Consider the group of integers modulo 2, $\mathbb{Z}_2 = \{\hat{0},\hat{1}\}$. Define the following action of $\mathbb{Z}_2$ on the closed unit disk $D^2=\{z\in \mathbb{C}: |z|\leq 1\}$: $$ \hat{0}\cdot z = z, \hat{1}\cdot z = -z$$ for all $z\in D^2$.
Prove that the resulting quotient $D^2/\mathbb{Z}_2$ is homeomorphic to $D^2$. Provide a complete argument, a proof based on only pictures is not enough.
For no reason, I don't know how this looks like in the picture. I think I have to show there is just one equivalenceclass, and because the equivalence classes form a partition, it is homeomorphic to $D^2$ itself. Am I thinking correct? who can help me?
Thanks!
HINT: Start with the disk $D^2$. Now imagine cutting two slits along the $x$-axis, one from $\langle -1,0\rangle$ to the centre and one from $\langle 1,0\rangle$ to the centre. Neither slit includes the centre: each has an open end at the centre. Imagine that the cutting is done so that $(0,1]\times\{0\}$ is just above the righthand slit, so that the part just below it has no actual top edge, and so that $[-1,0)\times\{0\}$ is just below the lefthand slit, so that the part just above it has no actual bottom edge. I’ve tried to suggest this in the (rather crude) figure below, in which I’ve opened up the slits so as to make their edges visible.
Now imagine pulling the point $A$ down and around to $B$, stretching the orange ‘semi-disk’ around into a full disk; keep it on top of the blue part. Then pull the point $C$ up and around to $D$, stretching the blue ‘semi-disk’ into a full disk; keep it beneath the orange part. You now have an object with three parts:
In polar coordinates, the first stretching operation takes an orange point $\langle r,\theta\rangle$ to the point $\langle r,2\theta\rangle$, and the second takes a blue point $\langle r,\pi+\theta\rangle$ to a point $\langle r,\pi+2\theta\rangle$.
You can now map this object to $D^2$ by the map $\pi$ that simply forgets about the colors: $\pi$ takes both the orange and the blue point at some $\langle r,\theta\rangle$ to $\langle r,\theta\rangle\in D^2$. If you check the details, you’ll see that if $p$ is in orange part, then $-p$ is in the blue part (and vice versa), and they end up in the same place after the two stretching operations, so the final map takes them to the same point of $D^2$. This means that if $h$ is the composition of the two stretchings with $\pi$, then $h$ is constant on each orbit of the action by $\Bbb Z_2$. This in turn means that $h$ induces a map from $D^2/\Bbb Z_2$ to $D^2$, and it’s not terribly hard to work out the details to show that this map is a homeomorphism.