TITLE: Real Projective Plane is Same as Identifying Antipodal Boundary Points of The $2$-Disc.
QUESTION [11 upvotes]: $\newcommand{\RP}{\mathbf RP}$
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.

REPLY [3 votes]: 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.