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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!

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    $\begingroup$ If $\pi$ is a double cover of curves with a ramification point $p\in X$ then the class of $p$ in cohomology seems to give negative answer. $\endgroup$
    – SashaP
    Nov 15 '16 at 22:57
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    $\begingroup$ Thanks for the anwer. I am interested in unramified cases. $\endgroup$
    – Lee
    Nov 16 '16 at 0:39
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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.

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As a concrete example, suppose $Y$ is a connected closed oriented 2-dimensional manifold and $X \to Y$ is a connected cover. Then $H_2(X) = H_2(Y) = H^2(X) = H^2(Y) = \Bbb Z$ by Poincare duality, and the map $H_2(X) \to H_2(Y)$ is multiplication-by-2 because the map is degree 2, so the dual map $H^2(Y) \to H^2(X)$ is multiplication-by-2. The image $2\Bbb Z$ must be contained in the subgroup $H^2(X)^\theta$, which forces $\theta$ to act trivially on $H^2(X)$. However, this means that the image of $H^*(Y)$ does not contain all of $H^*(X)^\theta$.

As abx states in the comments, there is a Hochschild-Serre spectral sequence $$ H^p(\Bbb Z/2; H^q(X)) \Rightarrow H^{p+q}(Y) $$ and the map $\pi^*: H^*(Y) \to H^*(X)$ is realized by the "edge morphism" that takes the entries in the column $p=0$. In particular, the image $\pi^*(H^2(Y)) \subset H^2(X)$ is the collection of elements in $H^0(\Bbb Z/2; H^2(X)) = H^2(X)^\theta$ that do not support any differentials in this spectral sequence. However, it is definitely possible to have differentials that land in $H^2(\Bbb Z/2; H^1(X))$ or (a quotient of) $H^3(\Bbb Z/2; H^0(X))$.

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Yes! It follows from the fact that $$ d\pi \circ d\theta=d\pi $$ whihc in turn follows from $\pi\circ \theta=\pi$.

I think you actually have an equality $H^2(X,\mathbb{Z})^{\theta}=\pi^*H^2(Y,\mathbb{Z}).$

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    $\begingroup$ The last sentence is not true, as you can see from the Hochschild-Serre spectral sequence. $\endgroup$
    – abx
    Nov 15 '16 at 19:13
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    $\begingroup$ Thanks for your quick answer but is it still true when the coeffiecient ring is $\mathbb Z$, not $\mathbb R$? $\endgroup$
    – Lee
    Nov 15 '16 at 19:20
  • $\begingroup$ Yes, it is. E.g. you can see that from the same spectral sequence and the fact that higher cohomologies of a finite group ($\mathbb Z/2\mathbb Z$ in your case) in an $\mathbb R$-vector space are 0 $\endgroup$
    – user42024
    Nov 16 '16 at 3:33

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