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 



compact support cohomology and finite morphism
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.


