Can a nontrivial presheaf have a trivial sheafification?

Question

I'm studying Vakil's Foundations of Algebraic Geometry, and working through the exercises in 2.4 about compatible germs as a method for constructing a sheafification of a given presheaf. Can this construction be vacuous? Can a nontrivial presheaf fail to have a nontrivial set of compatible germs? Can there fail to be a universal functor that maps a nontrivial presheaf to a nontrivial sheaf?

Note that, in this section of the book, Vakil goes back and forth between (pre)sheaves of sets and (pre)sheaves of abelian groups (including rings and $\mathscr{O}_X$-modules), so I would expect an answer to address either (pre)sheaves of abelian groups, or (pre)sheaves of sets more generally. In that sense, a trivial sheaf would be the zero sheaf, the sheaf that maps all open sets to the trivial group, or more generally, the one-point set.

Answer 1

If $\mathscr F$ is any presheaf (of abelian groups, or $\mathcal O_X$-modules) that is not a sheaf, and $\mathscr F \to \mathscr F^\#$ its sheafification, then the kernel $\mathscr K$ and cokernel $\mathscr C$ of $\mathscr F \to \mathscr F^\#$ have trivial sheafification. Indeed, exactness and idempotence of $(-)^\#$ show that sheafification turns the exact sequence $$0 \to \mathscr K \to \mathscr F \to \mathscr F^\# \to \mathscr C \to 0$$ into an exact sequence $$0 \to \mathscr K^\# \to \mathscr F^\# \stackrel\sim\to \mathscr F^{\#\#} \to \mathscr C^\# \to 0.$$ The easiest example is probably $X = \varnothing$ and $\mathscr F \neq 0$, for any sheaf on $X$ is trivial because of the empty covering $\{\}$ of $X$.

This is completely analogous to how the kernel and cokernel of $M \to M \otimes \mathbf Q$ are torsion for any abelian group $M$: the localisation $\mathbf{Mod}_{\mathbf Z} \rightleftarrows \mathbf{Mod}_{\mathbf Q}$ is also exact (and idempotent).

Answer 2

Let $X$ be a smooth manifold, $\Omega^k_X$ the sheaf of smooth $k$-forms on $X$. Let $H^k_{\mathrm dR}(-)$ denote the $k$th de Rham cohomology presheaf, that is the rule which associates to each open set $U\subseteq X$ the de Rham cohomology space $H^k_{\mathrm {dR}}(U)=Z^k_{\mathrm{dR}}(U)/B^k_{\mathrm{dR}}(U)$.

Then the sheafification $H^k_X:=\mathrm{Sh}(U\mapsto H^k_{\mathrm{dR}}(U))$ is the trivial zero sheaf for $k\ge 1$.

This can be seen in one of two ways, first for any $x\in X$, the stalk is $H^k_{X,x}=0$ since locally every closed $k$-form is exact, so any section $\phi\in H_{X}(U)$ satisfies $\phi_x=0$ for all $x\in X$. But if a section of a sheaf has zero germs everywhere, then it must be zero.

Or, it is known (proven in most texts on sheaf theory) that if $\mathcal F$ and $\mathcal G$ are presheaves on a space with $\mathcal G$ a subpresheaf of $\mathcal F$, then $\mathrm{Sh}(\mathcal F/\mathcal G)=\mathrm{Sh}(\mathcal F)/\mathrm{Sh}(\mathcal G)$, where on the left hand side, the quotient is taken in presheaves while on the right hand side the quotient is taken in sheaves. Thus $$ H^k_X=Z^k_X/B^k_X, $$ where $$ Z^k_X=\ker(\mathrm d:\Omega^k_X\rightarrow\Omega^{k+1}_X) = \mathrm{Sh}(U\mapsto Z^k_{\mathrm{dR}}(U)) \\ B^k_X=\mathrm d\Omega^{k-1}_X=\mathrm{Sh}(U\mapsto B^k_{\mathrm{dR}}(U)), $$with the latter being the image sheaf of $\mathrm d$, i.e. $B^k_X=\mathrm{Sh}(U\mapsto \mathrm d\Omega^{k-1}_X(U))$. By the Poincaré's lemma, we have $Z^k_X=B^k_X$ then, so the quotient sheaf is zero.