An action of $\mathbb{Z}_2$ on $\mathbb{R}^2$

Question

Let $M$ be the manifold $\mathbb{R}^2$ and $G$ be $\mathbb{Z}_2 = \{1,-1\}$, the cyclic group of order 2. Let $G$ act on $M$ by $(x,y) \mapsto (-x,-y)$. I need to show that the orbit space $M\vert G$ is homeomorphic to $\mathbb{R}^2$.

The way I see it is - the left half plane is half twisted (rotated by 180 degrees) about the negative $x$ axis and then pasted onto the right half plane. Then I paste the negative $y$ axis onto the positive $y$ axis giving me an infinite cone (with the origin as vertex). Now if I look at this cone is three dimensional space and project towards the base then I have $\mathbb{R}^2$.

Is the way I have seen it correct? If so can some one help me with how to write this explicitly with maps and all?

Thanks!

Answer 1

You can depict the result as follows: you just deal with the subspace of points with $x\geq 0$, and then you identify $(0,y) \ \text{with} \ (0,-y)$. From this description it should be clear that what you get is actually a copy of $\mathbb{R}^2$.

Answer 2

An alternative to Lano's (perfectly fine) answer.

Think of $\mathbb{R}^2$ as the cone on a circle $S^1$. Then in this picture, your $\mathbb{Z}_2$ action is the antipodal map on each circle, and acting trivially on the cone point.

It follows that we may think of the quotient space as the cone on $S^1/\mathbb{Z}_2$.

But $S^1/\mathbb{Z}_2 = \mathbb{R}P^1 = S^1$, so the quotient space is the cone on $S^1$. This is homeomorphic to $\mathbb{R}^2$ as claimed.