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Question:

In Hartshorne the additional glueing condition of a presheaf F to be a sheaf does not require for a open covering of the open set $U = \bigcup_I V_i$ that $V_i \cap V_j \neq \emptyset$. This seems wrong to me as for $U=V \cup W$ with $V \cap W = \emptyset $ for any $s_1 \in F(V)$, $ s_2 \in F(W)$ we have

$ s_1|_{V \cap W}= 0 = s_2|_{V \cap W} $

but clearly such a section $s \in F(U)$ s.t. $s|_V=s_1$ ans $s|_W =s_2$ does not need to exist (just think about sheaf of $C^\infty$ functions on a smooth manifold).

Then in Wells- Complex Manifolds one requires $V_i \cap V_j \neq \emptyset$ but with this logic the constant pre-sheaf ($U \mapsto A$, A abelian group) on a non connected top. space would be a sheaf, right?

Thanks in advance!

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    Its alright: Sections on disjoint open sets give rise to a section on the union (and no condition required). Why do you think that smooth functions should give a counterexample? – 2017-02-21
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    Oh ya right, smooth functions are no counterexample and all the other constructions I thought about were no sheaves... Thanks! – 2017-02-21

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