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Let $f \colon X \to Y$ be a map of schemes, $\mathcal{F}, \mathcal{G}$ two coherent sheaves on $Y$. I'm interested in conditions which guarantee an isomorphism $$f^{*} \mathcal{E}xt^i(\mathcal{F}, \mathcal{G}) \cong \mathcal{E}xt^{i}(f^{*} \mathcal{F}, f^{*} \mathcal{G}).$$

I know this is true for $f$ flat by [EGA III.12.3.4]. Moreover it seems to me that this is true for $f$ arbitrary when $\mathcal{F}$ is locally free and $i = 0$, essentially by the argument in [EGA I.]. I am also able to prove this in some other cases by ad hoc methods, but I could not find a general statement.

What are conditions which guarantee the existence of such an isomorphism? I am also interested in having some counterexamples at hand.

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In a pretty general situation there is an isomorphism of derived functors $$ Lf^* RHom(F,G) \cong RHom(Lf^* F,Lf^*G). $$ Further you can write down the spectral sequences converging to each side, (e.g. for the LHS it is quite simple --- $L_q f^* Ext^p(F,G)$, for the RHS it is more complicated). Now, if $f$ is flat, the both spectral sequences degenerate in the second term and give you the desired isomorphism. The same holds if $F$ is locally free. But the general statement is the above isomorphism of the derived functors.

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@Sasha: By pretty general situation are you referring to when you can resolve F by locally frees? Or are you thinking of something else? – babubba Sep 6 '10 at 21:45
Sasha, I'm sure I'm being dumb, but do you have a reference for that isomorphism between those derived functors in general? Residues and duality only asserts that statement when $f$ is flat as far as I can tell (Chapter II, Prop 5.8). I guess as John Calabrese remarks, one perhaps should be able to get away with F resolvable by locally frees? – Karl Schwede Sep 7 '10 at 2:38
John is right, it is sufficient to assume that $F$ is a perfect complex. Another alternative is to assume that $f$ has finite Tor-dimension. Maybe one can prove this even in a greater generality. – Sasha Sep 7 '10 at 17:42

Hi Andrea, I don't think one can prove much without flatness. Let's assume the simplest case, that $X=\text{Spec}(S)$, $Y= \text{Spec}(R)$, and $R\to S$ is a finite local homomorphism with $R$ regular. Then I claim what you want is equivalent to flatness.

Your condition amounts to $$\text{Ext}_R^i(M,N)\otimes_RS \cong \text{Ext}_S^i(M\otimes S,N\otimes S)$$ for $R$-modules $M,N$. There is a well-known result that the first $i$ such that $\text{Ext}_R^i(R/I,R)\neq 0$ is the length of the longest $R$-regular sequence in $I$. Let $M=R/m_R$, $N=R$, then by the Ext condition we can conclude that $m_RS$ contains a $S$-regular sequence of length equals to $\text{dim}\ S$. So $S$ is Cohen-Macaulay.

But then "miracle flatness" implies that $f$ is flat! This also provides counter-examples: if $S$ is not Cohen-Macaulay, choose $i$ to be $\text{depth}\ S$.

ADDED: for the sake of completeness, here is a class of examples to show the second nice situation ($\mathcal F$ is locally free) can't be generalized to much.

Let $Y=\mathbb A^n$, $X=V(f)\subset Y$ such that $f$ vanishes at the origin. Let $\mathcal F$ be locally free on $Y$ minus the origin, but not free at the origin. Let $\mathcal G$ be any torsion-free coherent sheaf on $Y$. Then I claim the condition you want (let's call it $(*)$) will not hold.

From the short exact sequence $0\to \mathcal G \to \mathcal G \to \mathcal G/(f)\to 0$ (the first map is multiplication by $f$, exact because $\mathcal G$ is torsion-free) one can take $\mathcal Hom(\mathcal F,-)$ to get a long exact sequence. Looking at such l.e.s, $(*)$ means precisely that the maps by multiplication by $f$: $$\mathcal Ext^i(\mathcal F,\mathcal G) \to \mathcal Ext^i(\mathcal F,\mathcal G) $$ must be injective for all $i>0$. But as all these ${\mathcal Ext}$ vanish away from the origin and $f$ vanishes at the origin, they have to be $0$. Now localize at the origin, take a minimal free resolution of the stalk of $\mathcal F$ to compute Ext and use Nakayama, one can show that $\mathcal F$ must also be free there, contradiction.

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