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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.

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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.

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  • $\begingroup$ $X$ is a covering of $X$. $\endgroup$ – Tsemo Aristide Feb 15 at 16:11
  • $\begingroup$ @TsemoAristide : thanks, you're right, I meant to say that it should hold for a basis of the topology. $\endgroup$ – Nicolas Hemelsoet Feb 15 at 16:18

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