Homeomorphism between $RP^2$ and $B^2$ quotient antipodal points on $S^1$ - difficulties in showing continuity I would like to prove that the real projective plane is homeomorphic to X where X is the quotient space obtained from the 2-ball in $R^2$ by identifying each point $x \in S^1$ (the unit sphere) with its antipode $-x$. I constructed the Map $(x,y) \rightarrow (x,y,\sqrt{1-x^2-y^2})$ going from X to the real projective plane. I showed injectivity and surjectivity but am stuck on the part of showing continuity for the map and its inverse.. I am trying to show that the presage of open sets is open but am not quite sure how to exactly show that. Would be grateful for any hints! 
 A: $\mathbb RP^2=\{[x]:x\in \mathbb R^3\}$, note that every lines $-\{0\}$ in $\mathbb R^3$ will be a "point" in $\mathbb RP^2$ and you can chose as set of representatives $S^2/\sim$, where $\sim$ is the antipodal relation. Now observe that $S^{2}_+=\{(x,y,z): x^2+y^2+z^2=1, z\ge0\}$ is homeomorphic to $E=\{(x,y,0):x^2+y^2\le1\}$. Just map $(x,y,z)$ to $(x,y,0)$. Now, obviously, $E\cong D^2=\{x\in \mathbb R^2:\|x\|_2\ \le \
1\}$.
Try to formalize better this idea, I  have spoken very informally.
A: Your idea is correct.
Let $\mathbb R^m_* = \mathbb R^m \setminus \{0\}$ and  $p : \mathbb R^{n+1}_*  \to \mathbb R P^n$ be the quotient map. Let $B^n \subset \mathbb R^n$ be the closed unit ball. Define
$$\phi : B^n \to \mathbb R^{n+1}_*, \phi(x) = (x,\sqrt{1 - \lVert x \rVert ^2}), $$
$$q = p \circ \phi : B^n \to \mathbb R P^n  .$$
The map $q$ is a surjection: Each $\eta \in \mathbb RP^n$ has a representative $y = (y_1,\dots, y_{n+1})\in \mathbb R^{n+1}_*$ such that $\lVert y \rVert  = 1$ and $y_{n+1} \ge 0$. But then $y = \phi(y_1,\dots,y_n) \in \phi(B^n)$ because $\lVert (y_1,\dots,y_n) \rVert \le \lVert y \rVert = 1$.
$q$ is a closed map since $B^n$ is compact and $\mathbb R P^n$ is Hausdorff. It is well-known that closed maps are quotient maps. Hence $q$ is a quotient map.
Define an equivalence relation on $B^n$ by $x \sim x'$ if $q(x) = q(x')$´and let $r : B^n \to P^n = B^n/\sim$ be the quotient map. Then $q$ induces a bijection $q': P^n \to  \mathbb R P^n$ which is a homeomorphism by the universal property of the quotient topology.
$x \sim x'$ means that $\phi(x) = t\phi(x')$ for some $t \in \mathbb R \setminus \{ 0 \}$. Since $\phi(x), \phi(x')$ have norm $1$, this means that $(x,\sqrt{1 - \lVert x \rVert ^2}) = \phi(x) = \pm\phi(x') = \pm(x',\sqrt{1 -  \lVert x' \rVert ^2})$. In other words, it means $x = \pm x'$ and $\sqrt{1 - \lVert x \rVert ^2} = \pm \sqrt{1 - \lVert x' \rVert ^2}$. The minus-sign is possible only when $\sqrt{1 - \lVert x \rVert ^2} = \sqrt{1 - \lVert x' \rVert ^2} = 0$, i.e. when $\lVert x \rVert  = \lVert x' \rVert =1$ which means $x,x' \in S^{n-1}$.
