The question is as in the title and comes from Justin Smith's book «Introduction to Algebraic Geometry» Appendix Proposition B.2.7. It is given without proof :
A flasque-separated presheaf is a sheaf
From the context, I expect that he made this statement with a sheaf of ring/module in mind.
Here, flasque means that the restriction maps are surjective, and separated means that the uniqueness property is respected. I tried to prove the statement for a while. Given sections $S_i\in \mathcal{F}(U_i)$ for some cover $U_i$ of some open set $U$, my idea has essentially been to use flasqueness to extend them in desirables ways, but I fail to get a global section that agrees with each $S_i$.
any help would be appreciated (is the statement even true ?). Thank you
This is false.
Let $X$ be any nonempty topological space, and let $\mathcal{F}$ be the presheaf on $X$ defined by $\mathcal{F}(\varnothing) = \{*\}$ and $\mathcal{F}(U) = \{0,1\}$ when $U \neq \varnothing$. The restriction maps are all identities, except for the maps $\mathcal{F}(U) \to \mathcal{F}(\varnothing)$, which are instead uniquely determined.
This presheaf is flasque, because every restriction is surjective by construction.
This presheaf is also separated, because any two distinct global sections will restrict to distinct local sections on any nonempty open set.
However, this presheaf is not always a sheaf. For example, take $X = \{0,1\}$ with the discrete topology. If $\mathcal{F}$ were a sheaf, then $\mathcal{F}(\{0\}) = \mathcal{F}(\{1\}) = \{0,1\}$ would force $\mathcal{F}(X) \cong \{0,1\} \times \{0,1\}$, contradicting the fact that $\mathcal{F}(X) = \{0,1\}$.
If you prefer a sheaf of rings or abelian groups, you can replace $\{*\}$ by the zero ring and $\{0,1\}$ by $\mathbb{Z}/2$.