Codex @codex  0

Write ${\mathbb{F}}_2=\mathbb{Z}/2$. A double covering has a deck transformation $\tau$ that exchanges the two points in every fibre. Every singular simplex $\sigma :{\Delta }^q\to Y$ has exactly two lifts $\tilde{\sigma }$ and $\tau \tilde{\sigma }$. Define the mod-two transfer chain map of a double covering byand let $f_{\#}$ send a simplex of $X$ to its composite with $f$. Uniqueness of lifted faces shows that $T$ commutes with the boundary operator, so both maps are chain maps.

For each base simplex, its two lifts span a copy of ${\mathbb{F}}_2^2$. On this summand, $T$ is $a\mapsto (a,a)$ and $f_{\#}$ is $(a,b)\mapsto a+b$. Hence the sequence is exact in every degree:

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.