# 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?

-
The question doesn't match the title: $D$ is a disc, not a sphere. –  Dan Shved Feb 20 '13 at 4:58
But they are not homeomorphic. –  user27126 Feb 20 '13 at 4:58
What about now? –  user44069 Feb 20 '13 at 5:01
What would be your definition of the projective plane? –  Gerry Myerson Feb 20 '13 at 5:04
This $D$ looks alright and it is homeomorphic to $\mathbb{R}P^2$, but still neither of them is a sphere. –  Dan Shved Feb 20 '13 at 5:13

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.

-
Can you please elaborate on the notation $x:y:\sqrt{1 -x^2 -y^2}$? What does it indicate? –  user44069 Feb 20 '13 at 5:33
@Stefan These are homogeneous coordinates on the projective plane. –  Dan Shved Feb 20 '13 at 5:40

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 $$f(x)=\frac{x}{\Vert x\Vert}.$$

-
Yes! Thank you very much! –  user44069 Feb 20 '13 at 5:02
No, what he wants is exactly what he asked... –  Chris Gerig Feb 20 '13 at 7:18
@ChrisGerig: See Loring Tu's An Introduction to Manifolds on Exercise 7.11 from Chapter 2. The exercise explicitly asks us to show that this map induces a homemorphism, where, in this case $$\overline{f}\big([x]\big)=\left[\frac{x}{\Vert x\Vert}\right].$$ –  Clayton Feb 20 '13 at 14:21

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

-
There is, they have different Euler characteristics, plus $\mathbb{R}P^2$ is not orientable. –  Dan Shved Feb 20 '13 at 4:59
One is a boundaryless manifold, another is not. –  user27126 Feb 20 '13 at 5:00
@muzzlator: Show me the homology calculation, because I disagree with you. Sanchez: Neither has boundary, my friend. Anyway, you start with $S^n/\sim$ and this is the same as the upper-hemisphere with antipodal points on the equator identified (and this is his "disk"!) –  Chris Gerig Feb 20 '13 at 7:14
@Chris When this answer was posted the question looked quite a bit different. Back then it was about the projective plane on one hand and either a sphere or a closed ball on the other hand. –  Dan Shved Feb 20 '13 at 11:07
Ah makes sense now! –  Chris Gerig Feb 20 '13 at 20:39