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:


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?

  • $\begingroup$ Just to stimulate interest, the sheaf P^++ is indeed the sheafification of P. So this expresses the sheafification endofunctor as the square of the separation endofunctor; pretty cool! $\endgroup$ – Andrew Critch Oct 15 '09 at 17:44
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    $\begingroup$ I'm under the impression that if you want an example not of the type in Anton's response, you need a Grothendieck topology which doesn't come from a topological space. $\endgroup$ – Reid Barton Oct 15 '09 at 20:01
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    $\begingroup$ Do you think so? I was hoping that there is a way to build a presheaf on a space that stays nonseparated as you keep restricting to finer and finer open covers, and that this will cause us to miss some sections when we apply +. $\endgroup$ – user332 Oct 15 '09 at 20:14
  • $\begingroup$ It's not so much that I believe it to be true as that I seem to recall reading it somewhere. But I don't remember where, sorry... $\endgroup$ – Reid Barton Oct 15 '09 at 23:05

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 $BC$.

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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.

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    $\begingroup$ This seems to work, but it wasn't really what I had in mind. I was hoping for something like a presheaf which stays nonseparated as we restrict to arbitrarily small open sets. Suppose we revise the question and assume that P(∅) = one point. Can we find an example then? $\endgroup$ – user332 Oct 15 '09 at 19:39
  • $\begingroup$ Is this example why some texts (such as Hartshorne) require a presheaf to send ∅ to the final object? That always bugged me. But I guess it simplifies the construction of the sheafification (when your Grothendieck topology comes from a topological space). $\endgroup$ – Reid Barton Oct 15 '09 at 19:44
  • $\begingroup$ I think the key point behind the + functor is that it allows us to do sheafification without referring to stalks, which is something we may not have for a general Grothendieck topology. As I recall, Hartshorne's construction of the sheafification is different and makes use of stalks in an essential way. In any case, I don't think that concerns about P(∅) tell the full story. I'm not satisfied with the above example because I suspect that there is another reason why P^+ can fail to be a sheaf which has nothing to do with P(∅) not being a point. $\endgroup$ – user332 Oct 15 '09 at 19:55
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    $\begingroup$ I don't know an example that doesn't use ∅ in this sneaky way. I think it's silly to assume that a presheaf sends ∅ to the final object; I prefer to just say a presheaf is a functor from the topology (thought of as a category) to some other category. For a sheaf, the sheaf axiom already implies that ∅ goes to the final object. $\endgroup$ – Anton Geraschenko Oct 15 '09 at 19:57
  • $\begingroup$ I believe if you translate Hartshorne's construction of his + functor, by replacing elements of stalks with equivalence classes of representing pairs (U,s), you get everyone else's + functor. Taking H^0 and passing to the limit of refinements is the same as taking compatible functions to the union of stalks. The compatibility for these functions becomes interesting exactly when you have a cover by disjoint open sets. (Note also the bizarre statement of Hartshorne Chapter 2, exercise 1.1.) $\endgroup$ – S. Carnahan Oct 16 '09 at 2:17

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.

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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.

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