# Does the induced map $\pi^*:H_{DR}^{k}(X)\rightarrow H_{DR}^{k}(\tilde{X})$ injective?

Let $\pi:\tilde{X}\rightarrow X$ be a covering mapping of differential manifolds,with $\tilde{X}$ is compact.

Does the induced map $\pi^*:H_{DR}^{k}(X)\rightarrow H_{DR}^{k}(\tilde{X})$ injective?

If not, under what conditions it will be injective ?

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If you have a morphism $\pi$ with finite generic fiber of cardinal $d$ between compact manifolds, then you have $\pi_* \circ \pi^* = d Id_{H^k(Y,\mathbb Z)}$, where $\pi_*$ is the Gysin morphism. In particular $\pi^*$ is injective.

Beside, one can still say something in the case where the generic fiber has positive dimension: if $\pi$ a surjective morphism between compact Kähler manifolds, then the induced map in cohomology is injective (this is proved in Hodge theory and Complex Algebraic Geometry by Voisin for instance, lemma 7.28 in the french version).

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It's easier to understand the injectivity when $X=\tilde{X}/G$ where $G$ is a finite group acting freely on $\tilde{X}$. The image of $\pi^*: \Omega^k(X)\to\Omega^k(\tilde{X})$ is the space $\Omega_G^k(\tilde{X})$ consisting of $G$-invariant forms, i.e., forms $\omega\in\Omega^k(\tilde{X})$ such that

$$g^*\omega=\omega,\;\;\forall g\in G.$$

Note also that the resulting map

$$\pi^*: \Omega^k(X)\to\Omega^k_G(\tilde{X})$$

is a bijection. Suppose $\omega\in\Omega^k(X)$ is a closed form on $X$ such that $\pi^*\omega$ is exact, i.e.,

$$\pi^*\omega= d\tilde{\eta}.$$

Set

$$\bar{\eta}:=\frac{1}{|G|}\sum_{g\in G} g^*\tilde{\eta}$$

By construction $\bar{\eta}$ is $G$-invariant and

$$d\bar{\eta}= \frac{1}{|G|} \sum_{g\in G} g^*d\tilde{\eta}= \frac{1}{|G|}\sum_{g\in G} g^* \pi^* \omega =\pi^* \omega$$

since $\pi^*\omega$ is $G$-invariant.

Since $\bar{\eta}$ is $G$-invariant, there exists $\eta\in \Omega^{k-1}(X)$ such that $\pi^*\eta=\bar{\eta}$. Hence

$$\pi^* \omega=d\pi^* \eta \Rightarrow \pi^*(\omega-d\eta)=0.$$

Since $\pi^*:\Omega^k(X)\to\Omega^k(\tilde{X})$ is injective we deduce $\omega=d\eta$ , i.e., a closed form on $X$ is exact if and only if its pullback to $\tilde{X}$ is exact.

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