Some question of sheaf generated by sections Let $\mathcal{F}$ be a sheaf of abelian groups on a topological space $X$.
Let $B= \cup_{U\subseteq X} \mathcal{F}(U)$ and let $A$ be the set of all finite subset of $B$. For each $\alpha \in A$,
let $\mathcal{F}_\alpha$ be subsheaf of $\mathcal{F}$ generated by the section in $\alpha$ (over a various open sets).
But I don't know why $\mathcal{F}_\alpha$ is a sheaf....
Is it true that for an open subset $U$, $\mathcal{F}_\alpha(U)$ be group generated by $\alpha \cap \mathcal{F}(U)$??
 A: $F_{\alpha}$ is a sheaf by definition ("let $F_{\alpha}$ be the sub*sheaf* generated by ..."). Here is some elabration of this definition:
The set $\alpha$ consists of finitely many (local) sections of $F$, say $s_1 \in F(U_1)$, $\dotsc$, $s_n \in F(U_n)$ with certain open subsets $U_i \subseteq X$. Then $F_\alpha$ is defined to be the smallest subsheaf $F' \subseteq F$ with the property that $s_i \in F'(U_i)$ for $1 \leq i \leq n$. This exists, since limits of sheaves exist, in particular arbitrary intersections of subsheaves of a sheaf, so that we can construct $F_\alpha$ as the intersection of all subsheaves $F'$ with the above property. It can be constructed more explicitly as follows: Every $s_i$ corresponds to a homomorphism of sheaves $s_i : \mathbb{Z}_{U_i} \to F$, where $\mathbb{Z}_{U_i}$ is the constant sheaf on $U_i$ extended by zero to $X$. They give a homomorphism of sheaves $\oplus_{i=1}^{n} \mathbb{Z}_{U_i} \to F$. Its image is $F_\alpha$. But actually this description involves two sheafifications (sheaf extended by zero; image sheaf), so one does not get a description of the sections for free.
But actually, a section $s \in F(W)$ belongs to $F_\alpha(W)$ if and only if there is an open covering $W = \cup_p W_p$ such that for each $p$ we can write $s |_{W_p}$ as a $\mathbb{Z}$-linear combination of $s_1|_{U_1 \cap W_p},\dotsc,s_n|_{U_n \cap W_p}$. In fact, this defines a subsheaf of $F$ and satisfies the definition of $F_\alpha$.
The covering is important here (otherwise the sheaf property breaks down). Therefore, 

Is it true that for an open subset $U$, $F_\alpha(U)$ be group generated by $\alpha \cap F(U)$

is not true. There are many counterexamples.
