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?
General Topology – Homeomorphism Between Real Projective Plane and Disc
algebraic-topologydifferential-geometrygeneral-topology
Related Solutions
Every point on the upper hemisphere is identified with a point on the lower one. The upper hemisphere is homeomorphic to a disc. What happens at the equator ( the boundary of this disc)?
edit: Let's do a proof. Let $D^2$ be the disc and $D^2/\sim$ the disc with opposite boundary points identified.
For every point $x\in S^2$ there are two points on the line through the origin and $x$. For all points outside the equator, there is a unique point on the upper hemisphere, but on the equator there are two points on the (closed) upper hemisphere. We can't directly define a map $S^2\rightarrow D^2$ this way, however we can map a point to $D^2/\sim$. The problems on the equator disappear when we take the equivalence relation. Thus we have a map
$S^2\rightarrow D^2/\sim$.
By definition of this map it sends antipodal points on $S^2$ to the same points on $D^2/\sim$. Hence it factors to a continuous map $\mathbb{RP}^2\rightarrow D^2/\sim$.
Now you have to show that this map is
- injective
- surjective
and that $\mathbb{RP}^2$ is compact, and $D^2/\sim$ is Hausdorff. Then you know that the map is a homeomorphism (see https://proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism).
There are some interesting constructive ways to visualize $\mathbb{R}\mathrm{P}^2$. I think the simplest way is as follow:
First, we prepare a Whitney umbrella, which is homeomorphic to the upper half of $S^2$ :
Let's check the points to quotient (be glued).
After gluing, the red and blue paths should match their start points and the end points respectively so that their antipodal points are matched.
In the picture of half $S^2$ their start points are far from each others' start points now, while Whitney Umbrella has the two start points meet together.
Now we can just bend the umbrella and glue the edge:
And get a cross-cap, which is homeomorphic to $\mathbb{R}\mathrm{P}^2$.
I cannot color it with two color since it is not "two sided" now.
Many books suggest Möbius strip for intuition. It is a little bit more difficult but more interesting to visualize that $\pi_1(\mathbb{R}\mathrm{P}^2)$ is isomorphic to $\mathbb{Z}_2$.
Best Answer
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.