Real Projective Plane is Same as Identifying Antipodal Boundary Points of The $2$-Disc.

Question

$\newcommand{\RP}{\mathbf RP}$ The real projective plane $\RP^2$ is defined as the quotient space $S^2/\sim$, where $\sim$ identifies the antipodal points of $S^2$.

I want to show that $\RP^2$ is homeomorphic to the quotient space of the closed $2$-disc $D^2$ obtained by identifying the antipodal points on the boundary circle of $D^2$.

This is not at all visually obvious to me.

An analogous problem is showing that $\RP^1=S^1$. Here, the situation is simple because we can think of $S^1$ as sitting inside $\mathbf C$ and consider the function $f:S^1\to S^1$ defined as $f(z)=z^2$ for all $z\in S^1$. The fibres of $f$ are precisely the antipodal points and we get an isomorphism $S^1/\sim \ \cong \ S^1$.

Again, the simpler case is also not visually clear to me.

Answer 1

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

1. injective
2. 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).

Answer 2

After discussing with a friend, here is a solution:

We think of $D^2$ as the closed upper half-sphere $\bar H^2_+$.

Consider the map $f:S^2 \mapsto \mathbb{RP}^2$ as $f(x)=[x]$, that is, a point in $\bar H^2_+$ is mapped to the corresponding line passing through origin.

Let $g=f|_{\bar H^2_+}$, that is, $g$ is the restriction of $f$ on the closed upper-half sphere.

So we have a continuous map $g$ on $\bar H^2_+$ whose fibres are precisely the one point sets $\{x\}$ whenever $x$ is in the upper-half sphere and $\{x, -x\}$ whenever $x$ is on the equator.

Thus we get a continuous map $\tilde g:\bar H^2_+/\sim\ \mapsto \mathbb{RP}^2$ which is bijective and continuous. This map is in fact a homeomorphism because it is a map from a compact space to a Hausdorff space.