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I know this question is absolutely trivial, but having self-studied the subject I feel extremely unsure on the basics. Do not hesitate to downgrade the question, if you feel it deserves so.

Given a morphism of complex of sheaves $\varphi:\mathcal{F}^\bullet\to \mathcal{G}^\bullet$ on a topological space $X$ (eventually as nice as needed), we have two induced morphisms, one in local cohomology, $\mathcal{H}^\bullet(\varphi):\mathcal{H}^\bullet(\mathcal{F}^\bullet)\to \mathcal{H}^\bullet(\mathcal{G}^\bullet)$, and one in global cohomology $H^\bullet(\varphi):H^\bullet(X;\mathcal{F}^\bullet)\to H^\bullet(X,\mathcal{G}^\bullet)$.

The question is: does $\mathcal{H}^\bullet(\varphi)=0$ imply $H^\bullet(\varphi)=0$?

I guess so, since there is a spectral sequence with $E_2$-page given by cohomology with cefficients in local cohomology abutting to global cohomology. But as I said, I do not trust myself too much on self-study..

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By local cohomology, I assume you mean the cohomology sheaves of the complex $\mathcal{F}^\bullet$. So then yes, your surmise is correct. You can use the spectral sequence as you said. –  Donu Arapura Nov 26 '10 at 21:46
    
Perhaps not (in view of Torsten's example). I suppose you should require $\phi$ to vanish in the derived category of sheaves.... –  Donu Arapura Nov 26 '10 at 22:10
    
What do you mean by local cohomology? –  Harry Gindi Nov 26 '10 at 22:37
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I just want to explain why the Donu's comment is wrong. This is because the spectral sequence converges not to the global cohomology, but to its factors with respect to a certin filtration. So, from vanishing of the map on local cohomology you can deduce its vanishing on factors of global cohomology. In other words you can deduce the fact that the filtration is shifted. –  Sasha Nov 27 '10 at 4:25
    
Hi Harry. I meant the cohomology sheaf. sorry for the inaccuracy. Sasha, that was a great comment! Absolutely obvious now that I see it, but it was it that made me see the light! if only books had more caveats like this! :) Just to see if I got it: in general it can only be said $\varphi:F^pH^{p+q}(\mathcal{F}^\bullet)\to F^{p+1}H^{p+q}(\mathcal{G}^\bullet)$, right? But then I would have vanishing of $\varphi$ on global cohomology in the exceptional case the spectral sequences abutting to hypercohomologies of $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$ would degenerate at $E_1$, right? –  domenico fiorenza Nov 27 '10 at 7:53

1 Answer 1

up vote 5 down vote accepted

The exact sequence $0\rightarrow\mathrm{Z}/p\rightarrow\mathrm{Z}/p^2\rightarrow\mathrm{Z}/p\rightarrow0$ ($p$ a prime say) gives a map $\mathrm{Z}/p\rightarrow\mathrm{Z}/p[1]$ in the derived category of abelian groups (which can be realised as a map of complexes). It clearly induces zero on cohomology. We can then take any topological space $X$ and pull this back to $X$ giving a map (in the derived category of sheaves on $X$) between constant sheaves again inducing zero in (local) cohomology. The maps $H^i(X,\mathrm{Z}/p)\rightarrow H^{i+1}(X,\mathrm{Z}/p)$ induced by this map are the Bockstein operations. Hence we can let $X$ be any space with non-zero Bockstein map. A particular case for $p=2$ are the real projective spaces of dimension at least $2$.

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Thanks. I guess with sheaves of vector spaces this cannot happen, and local triviality implies global triviality, right? –  domenico fiorenza Nov 26 '10 at 21:50
    
No. The same can be with vector spaces as well. For example consider an extension of $O$ by $O(-2)$ on $P^1$. It can be viewed as a map $O \to O(-2)[1]#. The map on local cohomology is zero, but on global is not. –  Sasha Nov 27 '10 at 4:27

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