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Let $X$ be a topological space and $\mathcal{B}$ a base of the topology (i.e. it is closed under intersection and every open set is the union of elements from $\mathcal{B}$). Any functor from $\mathcal{B}$ to, say abelian groups, extends to a sheaf in, at most, one way. I.e., if the functor is not already contradicting the sheaf condition, it will extend to a sheaf in a unique way.

Now, let us assume that $\mathcal{P}$ is a prebase. That is to say, we first have to take all the finite intersections of the subsets of $\mathcal{P}$ for it to become a base.

Assume we have a functor from $\mathcal{P}$ to, say abelian groups, and assume that it can be extended to a sheaf. (I.e., the sheaf condition is not already contradicted). Will this be unique? Or is the prebase too little information?

I think it is too little (I am fairly sure), BUT, we have an two additional condition(s).

(First): Assume that for all finite intersections $\bigcap\limits_{i\in I}U_i, U_i\in\mathcal{P}$, there exists a $V\in\mathcal{P}$, such that $V\subseteq \bigcap\limits_{i\in I}U_i$.

(Second): In particular, (I think this follows from the additional condition, if not, this condition STILL holds), we know the values at the stalks.

Is this enough to uniquely define a sheaf? At the very least, to define one in a canonical way?

My attempt was to try and cover the intersections with open subsets from the prebase and then cover their intersections etc. and continue. But I would need a strong finiteness condition for this to eventually stabilise, which I do not think is satisfied.

Any ideas? Cheers.

EDIT: The "delted" text reflects the answer by Simon Henry.

Edit 2: Question fully answered by Simon Henry. (Thanks). Edit 3: Thanks for the clarification under the answer. That too is a rather interesting construction that I was not aware of. I am not sure if it will be helpful for the immediate question, but is definitely a fact worth knowing! Cheers.

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  • $\begingroup$ You don't need to edit the question to mark it answered: as long as you mark the "tick" of an answer (which you did) the system will automatically signal that the question has been answered to the satisfaction of the OP. $\endgroup$ – Denis Nardin Mar 12 '18 at 14:08
  • $\begingroup$ K. So that means that I should not tick the "answer" button, if I feel like the question is only answered to 90%, so to speak? Also, what if one person answer 90%, then an other answers the remaining 10%, who should I consider to be the one who answered? It cannot be both can it? (if me answering essentially closes the question down)? $\endgroup$ – I.P Mar 12 '18 at 14:13
  • $\begingroup$ That's up to you. The "tick" button only signals that the question has been answered to your satisfaction. It does not forbid people to add further answers. You can be the only one to judge if an answer satisfies you. In practice edge cases like you describe are quite rare (altjhough I agree, it's kind of a pickle when they happen). $\endgroup$ – Denis Nardin Mar 12 '18 at 14:17
  • $\begingroup$ K. Then I will tend to thank the answerer "in person", so to speak, and promise them to tick it as an answer after the remaining part has been answered, or, once a few days go by. (Cause I think less people are going to look at a topic once it has been "answered"). $\endgroup$ – I.P Mar 12 '18 at 14:20
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What people usually call a base of the topology is a family $P$ such that if you have a finite set $U_i \in P$ then there is a covering of $\cap U_i$ by elements of $P$. you do not necessarily need $P$ to be stable under intersection.

This is stronger than the condition you are asking, but this is the correct condition for having this kind of property.

For a counter example under the condition you are asking, I believe essentially anythings that does not satisfies the "base" condition above would do, here is the simplest example: consider the space with four points $x,y,z,t$ and the following pre-base:

$U = \{ x ,y ,t\}$; $V = \{ z,y,t \}$; $W = \{ t\}$.

It satisfies your condition. The open are these three, $\emptyset$, the whole space $X$ and $Y=U \cap V = \{ y,t\}$.

A sheaf is the data of a diagram of groups $G(W)$,$G(V)$,$G(U)$ and $G(Y)$ with maps: $G(W) \leftarrow G(Y)$ ; $G(Y) \leftarrow G(U)$ ; $G(Y) \leftarrow G(V)$

and $G(\emptyset)$ and $G(X)$ are automatically defined by the sheaf condition $G(\emptyset)=\{0\}$ and $G(X) = G(U)\times_{G(Y)} G(V)$.

A sheaf in your sense, is omiting $G(Y)$, and only have a map $G(W) \leftarrow G(U) \coprod G(V)$.

any group $G(Y)$ that factor the maps above is a possible extension as a sheaf.

Also note that the value of the stalk at $y$ is $G(Y)$, so the value of the stalk is not determined by the value of $G$ on the pre-base as you claimed.


Another interesting possibility:

If you have a prebase of the topology, that you know the sections of your presheaf on every element of the prebase, that you know the stalk at every point, and how the set of sections are mapped to the stalk.

Then you can do the following construction: Lets call $X$ your base base space and $F$ your wannabe sheaf.

You Define $Et F$ to be the topological space whose points are the pair $(x \in X, f \in F_x)$.

For each section $s \in F(U)$ you have an open subset $V_{U,s} = \{x \in U, f = s_x \}$. And you take the topology generated by those. Then you have a continuous map from $Et F \rightarrow X$, mapping each pair $(x,f)$ to $x$.

If the data you started from comes from a sheaf, then this gives you the étale space of that sheaf. and hence you recover the full sheaf by looking at the locale sections of the map $Et F \rightarrow X$.

But it seems hard to say what kind of conditions the data you start from need to satisfies in order that this construction is well behaved (if you start from random data, then $Et F$ will not be étale over $X$, the section you start from might not be continuous etc...)

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  • $\begingroup$ Thank you very much. It made things a lot clearer for me. But if I may add, since I had also said it in the question (I will edit the question now, to reflect your answer), what if I also know all the stalks? This does not derive from my condition, but it too is satisfied in my case. Thank you again for the time you took to answer my question. $\endgroup$ – I.P Mar 12 '18 at 13:44
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    $\begingroup$ In the example above, knowing the value of th stalk give you value of $G(Y)$, but it does not tell you how it relates to the other $G(W)$ and $G(U)$ and $G(V)$. I suspect that if you know the value of each stalk it is because you have arbitrarily small neighborhood of each points in your prebase. This suggest that maybe the stronger condition I state in the begining is satisfied ? (If your prebase contains a foundamental system of neighborhood of each points then it automatically satisfies the condition for being a base ) $\endgroup$ – Simon Henry Mar 12 '18 at 13:58
  • $\begingroup$ Thank you again for taking the time to answer (I do mean that btw. time is one of the most precious things we have on this earth, so to spend it on helping other is very much appreciated). I will have to think about it. I will post whether the stronger condition you have mentioned is satisfied. But this definitely answers my question in full. Thank you! $\endgroup$ – I.P Mar 12 '18 at 14:01
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    $\begingroup$ You're welcome. Your edit made me think about a second answer : if you know the stalk at each points, the section on a prebase, and how the sections are maped to the stalk, then there is a canonical way to construct a sheaf : just define the étale space ! (take the disjoint union of each stalk and put a topology where any section gives an open subset, this will be a prebase again, and it seems it will be étale over the base). But it is very unclear to me under what kind of condition you can hope that the section of your new sheaf will be the one you started from. $\endgroup$ – Simon Henry Mar 12 '18 at 14:07
  • $\begingroup$ Oops sorry, all the arrow were backward in my orginal answer. I've edited (I have been working with cosheaves lately...) $\endgroup$ – Simon Henry Mar 12 '18 at 14:14

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