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In proving the formal function theorem, Grothendieck uses a rather technical lemma in EGA 0-III.13:

Lemma: Let $\mathcal{F}_n$ be an inverse system of sheaves on a space $X$ with surjective transition maps such that there is a basis on which the $\mathcal{F}_n$ all have no cohomology, and such that $H^\bullet(X, \mathcal{F}_n)$ satisfies the ML condition for each index. Then the natural map $$H^\bullet(X, \mathcal{F}) \to \varprojlim H^\bullet(X, \mathcal{F}_n)$$ is an isomorphism.

This is used to compare cohomology of the formal completion of a sheaf along a closed subscheme $Y$ with the limit of the pull-backs to the nilpotent thickenings $Y_n$. (The statement in EGA is actually a bit stronger, but in practice the sheaves are all quasi-coherent and at least for the application I have in mind, this suffices.)

I found the proof in EGA hard to digest: it uses the Godement resolution of a sheaf, flasqueness, as well as a result on inverse limits of complexes. Is there a cleaner, more natural approach to this result (e.g. using a spectral sequence possibly involving the derived functors of the inverse limit)?

Edit: OK, I just realized something: in the case where $X$ is separated (which is what matters for the fft), and the $\mathcal{F}_n$ are quasi-coherent, this is clear because we can just use a fixed Cech resolution (with respect to some affine cover) and then use the result in EGA on inverse limits of complexes. I'm still curious why the general result should be true though.

Edit 2 (and 3): Actually, my reasoning in the previous edit was incomplete. While it is then clear that, if the $\mathcal{F}_n$ are quasicoherent and $X$ separated, we have $H^\bullet(\mathfrak{A}, \mathcal{F}) = \varprojlim H^\bullet(\mathfrak{A}, \mathcal{F}_n)$, one should check that this is true for derived functor cohomology as well. But for this it suffices to show that the Cech cohomology is zero on any affine (for then one can "bootstrap" via the Cech-to-derived-functor spectral sequence as in EGA III.1 to see that derived functor cohomology of $\mathcal{F}$ is zero on any open affine, hence a Cech resolution by affines is sufficient), and this we have just checked. So formal arguments do handle the case of quasi-coherent sheaves. Sorry for the repeated edits!

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    $\begingroup$ It should be easy to modify Jannsen's paper on 'continuous etale cohomology' to get the formalism of the sort you want. On the other hand, proving that $lim^1$ is zero is more or less the same as the proof you've seen, I think. I'm not sure why you don't like the Godement resolution, if you don't mind a fixed Cech resolution. $\endgroup$
    – Minhyong Kim
    Commented May 17, 2011 at 16:21
  • $\begingroup$ Thanks! I'll take a look at that paper. I guess I was curious about a proof that would generalize to arbitrary sites; that was the main reason I was uneasy about the Godement resolution. (I guess if Jannsen does it for etale cohomology than presumably the Godement resolution cannot be used...) $\endgroup$
    – Akhil Mathew
    Commented May 17, 2011 at 16:30
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    $\begingroup$ I'm not sure why this question was voted down. It seems to be pretty clearly stated. $\endgroup$
    – S. Carnahan ♦
    Commented May 17, 2011 at 16:37
  • $\begingroup$ I agree. This is a precise question and leads to the sort of formalism that appears in Jannsen's paper; worth learning, I believe. By the way, Akhil, I think you can compute etale cohomology using a Godement resolution, unless there's some exotic situation I'm forgetting (perhaps some set-theoretic subtlety). $\endgroup$
    – Minhyong Kim
    Commented May 17, 2011 at 16:49
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    $\begingroup$ @Akhil: A very slightly different explanation is the following: if X is a topos, and F_n is an inverse system of abelian sheaves in X, then one always has an identification RGamma(X,Rlim_n F_n) =~ Rlim_n RGamma(X,F_n) (by comparing derivatives along the two evident paths Ab(X)^N ---> Ab(X) ---> Ab and Ab(X)^N ----> Ab^N ----> Ab). Now the two assumptions made in the lemma are exactly the ones needed to change each of the Rlim's in the above equality into an honest limits. $\endgroup$
    – Bhargav
    Commented Jun 11, 2012 at 20:40

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