Homemorphism from projective plane and Moebius strip

Question

I am looking for an explicit homeomorphism from

$ \mathbb{R} P^2=\mathbb{S}^2/(x \sim -x) $

to

$M/ \partial M$ with M the Moebius strip and $\partial M$ its boundary.

Answer 1

If you want to be explicit, then this depends on how you think of $M$. Since you think of it as having a boundary, perhaps you think of it as $M = ([0,1] \times [0,1])/((0,y) \sim (1,1-y))$. Then $M \setminus \partial{M} = ([0,1] \times {]0,1[})/((0,y) \sim (1,1-y))$. If so, then an explicit map is obtained as follows:

First, replace $[0,1]$ with $[0,\pi]$ using $x \mapsto \pi x$. Then replace $[0,\pi]$ with the top half of the unit circle in the complex plane using $\theta \mapsto \mathrm{e}^{\mathrm{i}\theta}$. Also, replace ${]0,1[}$ with ${]0,\infty[}$ using $y \mapsto y/(1-y)$. Finally, use $(u,r) \mapsto r u$ so that what used to be $[0,1] \times {]0,1[}$ is now the upper half of the complex plane, including the real axis, minus the origin. The identification $(0,y) \sim (1,1-y)$ is now $y/(1-y) \sim -(1-y)/y$ (for $y \in {]0,1[}$). Or taking $t$ to be any nonzero real number, it's $t \sim -1/t$.

Now think of $M \setminus \partial{M}$ as the entire complex plane (minus the origin) modulo $z \sim -1/\bar{z}$. This extends the above identification along the real line and also identifies the top half of the plane with the bottom half. If we interpret this formula liberally, it also identifies the missing origin with a missing point at infinity. If you throw in those points, then you get the Riemann sphere, which can be interpreted as $S^2$ using stereographic projection: $$ z \mapsto \left( \frac{2\,\Re{z}}{|z|^2+1}, \frac{2\,\Im{z}}{|z|^2+1}, \frac{|z|^2-1}{|z|^2+1} \right) .$$ The identification $z \sim -1/\bar{z} = -z/|z|^2$ indeed becomes $\mathbf{x} \sim -\mathbf{x}$. (Also, the missing points $0$ and $\infty$ become $(0,0,-1)$ and $(0,0,1)$, which are indeed identified.)

So in the end, given $(x,y) \in [0,1] \times {]0,1[}$, we have $\theta = \pi x$, then $u = \mathrm{e}^{\mathrm{i} \theta} = \mathrm{e}^{\pi \mathrm{i} x}$, also $r = y/(1-y)$, then $z = u r = \mathrm{e}^{\pi \mathrm{i} x} y/(1-y)$, which is identified with $-1/\bar{z} = -\mathrm{e}^{\pi \mathrm{i} x} (1-y)/y$, which finally becomes $$ \mathbf{x} = \left( \frac{2\,\Re{z}}{|z|^2+1}, \frac{2\,\Im{z}}{|z|^2+1}, \frac{|z|^2-1}{|z|^2+1} \right) = \left( \frac{2 \cos(\pi x) y (1-y)}{y^2 + (1-y)^2}, \frac{2 \sin(\pi x) y (1-y)}{y^2 + (1-y)^2}, \frac{y^2 - (1-y)^2}{y^2 + (1-y)^2} \right) ,$$ which is identified with $$ -\mathbf{x} = \left( \frac{-2 \cos(\pi x) (1-y) y}{(1-y)^2 + y^2}, \frac{-2 \sin(\pi x) (1-y) y}{(1-y)^2 + y^2}, \frac{(1-y)^2 - y^2}{(1-y)^2 + y^2} \right) .$$

While this is going on, the boundary $\partial{M} = [0,1] \times \{0,1\}$, with $(0,0) \sim (1,1)$ and $(0,1) \sim (1,0)$, collapses to $(0,0,-1)$ and $(0,0,1)$, which are identified. More specifically, $[0,1] \times \{0\}$ collapses to $(0,0,-1)$ and $[0,1] \times \{1\}$ collapses to $(0,0,1)$. So $\partial{M}$ collapses to a single point in $\mathbb{R}P^2 = S^2/(\mathbf{x} \sim -\mathbf{x})$.

Therefore, $M/\partial{M} \cong \mathbb{R}P^2$.