Is the sheafication of a "presheaf of $\mathcal{O}_X$-modules" an $\mathcal{O}_X$-module? Let $(X,\mathcal{O}_X$) be a ringed space. A presheaf of $\mathcal{O}_X$-modules is a presheaf $\mathcal{F}$ of abelian groups on $X$ such that $\mathcal{F}(U)$ is an $\mathcal{O}_X(U)$-module for each open $U\subseteq X$, and each restriction map of $\mathcal{F}$ is linear with respect to the corresponding restriction map of $\mathcal{O}_X$. More precisely, the latter condition says: if $V\subseteq U\subseteq X$ are open, then $\alpha(rm)=\beta(r)\alpha(m)$ for all $r\in\mathcal{O}_X(U)$ and $m\in\mathcal{F}(U)$, where $\alpha:\mathcal{F}(U)\to\mathcal{F}(V)$ and $\beta:\mathcal{O}_X(U)\to\mathcal{O}_X(V)$ are the restriction maps. 
Is the sheafication of a presheaf of $\mathcal{O}_X$-modules necessarily an $\mathcal{O}_X$-module? 

An $\mathcal O_X$-module structure on an abelian pre-sheaf is the same thing as a morphism of abelian pre-sheaves $\rho_F\colon \mathcal{O}_X\times F\to F$, making the associativity-diagram commute. Take associated sheaves and the induced morphism: $\rho^a_F\colon a(\mathcal{O}_X\times F)\to aF$ and observe that the natural morphism $a(\mathcal{O}_X\times F)\to \mathcal{O}_X\times aF$ is an isomorphism since it is an isomorphism on stalks and both are sheaves. Composing $\rho^a_F$ with the inverse, we get a morphism of abelian sheaves $\mathcal{O}_X\times aF\to aF$ which makes the associativity-diagram commute since it commutes on stalks. Thus, we find the induces module-structure on the associated sheaf simply by sheafifying the action morphism.
