Is knowing a sheaf equivalent to knowing all the stalks? We have a topological space $(X, \tau)$, and a sheaf of functions $F: \tau \rightarrow \textbf{Set}$ on $X$. Now I build a new sheaf of stalks of F, defined as $G(U) \equiv \prod_{x \in X} F_x$. I feel that this $G(U)$ ought to carry the same information as $F$ --- after all, what does does a sheaf contain other than the data at the stalks? However, I'm unable to think of how to either prove or disprove this!
Roughly, I believe that one can reconstruct $F$ from $G$ using the idea of compatible germs in $G$. The elements of $F(U)$ will be those tuples $(x_i)_{i \in I} \in G(U)$ such that the $x_i$ germs are compatible with each other.
Is this intuition correct? If so, why don't we just "define" a sheaf this way, rather than going through the pain of $\text{Sheaf} \rightarrow \text{Stalk}$?
 A: The sheaf comtains the data of how the stalks are "glued together".
There is actually a precise result to that effect : if you take a sheaf $F$, then there is an essentially unique space $Y$ together with a local homeomorphism $p:Y\to X$ such that on each open $U$, $F(U) \cong \{s: U\to Y$ continuous $\mid p\circ s_{\mid U} = id_U\}$
This space $Y$ is called the etale space of $F$ and it satisfies $p^{-1}(x)\cong F_x$ for each $x\in X$
Now you can see that a general element in $\prod_{x\in U}F_x$ is unlikely to be in $F(U)$ : if all of then were in it, it would mean that almost any map $s:U\to Y$ which satisfies $p(s(u)) = u$ would automatically be continuous !
So the obstruction to $F\to G$ being an isomorphism is this sort of continuity condition, and this somehow corresponds to gluing the stalks together "along the topology of $X$"
Let's find a concrete example : let $\exp : \mathbb C \to \mathbb C^*$, and consider $F$ on $\mathbb C^*$ such that for each $U$, $F(U)$ is the set of complex continuous logarithms on $U$ (so continuous $f: U\to \mathbb C$ with $\exp\circ f =id_U$). One can compute $F_x \cong \mathbb Z$ : all the posible logarithms of $x$ are just $+2k\pi$'s of one another
On the other hand if you fix a sufficiently small open set (one where there is a logarithm), then $F(U)\cong \mathbb Z$ as well ! So very far from $\prod_{x\in U}\mathbb Z$
You can see here that the difference between the two lies in the fact that for $F$, one requires the logarithm to be continuous
