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.

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...)

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: $G(W) \rightarrow G(Y)$ ; $G(Y) \rightarrow G(U)$ ; $G(Y) \rightarrow G(V)$

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

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

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.

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.