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Let $f:X\to Y$ be a morphism of schemes. If $\mathscr{F}$ is a sheaf over $X$ then I know that $$H^i(X,\mathscr{F})\cong H^i(Y,f_*\mathscr{F})\qquad\text{for all}\quad i\geq 0$$ whenever $j$ is a closed immersion or $\mathscr{F}$ is quasi-coherent and $f$ is an affine morphism between noetherian separated schemes.

I wonder if there's some similar relation for the inverse image.

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  • $\begingroup$ If $X\to Y$ is flat it will work. In general of course it won't work, think of the inclusion of a point in any scheme. Do you know about Grothendieck spectral sequence? $\endgroup$
    – Ahr
    Jan 15, 2021 at 18:02
  • $\begingroup$ @Ahr I don't expect it to work in general (even for the direct image it doesn't work in general). Do you have some reference for the fact that it works when $f$ is flat? Also no, I unfortunately do not know the Grothendieck spectral sequence. Would you mind telling me why its relevant here? $\endgroup$
    – Gabriel
    Jan 15, 2021 at 18:28
  • $\begingroup$ I talked a little bit too fast here, i tried to write down things but it's not exactly how I thought. The reason I'm mentionning Gortendieck spectral sequence, is that it makes you compute the deriviative of the composition of two functors (under mild hypothesis), here you want to compute the derivative of the composition of $f^*$ (with $f: X\to Y$) and $\pi_*$ (with $\pi$ the structura morphism to the point in the case of $k$-schemes), this is the same thing in the first case with $f_*$. $\endgroup$
    – Ahr
    Jan 17, 2021 at 9:09
  • $\begingroup$ When $f$ is affine, $f_*$ is exact and therefore the spectral sequence gives you the isomorphism you mention in the beginning, when $f$ is flat $f^*$ is exact, but it gets a bit more complicated, because $H^0(X, G)\to H^0(Y, f^*G)$ needs not be an isomorphism, it is easy to find conditions on $f$ (proper with connected fibers) and $G$ (locally free) for which it will work, but this is not enough to guarantee the result for $f$ flat. $\endgroup$
    – Ahr
    Jan 17, 2021 at 9:15

1 Answer 1

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Question: "I wonder if there's some similar relation for the inverse image."

Answer: Let $Y:=Spec(A)$ be a noetherian affine scheme and let $\mathcal{E}$ be an $\mathcal{O}_Y$-module. The isomorphism you write down above holds for any affine morphism of noetherian separated schemes and $\mathcal{F}$ quasi coherent.

Let us assume there is an isomorphism

$$ H^i(X, f^*\mathcal{E})\cong H^i(Y, \mathcal{E}).$$

The right hand since is zero for $i\geq 1$ since $Y$ is affine (Hartshorne III.3.5).

It seems the left hand side is not zero in general, but I do not have an immediate example. If $E$ is a non-trivial finite rank projective $A$-module, it follows $f^*\mathcal{E}$ will be a non-trivial locally free sheaf on $X$ and this should have non-zero cohomology groups in general if $X$ is non-affine.

Example: If $X:=Spec(B)$ there is a canonical map

$$ \rho: H^0(Y, \mathcal{E}) \cong E \rightarrow E\otimes_A B\cong H^0(X, f^*\mathcal{E})$$

defined by

$$\rho(e):=e\otimes 1$$

and the map $\rho$ is seldom an isomorphism (even when $f$ is flat).

Example: Let $A:=k$ a field and $dim(B) \geq 1$. It follows $B\cong \oplus_{i \in I}ke_i$ with $\# I=\infty$

$$E\otimes_A B\cong \oplus_{i\in I} Ee_i \neq E.$$

The map $f$ is flat since $B$ is a flat $A$-algebra. Hence the map $\rho$ is never an isomorphism for $X$ affine of positive dimension over a field $k$.

Example: If $B:=A/I$ we get a canonical map

$$\rho: E \rightarrow B\otimes_A E\cong E/IE=E$$

since $IE=(0)$. Hence when $f$ is a closed immersion the map $\rho$ is always an isomorphism.

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  • $\begingroup$ I surely do not expect this to work in general. Neither do I expect the two cohomologies to be isomorphic. Maybe they are related in some weaker sense... $\endgroup$
    – Gabriel
    Jan 15, 2021 at 18:29

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