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.

  • 1
    $\begingroup$ 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

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 $x$.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.