I am currently working on surjective morphisms of sheaves and trying to understand the subtleties connected to the need for sheafification of the presheaf image.
By definition (e.g. in Hartshorne) $\beta: \mathcal{F} \to \mathcal{F}''$ is surjective iff $im(\beta) = \mathcal{F}''$.
In some posts, I read that this is equivalent to: For every open $U\subset X$ there is an open cover ${U_i}$ of $U$ s.t. there are $t_i \in \mathcal{F}(U_i): \beta(U_i)(t_i) = s|_{U_i}$.
I do not know how to prove this equivalent definition of surjectivity; I think that the sheafification of the presheaf image is crucial, however, I cannot find a way to explicitly show it.
Assume $\beta$ is surjective and $U$ is open in $X$. Let $s\in \mathcal{F}’’(U)$. If $p\in U$ the induced map $\beta_p$ on stalks is surjective, so there is a germ $t_p\in \mathcal{F}_p$ such that $\beta_p(t_p) = s_p$. Let the pair $(V, t)$ represent germ $t_p$. There is an open neighborhood $U_i$ of $p$ contained in $U\cap V$ such that $\beta(U_i)(t|_{U_i}) = s|_{U_i}$. Hence $U$ is covered by open sets $U_i$ for which there are sections $t_i\in \mathcal{F}(U_i)$ satisfying $\beta(U_i)t_i = s|_{U_i}$.
Conversely, suppose the covering condition holds. If $U$ is open in $X$ and $s\in \mathcal{F}’’(U)$, then there is a cover $\{U_i\}$ of $U$ and sections $t_i\in \mathcal{F}(U_i)$ such that $\beta(U_i)(t_i) = s|_{U_i}$. If $p\in U$, then $p\in U_i$ for some $i$, and $\beta_p((t_i)_p) = s_p$. Thus $\beta_p$ is surjective. Since $p$ was arbitrary, $\beta$ is surjective.