The questions asks us to show that if $M$ is a closed orientable 4-manifold such that $H_2(M)$ is rank $1$, then $M$ does not admit a free action of $\mathbb{Z}/2$.
My attempt has been to suppose $M$ has a free action of $\mathbb{Z}/2$. So there is a homeomorphism $\phi:M\rightarrow M$ satisfying $\phi^2=\textrm{id}$. I've looked at the duality isomorphism $H^2(M)\rightarrow H_2(M)$ and played around with identities like $$ \phi_{\ast}(\phi^{\ast}\alpha\cap[M])=\alpha\cap\phi_{\ast}[M]=\pm\alpha\cap[M] $$ trying to get at some kind of contradiction. But I suspect I need to incorporate the freeness assumption for the action $\mathbb{Z}/2$. I understand that this action will be properly discontinuous and so if $M$ were path connected then the quotient map $M\rightarrow M/(\mathbb{Z}/2)$ would be a covering space and $M/(\mathbb{Z}/2)$ would inherit the structure of a manifold. The trouble is that $M$ isn't necessarily path-connected. Can anyone suggest a way of procedding?
Let $\phi\colon M\to M$ be a homeomorphism with $\phi^2=\text{id}$. You can use the Lefschetz fixed-point theorem: If $$\Lambda_\phi=\sum_{k\geq0}(-1)^k \text{Tr}(\phi_*|_{H_k(M;\mathbb{Q})})$$ is non-zero then $\phi$ has a fixed-point. Now we have $H_k(M;\mathbb{Q})=H_k(M;\mathbb{Z})\otimes \mathbb{Q}$ and by Poincaré duality together with universal coefficients $H_k(M;\mathbb{Q})\cong H_{4-k}(M;\mathbb{Q})$. The map $\phi$ induces involutions on all these vector spaces, so it is diagonalizable with eigenvalues $\pm1$. With these information you can calculate $\Lambda_\phi\mod 2$ and show that it is non-zero.