Let $F=(F\\_n)\\_n$ be an $\ell$-adic sheaf on $X\\_{et}$, for a variety $X$ over an algebraically closed field $k$ of characteristic not equal to $\ell$. Does the presheaf sending $U$ to $H^i(U,F):=\lim\\_n H^i(U,F\\_n)$ sheafify to zero?



sheafifying a projective limit of presheaves
CORRECTED ANSWER: I believe that the answer is no, at least in some contexts. 


For example, suppose that
$X = $Spec $k$, with $k$ a field, and $F = {\mathbb Z}\\_{\ell}(1)$. Then
$U = $Spec $l$ for some finite separable extension $l$ of $k$, and $H^1(U,F)
= \ell$-adic completion of $l^{\times}$, which I will denote by $\widehat{l^{\times}}$.


Thus the stalk of the presheaf $U \mapsto H^1(U,F)$ (and hence of the associated sheaf)
at the (unique) geometric point of $X$ is the direct limit over $l$ of $\widehat{l^{\times}}.$


This direct limit need not vanish. For example, if $k$ is finite, then so is $l$,
and $\widehat{l^{\times}}$ is just the $\ell$-Sylow subgroup of $l$. Thus the stalk
in this case is just $\bar{k}^{\times}[\ell^{\infty}],$ the group of $\ell$-power roots of unity in $\bar{k}$.


This fits with a certain intuition, namely that one has to go to smaller and small etale neighbourhoods to trivialize $F\_n$ as $n$ increases, and hence one can't kill of cohomology
classes in $H^i(U,F)$ just by restricting to some $V$. 



I think that the answer is yes. Here is a proof (hopefully blunder-free):


It is true for the presheaf $U \mapsto H^i(U,F\\_1).$ In other words, if we fix $U$,
then for each element $h \in H^i(U,F\\_1)$ and each geometric point $x$ of $U$,
there is an etale n.h. $V$ of $x$ such that $h\\_{| V} = 0.$ Since $H^i(U,F\\_1)$
is finite dimensional, there is a $V$ that works for the whole of $H^i(U,F\\_1)$ at once.


I claim that then $H^i(U,F\\_n)$ restricts to $0$ on $V$ as well.


To see this, consider the exact sequence $0 \to F\\_n \to F\\_{n+1} \to F\\_1 \to 0.$
Applying $H^i(U,\text{--})$ to this yields a middle exact sequence
$H^i(U,F\\_n) \to H^i(U,F\\_{n+1}) \to H^i(U,F\\_1).$ Applying $H^i(V,\text{--})$
yields a middle exact sequence
$H^i(V,F\\_n)\to H^i(V,F\\_{n+1}) \to H^i(V,F\\_1).$ Restriction gives a map from the
first of these sequences to the second. It is zero on the two outer terms, by induction
together with the case $n = 1$ proved above, and so is zero on the inner term. 


This shows that restricting from $U$ to $V$ kills $H^i(U,F\_n)$ for all $n$, and hence
$H^i(U,F)$, as required.


EDIT: As was noted in the comment below, this proof assumes that $F$ is 
${\mathbb Z}\_{\ell}$ -flat. Let me sketch an argument that hopefully handles the general case:


Put $F$ in a short exact sequence $0 \to F\\_{tors} \to F \to F\\_{fl} \to 0.$ The same kind
of argument as above reduces us to checking $F\\_{fl}$ and $F\\_{tors}$ separately. The above proof handles the case of $F\\_{fl}$, while $F\\_{tors} = F\\_{tors,n}$ for some large enough $n$,
and so the projective limit collapses in this case and there is nothing to check. 


(Note: I am assuming some basic kind of finiteness assumption on $F$ here, so that the above
makes sense. Constructibility should be enough.)



