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(This is 20.7.B in Ravi Vakil's notes)

Suppose $f:Y \to Z$ is any morphism, and $\pi: X\to Y$ is quasicompact and quasiseparated. Suppose $\mathcal{F}$ is a quasicoherent sheaf on $X$. Let

$$\require{AMScd}\begin{CD} W @>{f'}>> X \\ @V{\pi'}VV @VV{\pi}V \\ Z @>>{f}> Y \end{CD}$$

be a fiber diagram. Describe a natural morphism $f^*(R^i\pi_*\mathscr{F}) \to R^i\pi'_*(f')^* \mathscr{F}$ of sheaves on $Z$.

I'm not sure how to work with higher direct image sheaves at all, so help would be appreciated.

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  • $\begingroup$ I fixed your commutative diagram and adjusted your notation to match Vakil's. $\endgroup$ – Zhen Lin Jul 27 '12 at 11:59
  • $\begingroup$ Thanks. I'm not familiar with latexing diagrams. $\endgroup$ – only Jul 27 '12 at 12:01
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When the bases are affine, say $Y=Spec(A)$ and $Z=Spec(B)$. Then you can work out or check somewhere (maybe in Hartshorne, I can't remember) that the higher direct images are just the coherent sheaves associated to taking cohomology. Thus the left hand side becomes $f^*(H^i(X, \mathcal{F}))=H^i(X, \mathcal{F}) \otimes_A B$ (tildes weren't working, so imagine they are there).

The right hand side becomes $H^i(W, (f')^*\mathcal{F})$. Thus the natural map is the one that comes from the one induced by pullback on sheaf cohomology via $f'$, since $f':W\to X$.

Now using quasi-coherent you can reduce to this case.

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