Finite pushforward commute with taking cohomology

Question

Let $f: X \to Y$ be a finite morphism of schemes. How one can show that $f_*H^i(G) \cong H^i(f_* G)$ for any $G \in D(X)$ and any $i \in \mathbb{Z}$? In english, $G$ is a complex of quasi-coherent sheaves on $X$, $H^i$ means taking cohomology of this complex (but not hypercohomology!)

Answer 1

Since $f$ is finite, $f_*$ is exact. (This is more or less exercise III.8.2 in Hartshorne -- namely $R^1f_* = 0$. More generally, this property is true for affine morphisms.)

Taking (co)homology of a complex commutes with exact functors since cohomology is computed using kernels and cokernels (which exact functors preserve).