Defining a sheaf from its values on a prebase (plus little more structure)

Question

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.

Answer 1

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.

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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...)