Pushing forward sheaves and the result on sheaf cohomology

Question

Let $f:X \rightarrow Y$ be a continuous map of topological spaces and let $\cal{F}$ be a sheaf on $X$. Is there an obvious map $H^\ast(Y,f_\ast \mathcal{F} ) \rightarrow H^\ast (X,\cal{F})$?

Answer 1

There is always a natural map $$\Gamma(Y, \mathcal{G}) \to \Gamma(X, f^{-1}(\mathcal{G}))$$ for any sheaf $\mathcal{G}$ on $Y$. However, these are both left-exact functors in the sheaf $\mathcal{G}$ (since $f^{-1}$ is an exact functor). By general (Tohoku) nonsense, there is an induced natural transformation of $\delta$-functors $H^*(Y, \mathcal{G}) \to H^*(X, f^{-1}(\mathcal{G}))$.

Take $\mathcal{G} = f_*(\mathcal{F})$ now. We get a map $$H^*(Y, f_*(\mathcal{F})) \to H^*(X, f^{-1} f_*(\mathcal{F})).$$ However, the adjunction between push-forwards and inverse images produces a map $f^{-1} f_*(\mathcal{F})) \to \mathcal{F}$. This induces a morphism in cohomology $H^*(X, f^{-1} f_*(\mathcal{F})) \to H^*(X, \mathcal{F})$. Composing the two maps just described gives the map that you ask for.