Sheafs with $F(X)=\emptyset$

Question

In the book "A Gentle Introduction to Homology, Cohomology, and Sheaf Cohomology" by Jean Gallier, the author states on page 214 that if F is a sheaf on a topological space X and for an open set U, $F(U)=\emptyset$ (which is only possible if it is a sheaf of sets or some other category including the empty set), then obviously since there must be restriction maps, $F(X)=\emptyset$ and therefore the author claims that $F(A)=\emptyset$ for all open sets A in X, explaining that this is also because of the restriction maps and cites the book by Godement, but I don't understand why this should be the case? For example, nontrivial principal fibre bundles do not allow global sections, but certainly have local ones. Thank you in advance for your help.

Answer 1

I agree with your counter-example. Here is another one : let $X = \{a,b\}$ where the topology has open set $\{X, \{b\}, \emptyset\}$. The sheaf $F$ defined as $F(X) = \emptyset, F(\{b\}) = F(\emptyset) = \{*\}$ is a counter-example.

The argument is true if instead for a basis $U_i$ of the topology we have $F(U_i) = \emptyset$ for all $i \in I$, and this is maybe what the author had in mind.