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Lat $A$ be a set and $\underline{A}$ the associated constant Zariski sheaf on the category $Sm/S$ of schemes which are smooth over $S$ for a fixed base scheme $S$. Is $\underline{A}$ already a (constant) sheaf for the Nisnevich topology on $Sm/S$?

I ask this because constant Zariski sheaves are easier to describe, which only depend on connected components.

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    $\begingroup$ Did you try to prove this by hand? $\endgroup$ – Will Sawin Mar 30 at 15:25
  • $\begingroup$ First, I don't know the answer. A possible way to give an answer is to prove that on sections over a Nisnevich distinguished square, it gives a pullback diagram. Roughly this corresponds to analyzing the connected components, but I don't succeed with it. $\endgroup$ – Lao-tzu Mar 30 at 15:28
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Yes. This is fine for every topology in which covers are collections of morphism that are open for the Zariski topology and surjective on points.

To see this, because $\underline{A}$ satisfies the sheaf condition for disjoint unions, it suffices to show for $f: Y \to X$ open and surjective on points, $\underline{A}(X) \to \underline{A}(Y) \substack{ \to \\ \to} \underline{A}( Y\times_X Y)$ is a pullback square.

To do this, it is helpful to note that $\underline{A}(X)$ is the set of disjoint $A$-indexed open covers of $X$.

Given a disjoint $A$-indexed open cover $(F_a)_{a \in A}$ of $Y$ in $\underline{A}(Y)$, look at the image $f(F_a)$ of each set in $X$. This gives an $A$-indexed open cover of $Y$. We must check that if $(F_a)_{a \in A}$ satisfies the gluing condition, the cover of $X$ is disjoint.

In other words we must check that if $x \in X$, $y_1,y_2$ lie in the fiber of $Y$ over $X$, and $y_1 \in F_{a_1}$, $y_2 \in F_{a_2}$, then $a_1=a_2$. This follows from the existence of a point in $Y \times_X Y$ that maps to $y_1$ and $y_2$, which follows from the fact that $\operatorname{Spec} \kappa(y_1) \times_{ \operatorname{Spec} \kappa(x)} \operatorname{Spec} \kappa(y_2)$ is nonempty, where $\kappa(x)$ denotes the residue field at $x$.

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  • $\begingroup$ You said "it is helpful to note that $\underline{A}(X)$ is the set of disjoint $A$-indexed open covers of $X$", do you mean that $\underline{A}(X)$ is in fact $A$-copies disjoint union of $X$? $\endgroup$ – Lao-tzu Mar 30 at 15:59
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    $\begingroup$ @Lao-tzu No, $\underline{A}(X)$ is the set of maps from the underlying set of $X$ to $ A$ that are continuous for the discrete topology of $A$ and the usual topology on $X$. In other words the inverse image of each element of $A$ is an open subset of $X$, and these open sets are disjoint. $\endgroup$ – Will Sawin Mar 30 at 16:21
  • $\begingroup$ In fact, $\underline{A}$ also has the sheaf property with respect to fpqc, h, and V coverings. $\endgroup$ – Johan Mar 30 at 16:47
  • $\begingroup$ @Will Sawin A great answer! I would never take this perspective on constant sheaves before I see your unique proof. $\endgroup$ – Lao-tzu Mar 30 at 18:01
  • $\begingroup$ @Johan What do you mean by V coverings? $\endgroup$ – Lao-tzu Mar 30 at 18:02
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Please allow me to rewrite Will's fantastic answer below, which I hope to be easier to understand and for others' convenience.

Let $(\mathscr{C}, \tau)$ be a Grothendieck site, where $\mathscr{C}$ is a category of schemes and $\tau$ is a topology on $\mathscr{C}$ finer than the Zariski topology, all of whose covers are collections of morphisms that are open for the Zariski topology and surjective on points. Lat $A$ be a set and $\underline{A}$ the associated constant sheaf on $\mathscr{C}$ with the Zariski topology. Then $\underline{A}$ already a (constant) sheaf on $(\mathscr{C}, \tau)$.

Proof. Let $Y\xrightarrow{f}X$ be a $\tau$-cover on the site $\mathscr{C}$. Note that $\underline{A}(X)$ is the set of locally constant functions with values in $A$, hence can be identified with all disjoint $A$-indexed open covers of $X$ (given by mapping $v\in\underline{A}(X)$ to the family $\{v^{-1}(a)\}_{a\in A}$). enter image description here

Given any $u\in\underline{A}(Y)$ with $up_1=up_2$, we want to find a (unique) locally constant function $v\in\underline{A}(X)$ with $vf=u$. Of course, we have to define $v$ by letting $v^{-1}(a)=f(u^{-1}(a))$ (note that $f$ is an open map), provided that it is well-defined, that is, $f(u^{-1}(a))\cap f(u^{-1}(b))=\varnothing, \forall a\ne b\in A$.

We show this now: Otherwise, there would exist $x\in X$ and $y_1, y_2\in f^{-1}(x)$ with $u(y_1)=a, u(y_2)=b$. By scheme theory, there exists $z\in Y\times_XY$ with $p_1(z)=y_1, p_2(z)=y_2$, thus $a=u(y_1)=up_1(z)=up_2(z)=u(y_2)=b$, a contraction.

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