Showing that $\mathscr{F}(U\sqcup V)=\mathscr{F}(U)\times \mathscr{F}(V)$

Question

Let $X$ be a topological space and $\mathscr{F}$ be a presheaf on $X$. If $U$ and $V$ are disjoint, then $$\mathscr{F}(U\cup V)=\mathscr{F}(U)\times \mathscr{F}(V).$$

Does anyone have a suggestion on where I can start?

Answer 1

As Claudius correctly says, this is false for a presheaf. Indeed, simply take the presheaf $\mathcal{F}$ assigning to each open $U$ a fixed set $A$ of cardinality $2$. Then $\mathcal{F}(U\sqcup V)=A$ while $\mathcal{F}(U)\times \mathcal{F}(V)=A\times A$. For reasons of cardinality, these are not isomorphic as sets. Similar counterexamples using sheaves of rings can be constructed.

Hint: If you assume that $\mathcal{F}$ is a sheaf, then you can take as an open covering of $U\sqcup V$ the set $\{U,V\}$ and note that $s\in\mathcal{F}(U\sqcup V)$ is uniquely determined by $s|_U$ and $s|_V$ using the sheaf axioms. This will allow you to define an isomorphism $\mathcal{F}(U\sqcup V)\cong \mathcal{F}(U)\times \mathcal{F}(V)$ as you should check.