$\newcommand{\RP}{\mathbf RP}$
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.
Best Answer
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
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).