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?
Best Answer
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.