Proving a homeomorphism between $RP^n$ and $D^n/{\sim}$ How to prove that $RP^n$ is homeomorphic to quotient space $D^n/{\sim}$ ,where  $x\sim -x$ for $x\in\partial D^n$.
Definition: $RP^n=R^{n+1}\smallsetminus\{0\}/{\sim}$ where $\sim$ is defined as :
$x\sim\lambda x$ for all $\lambda\in R^x,$ $x\in R^{n+1}\smallsetminus\{0\}.$
$R^x$ is the productive group of scalars that are not $0.$
It is a little weird to me to build a homeomorphism like this.
 A: You can find almost everything you need in my answer to How to show this property of $\mathbb{RP}^{n}$. Let us adapt the arguments to the situation in your question.
We can identify $D^n$ with the closed upper hemisphere $S^n_+ = \{(x_1,\ldots,x_{n+1}) \in S^n  \mid x_n \ge 0 \}$ of the unit sphere $S^n \subset \mathbb R^n$. An explicit homeomorphism is given by $h : D^n \to S^n_+,  h(x_1,\ldots,x_n) = $ $(x_1,\ldots,x_n,\sqrt{1 -(x_1^2+\ldots+x_n^2)})$. Its inverse is $\pi : S^n_+ \to D^n, \pi(x_1,\ldots,x_{n+1}) = (x_1,\ldots,x_n)$. We have $\partial D^n = S^{n-1}$ and $h(S^{n-1}) =S^{n-1}_0 = S^{n-1} \times \{0\} \subset S^n_+$. Define an equivalence relation on $S^n_+$ by $z \equiv -z$ for $z \in S^{n-1}_0$, i.e. by identifying antipodal points on $S^{n-1}_0$. Then $h(x) \equiv h(x')$ iff $x \sim x'$. This means that $h$ induces a homeomorphism $h' : D^n/\sim \phantom{} \to S^n_+/\equiv$ .
The quotient map $p : \mathbb R^{n+1} \setminus \{0\} \to \mathbb RP^n$ restricts to a surjective map $q : S^n_+ \to \mathbb RP^n$.  Since $S^n_+$ is compact and $\mathbb RP^n$ is Hausdorff, $q$ is closed map and therefore a quotient map.  This gives an alternative description of $\mathbb RP^n$: It is the quotient space obtained from $S^n_+$ by identifying all fibers $q^{-1}(y) \subset S^n_+$, $y \in \mathbb RP^n$, to the single point $y$. What are these fibers? If $x \in q^{-1}(y)$, then $q^{-1}(y) = l(x) \cap S^n_+$, where $l(x)$ denotes the line through $x$ and $0$. Therefore, if $q^{-1}(y)$ contains a point $x \in S^n_+ \setminus S^{n-1}_0$, we have $q{-1}(y) = \{x\}$. If $q^{-1}(y)$ contains a point $x \in S^{n-1}_0$, then $q{-1}(y) = \{x,-x\}$. In other words, $q$ identifies precisely the equivalence classes with respect to $\equiv$ to points in $\mathbb RP^n$. Therefore $q$ induces a homeomorphism $q' : S^n_+/\equiv \phantom{} \to \mathbb RP^n$.
