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Let X be a smooth complex algebraic variety and $\pi: X \to Y$ a finite morphism. Is it true that

$H_c^k(X, \mathbb{Q})=H_c^k(Y, \pi_\ast\mathbb{Q})$?

Here $H_c$ stands por cohomology with compact support. If so, one does one prove it? Thanks

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Yup, because finite maps are proper. – Donu Arapura Jun 9 '13 at 21:28
up vote 2 down vote accepted

Yes, this is true.

In this case, the functor $\pi_\ast$ is exact and equal to the pushforward with compact support $\pi_!$. Denote by $p$ be the unique map from $Y$ to a point and let $p_!$ be the derived functor of pushforward with compact support. So

$$ H^k_c(Y,\pi_\ast \mathbb Q) = H^k(p_! (\pi_! \mathbb Q)) = H^k ((p \circ \pi)_! \mathbb Q) = H^k_c(X, \mathbb Q), $$

as required.

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