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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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It would be best if you wrote down in the question the argument in detail that you are thinking of, and then we can answer something useful. Otherwise, the answer is yes for sufficiently general meanings of the word "usual" and this does not help you or anyone! –  Mariano Suárez-Alvarez Nov 27 '11 at 18:35
    
(It is usually best if titles are not entirely composed of TeX formulas, by the way) –  Mariano Suárez-Alvarez Nov 27 '11 at 18:42
    
I have edited my question accordingly. –  user5262 Nov 27 '11 at 18:45

1 Answer 1

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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Left adjoints are right exact, so if $F$ is not exact then it must fail to be left exact. So there is a monomorphism that is sent to a non-monomorphism $F(m)\colon F(A) \to F(B)$ under $F$. If there are enough injectives then a morphism is monic if and only if all embeddings of the domain into an injective lift over it, so there is $F(A) \rightarrowtail I$ with $I$ injective such that $F(m)$ doesn't lift. By adjointness $G(I)$ cannot lift over $m$, so $G(I)$ isn't injective. The other direction is proved in Weibel, somewhere in chapter II, I think. –  t.b. Sep 8 '12 at 18:02
    
Thank you, I proved the other direction on my own, but I just found it in Weibel, I'll edit my post accordingly. –  John Stalfos Sep 8 '12 at 20:17

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