Let $\pi:X \rightarrow Y$ be a double cover between compact manifolds $X$, $Y$ and $\theta$ be the deck transformation. Let $H^2(X, \mathbb Z)^\theta$ be a group of $\theta^\*$-invariant elements in $H^2(X, \mathbb Z)$.


My Question is:


Is $H^2(X, \mathbb Z)^\theta$ a subset of $\pi^\* (H^2(Y, \mathbb Z) )$?


You can assume that $X$ is a simply-connected projective variety if necessary.


Thanks!



Cohomologies of double covers
Using spectral sequences is a bit heavy here. Given a map $s\colon\Delta\_p\to Y$, there are precisely two different lifts to $X$, and we can add them to get a singular chain in $C\_p(X)$. This construction extends linearly to give a chain map $\tau\colon C\_\*(Y)\to C\_\*(X)$ (called the transfer), with $\pi\_\*\circ\tau=2$ and $\tau\circ\pi\_\*=1+\theta\_\*$. This induces a map $\pi\_!\colon H^\*(X)\to H^\*(Y)$ with $\pi^\*\pi\_!=1+\theta^\*$ and $\pi\_!\pi^\*=2$ and $\theta^\*\pi^\*=\pi^\*$. Using this we get $2\ker(\pi^\*)=0$ and $2H^\*(X)^{\theta}\leq\text{img}(\pi^\*)\leq H^\*(X)^{\theta}$. Thus, if we invert $2$ then $\pi^\*$ gives an isomorphism from $H^\*(Y)$ to $H^\*(X)^{\theta}$. The example $S^3\to\mathbb{R}P^3$ shows that this does not work without inverting $2$, even if $X$ and $Y$ are oriented.


