Cohomology of covering spaces

Question

Let $\pi: X\to Y$ be a finite covering map between compact differentiable manifolds such that $Y=X/G$. It is well-known that $$ H^q(Y,\mathbb R)= H^q(X,\mathbb R)^G. $$ Can we expect the converse ? Precisely, can we expect that $H^q(X, \mathbb R)$ could be recovered from $H^\bullet(Y,\mathbb R)$ and $G$? Can we expect this if further conditions are assumed?

Answer 1

In general, no.

Example: Let $X_1 = \mathbb{RP}^2\times S^2$ and $X_2 = (S^2\times S^2)/\sim$ where $(x, y) \sim (-x, -y)$. Both $X_1$ and $X_2$ have $Y = \mathbb{RP}^2\times\mathbb{RP}^2$ as a $G = \mathbb{Z}_2$ quotient, but $H^*(X_1;\mathbb{R}) \not\cong H^*(X_2; \mathbb{R})$. For example, $H^4(X_1; \mathbb{R}) = 0$ because $X_1$ is non-orientable, while $H^4(X_2; \mathbb{R}) \cong \mathbb{R}$ because $X_2$ is orientable.