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Let $X$ be a "reasonable" scheme (I am particularly interested in smooth algebraic varieties over a field). Let $Zar_X$ denote the (small) Zariski site of (open subschemes of) $X$ and $Et_X$ denote the (small) etale site of (etale schemes over) $X$. There is a natural map of sites $\pi:Et_X\to Zar_X$ (the direction of the site map is opposite to that of the functor on the categories of open sets).

Hence to any Zariski sheaf $A$ on $X$ one can assign its inverse image $\pi^*A$, which is an etale sheaf on $X$. There is also the adjoint functor $\pi_\ast$. It appears that $\pi_*\pi^*A=A$, so the functor $\pi^*$ is fully faithful. What is its essential image?

Given an etale sheaf $B$ on $X$, for any scheme point $i_x:x\to X$ there is the inverse image $i_x^*B$, which is an etale sheaf over $x$. Since $x$ is the spectrum of a field (namely, the residue field $k(x)$ of $X$ at $x$), an etale sheaf over $x$ can be viewed as a discrete module over the absolute Galois group $G_x$ of the field $k(x)$.

When the sheaf $B$ has the form $B=\pi^*A$, all the $G_x$-modules $i_x^*B$ are trivial (in the sense that the action of $G_x$ is trivial). Is the converse true?

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The "obvious" (i.e. the inclusion) map of sites goes in the opposite direction, no? –  Harry Gindi Nov 5 '10 at 19:32
    
It's explained in the question. The category of open subsets of a topological space depends contravariantly on the topological space. Same with sites. –  Leonid Positselski Nov 5 '10 at 19:38
    
@Leonid: I don't think so. The map of topoi goes in that direction (it is an inclusion map, since every étale sheaf is a fortiori a Zariski sheaf). The map of sites goes in the opposite direction. –  Harry Gindi Nov 6 '10 at 2:50
    
I don't know what toposes are and cannot discuss things in their terms. For sheaves of abelian groups, the functor from etale sheaves to Zariski sheaves that you describe is called the direct image. It is distinguishable from an inverse image in that the direct image is not exact. In the case of the spectrum of a field, etale sheaves can be identified with G-modules (G being the absolute Galois group), Zariski sheaves are abelian groups, and the functor from etale sheaves to Zariski sheaves is the functor of G-invariants. –  Leonid Positselski Nov 6 '10 at 13:35

1 Answer 1

I don't think so. Let $X$ be $\mathbb A^1_{\mathbb C}$ with two points glued together, and let $Y$ be the standard double étale cover, obtained by identifying two copies of $\mathbb A^1$. I claim that the étale sheaf defined by $Y$ does not come from the Zariski topology. This follows from the fact the stalk at the node of the restriction of the sheaf to the Zariski topology is empty, while the stalk of the étale sheaf consists of two points.

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Sorry, what is meant by "etale cover obtained by identifying two copies of $\mathbb{A}^1$? –  Leonid Positselski Nov 5 '10 at 18:23
    
I think I've understood your example now, thank you. Do you know a counterexample with $X$ being a regular scheme? –  Leonid Positselski Nov 5 '10 at 18:32
    
I think the answer is positive for regular schemes. I will try to post a proof tomorrow. –  Angelo Nov 5 '10 at 22:10
    
By the way, sorry, I had not noticed that you were asking specifically about smooth varieties. –  Angelo Nov 5 '10 at 22:14

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