Let $\mathcal{X}$ be a topological space and $\mathcal{F}$ the constant presheaf, that assigns to each open set $\mathcal{U}$ of $\mathcal{X}$ the set $A$. The restriction map is the identity $A \rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf. In particular, let $\emptyset = \cup_i \mathcal{U}_i, \, \mathcal{U}_i=\emptyset$, be a covering of the empty set. Take two sections over $\emptyset$, i.e. take $s,s' \in \mathcal{F}(\emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $\mathcal{U}_i$ must be equal. I don't understand this. For example, since $\mathcal{U}_i \subset \emptyset$, then we have a restriction map $\mathcal{F}(\emptyset) \rightarrow \mathcal{F}(\mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$? Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{\mathcal{U}_i}$, right?
Thanks.