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.
Unfortunately, the result is not true in dimensions of the form $4k$. That is, in each dimension of the form $4k$, there is a $Z_2$ action on $S^{4k}$ with quotient not homeomorphic to $\mathbb{R}P^{4k}$. (I don't know what happens in other dimensions. I suspect that Ricci flow techniques in dimension $3$ would show that the the only $Z_2$ quotients of $S^3$ is the standard $\mathbb{R}P^3$. I'm far from an expert in this area, though.)
More specifically, in
Ruberman shows that there is a non-smoothable $4$-manifold $X$ which is homotopy equivalent, but not homeomorphic to $\mathbb{R}P^4$.
The universal cover $\tilde{X}$ of $X$ must therefore be homotopy equivalent to $S^4$. By Freedman's classification, we know $\tilde{X}$ is homeomorphic to $S^4$. Now, the deck group (which is isomorphic to $\pi_1(X)\cong \pi_1(\mathbb{R}P^2)\cong \mathbb{Z}_2$) acts freely on $\tilde{X}\cong S^4$.
Further, in
Fintushel and Stern find examples of smooth $4k$ manifolds (with $k>1$) which are homotopy equivalent, but not homeomorphic to $\mathbb{R}P^{4k}$. One can repeat the argument above (using Smale's proof of the Poincare conjecture in high dimensions instead of Freedman's result) to find the involutions on $S^{4k}$.