I'm reviewing my basic sheaf theory and the following question came up:
Let $F$ be an abelian group, $X$ a topological space, and let $\mathbb{F}$ be a presheaf of abelian groups on $X$ such that $\mathbb{F}(U) = F$ for all open subsets $U \subset X$. Does it follow that $\mathbb{F}(U \to V) = \mathrm{id} : F \to F$ for all $V \subset U$?
So far I've explored by considering the triple $\emptyset \subset U \subset X$ where $U$ is an arbitrary open subset of $X$, and considering the composition $\mathbb{F}(U \to \emptyset) \circ \mathbb{F}(X \to U) = \mathbb{F}(X \to \emptyset)$, but haven't gotten anywhere.
Appreciate any help.