Homeomorphism between real projective plane and disc Let $D = \{\mathbf{x} \in \mathbb{R}^2 \ | \ \|\mathbf{x}\| \le 1\}$ and let $\mathbb{R}\mathbb{P}^2$ be the real projective plane Let $X = D/\sim$ where $\sim$ identifies antipodal points in the boundary of $D$. I am interested in finding explicitly a homeomorphism between X and the projective plane. Could anyone please help me with this one?
 A: OK, if $D$ is a 2-dimensional closed ball and $X = D / \sim$, where $\sim$ identifies antipodal points on the boundary, then there is a homeomorphism $\varphi \colon X \to \mathbb{R}P^2$. Here is how we define it:
$$
\varphi(x,y) = x:y:\sqrt{1-x^2-y^2}
$$
where $(x,y)$ are coordinates of a point in $D$. Notice that if $(x,y)\sim(x',y')$ then $\varphi(x,y)=\varphi(x',y')$, so $\varphi$ is well defined. You can check yourself that this is a homeomorphism.
Basically, what I've done here is compose the natural homeomorphism $\mathbb{R}P^2 \to (S^2 / \approx)$, where $\approx$ identifies antipodal points on the sphere, with a homeomorphism $(S^2 / \approx) \to X$ which simply projects the upper half-sphere to the $xy$-plane.
A: I think perhaps what you want is that $\Bbb RP^n$ is homeomorphic to $S^n/\sim$ where we say $x\sim y$ if $y=tx$ for some $t\in \Bbb R$ (we're identifying antipodal points). You can check that a homeomorhism is induced by the map $$f(x)=\frac{x}{\Vert x\Vert}.$$
A: They aren't homeomorphic.  You can look at their homology to see this.  (There is probably an easier way, this seems like overkill).
Edit:  And yes, as a commenter has pointed out, the $D$ you described is a closed ball, and this is contractible so that's not homeomorphic to $\mathbb{R}\mathbb{P}^2$ either
