# Sheafs with $F(X)=\emptyset$

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.

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.
• $X$ is a covering of $X$. – Tsemo Aristide Feb 15 at 16:11