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 $H^\ast(Y,f_\ast \mathcal{F} ) \rightarrow H^\ast (X,\cal{F})$?
Pushing forward sheaves and the result on sheaf cohomology
1 Answers
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.
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0Thanks this is very clear. – 2011-02-09