Let $(X,\mathcal{O}_X)$ be a ringed space and $\mathcal{F}$ be an $O_X$-module. Then for $x \in X$, $\mathcal{F}_x$ has a natural structure of an $O_{X, x}$-module.
Question: Is there some abstract nonsense argument how to see that? By that I mean an argument interpreting stalks as direct limits and using their universal properties.
I tried to find one, but usually when one needs a nice description of an action of a ring $R$ on a module $M$, one describes it as a ring homomorphism $R\rightarrow \mathrm{End}_{\mathbb{Z}}(M)$. This seems of no use here, since I cannot see how can something like $\mathrm{End}_\mathbb{Z}(-)$ be used as a functor.
Thanks in advance for any help.
Taking stalks is a functor, so if a ring $A$ acts on $\mathcal F$, then $A$ acts on the stalk $\mathcal F_x$.
Now if $\mathcal F$ is an $\mathcal O_X$-module, then $A := \mathcal O_X(X)$ acts on $\mathcal F$.
Applying this with $X$ replaced by a basis of n.h.'s $U$ of $x$, we find that each $\mathcal O_X(U)$ acts on $\mathcal F_x$, and thus that the stalk $\mathcal O_{X,x}$ does.
Is this nonsensical enough?