$\Bbb RP^2$ as the union of a Möbius band and a disc I spent some time trying to understand that the projective plane, $\Bbb RP^2$, can be viewed as the union of a Möbius band and a disc. I consider this using Homogeneous coordinates. But I still hope to really 'see' this fact in some way. 
 A: In general you have $\mathbb RP^n = \mathbb R^n \cup \mathbb RP^{n-1}$. To see this consider the embedding $i: \mathbb R^n \to \mathbb RP^n$ given by $i(x_1,x_2, \dots, x_n) = [1,x_1,x_2, \dots, x_n]$. The complement of $i(\mathbb R^n)$ in $\mathbb RP^n$ is $$\{[0,x_1,x_2, \dots, x_n]\;|\; (x_1,x_2, \dots, x_n) \in \mathbb R^n \setminus \{0\} \} \cong \mathbb RP^{n-1}.$$
Now in your special case we have $\mathbb RP^2 = \mathbb R^2 \cup \mathbb RP^1$, where $\mathbb RP^1$ is readily identified with $S^1$. Now consider the circles given by $[1,r\cos\phi, r\sin \phi]$. If you let $r\to \infty$, the circle will double cover $\mathbb RP^1 \cong S^1$. This gives the desired decomposition $$\mathbb RP^2 = \{[1,r\cos\phi, r\sin \phi]\;|\;0\leq r \leq 1 \text{ and } \phi \in [0,2\pi) \} \cup \{[r,\cos\phi, \sin \phi]\;|\;0\leq r \leq 1 \text{ and } \phi \in [0,2\pi) \},$$
where the two components are identified with the closed disk $D^2$ and the moebius band $M$, with common boundary $S^1$.
A: http://en.wikipedia.org/wiki/Real_projective_plane
Look at the first two pictures on the right. Imagine that you cut out a vertical rectangle in the middle of the projective plane square. The cutout piece is a Möbius strip.
Now glue first the edges B in the right direction of the two remaining pieces and then the parts of edge A and realize that this corresponds to a disc.
A: $RP^2$ is the sphere $S^2$ where you identify $x \sim -x$. If $D$ is a disc then $RP^2 \setminus D$ is the space $S^2 \setminus (D \cup -D)$ where you identify $x \sim -x$. But $S^2 \setminus (D \cup -D)$ is a cylinder. I let you see that if you identify a cylinder by $x \sim -x$ then you have a Moebius strip (I am sorry if the notations are not clear, but I couldn't find better).
