What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf?

Question

There is a standard way to construct the sheafification of a presheaf on a Grothendieck topology which involves matching families. Details may be found here:

http://ncatlab.org/nlab/show/matching+family

In short, there is a functor + sending presheaves to separated presheaves and then separated presheaves to sheaves. So P^++ is always a sheaf.

Gelfand/Manin's Methods of Homological Algebra has a wrong proof that P^+ is a sheaf, and I have seen in several places a proof that P^++ is a sheaf. However, it seems that for any presheaf P I run into, P^+ is already a sheaf.

Does anyone know an example of a presheaf P where P^+ is not a sheaf i.e. where you actually need to apply the functor + twice to get a sheaf?

Answer 1

I think this works:

Consider a topological space consisting of 4 points $A$, $B$, $C$, $D$, where the topology is given by open sets $ABC$, $BCD$, $B$, $C$, $ABCD$, $\emptyset$.

Then let the presheaf $\mathcal{F}$ be given by: $$\mathcal{F}(ABC)=\mathbb{Z}$$ $$\mathcal{F}(BCD)=\mathbb{Z}$$ $$\mathcal{F}(BC)=\mathbb{Z}$$ $$\mathcal{F}(ABCD)=\mathbb{Z}$$ $$\mathcal{F}(B)=\mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}(C)=\mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}(\emptyset)=0$$

where all restrictions are what you expect (identity in the case of $\mathbb{Z} \to \mathbb{Z}$ and canonical surjection in the case $\mathbb{Z} \to \mathbb{Z}/2 \mathbb{Z}$).

Then if we we get $\mathcal{F}^+$ is given by:

$$\mathcal{F}^+(ABC)=\mathbb{Z}$$ $$\mathcal{F}^+ (BCD)=\mathbb{Z}$$ $$\mathcal{F}^+ (BC)= \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}^+ (ABCD)=\mathbb{Z}$$ $$\mathcal{F}^+ (B)= \mathbb{Z}/2\mathbb{Z} $$ $$\mathcal{F}^+ (C)=\mathbb{Z}/2\mathbb{Z}$$ $$\mathcal{F}^+ (\emptyset)=0$$

where the map from $\mathcal{F}^+ (BCD)$ to $\mathcal{F}^+ (BC)$ is given by taking the canonical surjection on both copies, and other restrictions are obvious. Then note that if we take 1 over $BCD$ and 3 over $ABC$, these two are compatible over $BC$ but they do not patch.

The key point is that being compatible over a refinement is not the same thing as being compatible. That is, the way the plus construction works is by taking $F^+$ of a space to be some direct limit over open covers of guys on the covers which are compatible on intersections. If we had said instead take direct limit over open covers of guys on the covers which compatible on some refinement of the intersection, then applying just once probably works.

So in our example, 1 and 3, over $ABC$ and $BCD$, in our original presheaf were compatible on a refinement of $BC$ but not on $BC$.

Answer 2

I haven't checked carefully that I mean the same thing by + as you do, but I think the following example works.

Take X={p,q} with the discrete topology, let S be any set with |S|>1, and let F be the constant presheaf which returns S for any open subset of X (and all restrictions are identity maps). In particular, F(∅)=S.

Then it is easy to see that F⁺⁺(X)=SxS, but I claim that F⁺(X)=S. To see this, suppose you have two sections s∈S=F({p}) and s'∈S=F({q}). These section "agree on intersections" only if their restrictions in F({p}∩{q})=F(∅)=S agree (i.e. only if s=s').

Note that F⁺(∅) is a one point set because ∅ is covered by the empty cover (a covering by no sets at all, not even ∅), and any two sections of F(∅) agree on this cover, so when you take F⁺⁺, this problem doesn't happen again.

Answer 3

Anton's example can be modified to avoid the "empty set". Here's one way: let X be the category associated to the partially ordered set

a ≤ b ≤ c ≤ e

b ≤ d ≤ e

Give this the minimal topology in which c and e cover d and a covers b. Let F be the presheaf with F(a) = 1, F(b) = F(c) = F(d) = S, F(e) = ∅. Then F⁺(b) = 1, F⁺(c) = F⁺(d) = F⁺(e) = S. This is not a sheaf, since F⁺⁺(e) = S x S.

Of course, all I have done is to introduce an object a to play the role of the empty cover in Anton's example. It's probably worth remarking, however, that what makes the empty set empty (from the point of view of sheaves) is that it is covered by the empty cover. In fact, the usual topology on the category of open subsets of {p,q} can be modified so that the empty subset is not covered by the empty cover. With respect to this topology, the presheaf F that Anton defined is already a sheaf.

Answer 4

An example is given in MacLane's "Sheaves in Geometry and Logic." Consider the constant presheaf on a space $X$ with $P(U) = S$ where $S$ is a set with more than one element and restriction maps are identities. The plus construction doesn't change anything except that $P(0) = 0$, and one can show easily that this is not a sheaf.

In general it is true that the plus construction turns separated presheaves into sheaves and any presheaf into a separated presheaf; hence ++ = sheafification.