Free $\mathbb{Z}_{2}$ action on the plane

Question

Motivated by the following question we ask:

Is there a free action of $\mathbb{Z}_{2}$ by homeomorphism on $\mathbb{R}^{2}$?

Lie groups with no free $\mathbb{Z}/2\mathbb{Z}$ action

Answer 1

We have an integer invariant of topological spaces, called the euler characteristic. This has two special properties, namely that for a covering map $X\to Y$, with fiber $F$, we have $\chi(X)=\chi(Y)|F|$. In this case, if the action is free, the map $\mathbb{R}^2\to \mathbb{R}^2/\{gx=x\}=Q$ is a covering map with fiber $\mathbb{Z}/(2)$, so that we have that $\chi(\mathbb{R}^2)=2\chi(Q)$, so that $\chi(\mathbb{R}^2)$ is even. But you can easily compute that $\chi(\mathbb{R}^2)=1$, so we have our contridiction!