TITLE: Is there only one free (continuous) action of $\mathbb{Z}_2$ on $S^2$?
QUESTION [5 upvotes]: We all know that the antipodal map is a free action of $\mathbb{Z}_2$ on $S^2$. Considering $\mathbb{Z}_2 = \{1, -1\}$, a free action may be viewed as a map $f : S^2 \rightarrow S^2$, i.e. the action of $-1$, with no fixed points, such that $f \circ f = Id_{S^2}$. We then readily see that $deg(f) = \pm 1$. Since $f$ does not have fixed points, it cannot have degree 1, and hence $f$ is homotopic to the antipodal map.
My question is: does this imply that the quotient of $S^2$ by this action is homeomorphic to $\mathbb{RP}^2$? Actually, I know this is true, because the quotient must be a closed surface with fundamental group $\mathbb{Z}_2$, so we can use the classification of closed surfaces. But I would like to be able to generalize this to spheres of any even dimension, so I wanted to see if I could get an answer that doesn't use that.

REPLY [2 votes]: The quotients of such actions are called "Fake real projective spaces" and you find their classification in http://www.map.mpim-bonn.mpg.de/Fake_real_projective_spaces