# Is any constant Zariski sheaf already a Nisnevich sheaf?

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.

• Did you try to prove this by hand? – Will Sawin Mar 30 at 15:25
• 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. – Lao-tzu Mar 30 at 15:28

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 $$x$$.

• 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 $X$? – Lao-tzu Mar 30 at 15:59
• @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. – Will Sawin Mar 30 at 16:21
• In fact, $\underline{A}$ also has the sheaf property with respect to fpqc, h, and V coverings. – Johan Mar 30 at 16:47
• @Will Sawin A great answer! I would never take this perspective on constant sheaves before I see your unique proof. – Lao-tzu Mar 30 at 18:01
• @Johan What do you mean by V coverings? – Lao-tzu Mar 30 at 18:02

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 $$\{v^{-1}(a)\}_{a\in A}$$). 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.