Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 113 / 4 / ii / Solution

For the closed inclusion $f$, the direct image sheaf has stalk ${\mathcal{F}}_x$ at $x\in X$ and zero stalk at points outside $X$: outside the closed set there is an open neighbourhood disjoint from $X$. Hence $f_{\ast }$ is exact, since exact sequences of sheaves are detected on stalks.

Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Direct image preserves flasqueness, because restrictions are the restrictions on inverse-image opens. Exactness makes $f_{\ast }{\mathcal{I}}^{\bullet }$ a flasque resolution of $f_{\ast }\mathcal{F}$. MoreoverThe complexes computing sheaf cohomology are identical. ThereforeThis is sheaf cohomology under a closed inclusion; the closedness hypothesis is used in the exactness of direct image.