between $\mathbb{P}^n$ and $D^n / R$ with the equivalence relation $R$ on $D^n$ given by: $(x,y) \in \mathbb{R}$ iff $x=y$ or ($(x,y \in S^{n-1}$ and $x = -y$). I know that this is the same as proving that $S_n$ positive is homeomorphic to $\mathbb{P}^n$, but I don't know how to prove this.

Answer 1

Define $i : D^n \to S^n, i(x) = (x,\sqrt{1-\lVert x \rVert^2})$. This map embeds $D^n$ as the closed upper hemisphere of $S^n$.

You know that $\mathbb P^n$ is the quotient space of $S^n$ obtained by identifying antipodal points. Let $p : S^n \to \mathbb P^n$ denote the quotient map. The map $H = p \circ i : D^n \to \mathbb P^n$ has the property $H(x) = H(y)$ if and only if $(x,y) \in R$. Moreover, for each $x \in S^n$ we have $x \in i(D^n)$ or $-x \in i(D^n)$, hence $H$ induces a continuous bijection $h : D^n/ R \to \mathbb P^n$. But $D^n/ R$ is compact (as a quotient space of a compact space) and $\mathbb P^n$ is Hausdorff, thus $h$ is a homeomorphism.