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Given a closed immersion $i: Z \hookrightarrow X$ a coherent sheaf $\mathcal{F}$ on $X$ and a coherent sheaf $\mathcal{G}$ on $Z$, do we have $\mathrm{Ext}^n(\mathcal{F}, i_*\mathcal{G}) = \mathrm{Ext}^n(i^*\mathcal{F}, \mathcal{G})$? For $n = 0$ it is the usual adjunction, so can we deduce it by the usual "universal $\delta$-functor" argument?

Consider the exact $\delta$-functors $\mathrm{Coh}(Z) \to (Ab), \mathcal{G} \mapsto \mathrm{Ext}^n(\mathcal{F}, i_*\mathcal{G})$ and $\mathcal{G} \mapsto \mathrm{Ext}^n(i^*\mathcal{F}, \mathcal{G})$ (exact since $i_*$ is exact). They coincide for $n = 0$, so we just have to check if they are both effacable to coincide for every $n$.

Could we also derive this using derived categories?

Edit: It seems to be wrong: Take $i$ the inclusion of a closed point, then the RHS is always trivial, but I don't think the LHS is. E.g. $\mathrm{Ext}^1(k(x),k(x)) = T_x$ the Zariski tangent space. So where does the above "argument" go wrong?

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    I have edited my que$s$$t$ion accordingly.2011-11-27

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I feel somewhat weird answering your question almost a year later, but here it goes:

I've edited this a bunch of times, and I think the issue is that the left functor, $\mathrm{Hom}(\mathscr{F},i_*(-))$ is not effaceable and not universal. $i_*$ doesn't necessarily preserve injectives, in fact it preserves injectives iff $i^*$ is exact,(If $(\mathscr{F,G})$ is an adjoint pair of additive functors, then $\mathscr{G}$ preserves injectives if and only if $\mathscr{F}$ is exact, the 'only if' when we have enough injectives. See Weibel, An introduction to homological algebra, Proposition 2.3.10 for proof of the 'if' case, and the comment by t.b. below for the 'only if' case).

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    Thank you,$I$proved the other direction on my own, but I just found it in Weibel, I'll edit my post accordingly.2012-09-08