Let $M$ be the Möbius strip and $C$ it's boundary circle. Prove that $M/C$ is homeomorphic to $\Bbb RP^2.$
I know that $\Bbb RP^2 \approx D^2/x \sim -x, x \in \partial D^2.$ So if we can able to get a continuous surjective map from $M$ to $D^2$ or at least from $M$ to $\Bbb R P^2$ which takes all the points of the boundary circle $C$ to a single point then by universal property of quotient topology we are through. Would anybody please help me finding an explicit homeomorphism from $M$ onto $D^2$ or at least onto $\Bbb R P^2\ $?
Thanks in advance.
We can identify $\mathbb{RP}^2$ as $I^2/\sim_A$ where $I^2$ denotes the unit square and $(0,t)\sim_A (1,1-t)$ and $(t,0)\sim_A (1-t,1).$
A mobius strip is $M=I^2/\sim_B$ where $(0,t)\sim_B (1,1-t)$
So, the quotient map , $q: I\to I^2/\sim_A$ gives rise to a map $q':I^2/\sim_B\to I^2/\sim_A$ by universal property of quotient map.
The boundary circle of the mobius strip can be identified with $S^1.$ Therefore by applying universal property of quotient map, we get a homeomorphism , $\phi: M/x\sim(-x)\to \mathbb{RP}^2$ from $q'$ where $x\in S^1.$
Since, $\mathbb{RP}^1$ is homeomorphic to $S^1$ therefore we can say that, $M/C$ is homeomorphic to $\mathbb{RP}^2.$