 Solution

Assume for contradiction that the involution $\varphi$ is fixed-point-free. The finite group action is then a covering space action, so the quotient mapis a double covering and $Y$ is an $n$-manifold.

Apply the long exact sequence in homology to the transfer chain map of a double covering. Since ${\mathbb{R}}^n$ is contractible, its positive-dimensional mod-two homology vanishes. The degree-zero portion, together with the fact that $q_{\ast }:H_0({\mathbb{R}}^n;{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism, givesIn every higher degree the same exact sequence givesThus $H_i(Y;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ for every $i\ge 0$. This contradicts homology above the dimension of a manifold, which gives $H_i(Y;{\mathbb{F}}_2)=0$ for $i>n$. Therefore $\varphi$ has a fixed point.