(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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.