As the title says, if $f: X \to Y$ is a continuous map, when do we have an isomorphism in sheaf cohomology $$H^k(X, \mathcal{S}) \cong H^k(Y, f_* \mathcal{S})?$$ Under what conditions on the map $f$ and the sheaf $\mathcal{S}$ is this true?
The reason I am curious is because I know that if $i: S \hookrightarrow X$ is the inclusion of a closed subspace of a (locally compact?) topological space $X$, $j: X \setminus S \hookrightarrow X$ the inclusion of the complement, and $\mathcal{F}$ is an abelian sheaf on $X$, then we have a short exact sequence of sheaves $$0 \to j_!j^* \mathcal{F} \to \mathcal{F} \to i_* i^* \mathcal{F} \to 0.$$ In this situation I am curious if we have an isomorphism in sheaf cohomology $$H^k(S, i^* \mathcal{F}) \cong H^k(X, i_* i^* \mathcal{F}).$$ In his answer Liviu Nicolaescu seems to imply that this is the case if $\mathcal{F}= \underline{A}$ is a constant sheaf for an abelian group $A$, since $H^k(X, i_*i^* \underline{A}) \cong H^k(S, A)$ where the right hand side is singular cohomology.
Posting my comment as an answer: The relationship between $H^k(X,\mathcal S)$ and $H^k(Y, f_* \mathcal S)$ is encoded by the Leray spectral sequence. A sufficient condition for them to be isomorphic is the vanishing of the higher direct image sheaves $R^i f_* \mathcal S$ ($i>0$).