Does the push forward of the de Rahm complex remain exact? I am trying to understand the proof of the theorem of Artin-Grothendieck (theorem 3.1.13. in 'Positivity in algebraic geometry' by Lazarsfeld). Right in the beginning of the proof, he chooses a finite map $f: X \rightarrow \mathbb{C}^n.$ Here $X$ is some affine variety of dimension $n$, considered as complex analytic space. However smooth (i.e. a complex manifold) would totally suffice for me. He then claims that
$$ H^p(X, \underline{\mathbb{R}}) = H^p(\mathbb{C}^n, f_*\underline{\mathbb{R}})$$
(he does it more generally for constructible sheaves but again this would totally suffice for me). I do not really understand why this shall be true. After all, $\underline{\mathbb{R}}$ is not a coherent sheaf (it is not even a sheaf of $\mathcal{O}_X$-modules), so we can not invoke GAGA + vanishing of higher direct images in the algebraic setting. Do you have any hints for me?
Moreover, I started to wonder if in this situation the push forward of the de Rahm complex on $X$ is still exact on $\mathbb{C}^n$ (of course this would yield the result above). If I try to picture it, then I think that the preimage of sufficiently small balls down stairs should still be contractible upstairs (or at least copies of such sets). However, sadly I do not have much knowledge of branched covers of manifolds and my intuiton already failed me several times in this proof.
Any help, hints or references would really be appreciated!
 A: This may be "taking a sledgehammer to a mosquito," but the Leray spectral sequence for $f$ degenerates since the higher cohomology of the constant sheaf will vanish along the fibers (which are finite).
(More explanation if it helps --  you have a spectral sequence
$$
H^p(\mathbb{C}^n, R^qf_*\mathbb{R}) \Rightarrow H^{p+q}(X, \mathbb{R}),
$$
but for $q > 0$ the terms on the left vanish, so what you really have is an isomorphism
$$
H^p(\mathbb{C}^n, f_*(\mathbb{R})) = H^p(X, \mathbb{R}).
$$
As mentioned by @AGLearner in a comment, the appeal to the Leray spectral sequence is overkill -- given a map $f: X \to Y$ you can prove directly that the natural map
$$
H^p(Y, f_*\mathcal{F}) \to H^p(X, \mathcal{F}).
$$
is an isomorphism whenever the higher direct images of $\mathcal{F}$ vanish.
For deducing the vanishing of the higher direct images in this particular case, we just need that the fibers are discrete, so they have no higher singular cohomology (maybe we are appealing to something like Ehresmann's lemma implicitly here).
