Is the sheafication of a "presheaf of $\mathcal{O}_X$-modules" an $\mathcal{O}_X$-module?

Question

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?

Answer 1

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.

Answer 2

Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules, and let $$K=\coprod_{P\in X}\mathcal{F}_P$$ be the disjoint union of the stalks of $\mathcal{F}$. Now let $\mathcal{H}$ be the $\mathcal{O}$-module given by $$\mathcal{H}(U)=\{f:U\to K\mid \forall P\in U,\ f(P)\in\mathcal{F}_P\}$$ for all open $U\subseteq X$, with restriction maps as in the usual sense. Then we identify $\mathcal{F}$ with a subpresheaf $\mathcal{G}$ of $\mathcal{H}$. In particular, we take $$\mathcal{G}(U)=\{f:U\to K \mid \mbox{$\exists s\in \mathcal{F}(U)$ such that, $\forall P\in U$, $f(P)=s_P$}\}$$ for all open $U\subseteq X$; $s_P$ is the germ of $s$ at $P$. The sheafication of $\mathcal{G}$ is given by \begin{align*}\mathcal{G}^+(U) = \{f:U\to K \mid \mbox{$\forall$ $P\in U$, $\exists V\in N_U(P)$ and $t\in\mathcal{F}(V)$ such that, $\forall Q\in V$, $f(Q)=t_Q$}\}\end{align*} for all open $U\subseteq X$; $N_U(P)$ is the set of all open neighbourhoods of $P$ in $U$. Since $\mathcal{G}^+$ is a subsheaf of $\mathcal{H}$, it suffices to show that $\mathcal{G}^+(U)$ is an $\mathcal{O}(U)$-submodule of $\mathcal{H}(U)$ for all open $U\subseteq X$. Fix an open $U\subseteq X$. Observe that the $0$ function is in $\mathcal{G}^+(U)$. Now let $r\in\mathcal{O}(U)$, let $f$ and $g$ be in $\mathcal{G}^+(U)$, and fix $P\in U$. Then there exist $V$ and $W$ in $N_U(P)$, $s\in\mathcal{F}(V)$ and $t\in\mathcal{F}(W)$ such that $f(Q)=s_Q$ for all $Q\in V$, and $g(R)=t_R$ for all $R\in W$. Therefore, $(rf+g)(Q)=rs_Q+t_Q=(rs+t)_Q$ for all $Q\in V\cap W$, and thus $rf+g\in \mathcal{G}^+(U)$. Hence, $\mathcal{F}^+\simeq\mathcal{G}^+$ is an $\mathcal{O}$-module.