TITLE: Prove that a presheaf is a sheaf
QUESTION [5 upvotes]: Let $X$ be a variety. Show that if $X$ is irreducible, then the constant abelian presheaf $\mathcal{F}$ with $\mathcal{F}(U)=\mathbb{Z}$ for every nonempty open subset $U\subseteq X$ and $\mathcal{F}(\emptyset)=0$ is a sheaf. Any leads? What does the word "constant" means here?

REPLY [4 votes]: Hint: Since $X$ is irreducible, any two non-empty open set always intersects. And for any inclusion $V \subseteq U$ of non-empty open sets, the restriction map $\mathcal{F}(U) \rightarrow \mathcal{F}(V)$ is the identity map.
Sheaf Property: (1) Let $U \subseteq X$ be open and let $\{U_i\}$ be an open cover of $U.$ Suppose $s \in \mathcal{F}(U)$ and $s|_{U_i} = 0, \forall i.$ The restriction maps are identity maps. So we have $s = 0.$
(2) Let $U \subseteq X$ be open and let $\{U_i\}$ be an open cover of $U.$ Suppose for each $i,$ we have $s_i \in \mathcal{F}(U_i)$ such that $s_i|_{U_I \cap U_j} = s_j|_{U_i \cap U_j}, \forall i, j.$ Since $X$ is irreducible, $U_i \cap U_j \neq \emptyset, \forall i, j$ and the restriction maps $\mathcal{F}(U_i) \rightarrow \mathcal{F}(U_i \cap U_j)$ are identity maps, this shows that $s_i = s_j \in \mathbb{Z}, \forall i, j.$ Take $s \in \mathcal{F}(U)$ to be $s_i.$