Sheaf of rings on a topological space and functoriality.

Question

I'm trying to reconcile an issue I'm having between the definitions of a presheaf on a general category and the notion of a sheaf of rings on a topological space.

Definition: An $\mathsf{S}$ valued presheaf on a category $\mathsf{C}$ is a contravariant functor $F: \mathsf{C} \rightarrow \mathsf{S}$.

Definition: A sheaf of rings $\mathscr{F}$ on $X$ consists of the following data.

1. For every open set $U$ of $X$, a ring $\mathscr{F}(U)$, and $\mathscr{F}(\varnothing) = 0$.
2. For every inclusion $U \subseteq V$ a ring homomorphism $\rho_{VU}: \mathscr{F}(V) \rightarrow \mathscr{F}(U)$ such that $\rho_{U,U} = 1_{\mathscr{F}(U)}$ and if $U \subseteq V \subseteq W$ then $\rho_{W, U} = \rho_{V,U} \circ \rho_{W,V}$.
3. (Locally zero is zero.) Suppose $U = \cup_i U_i$. If $s \in \mathscr{F}(U)$ is such that $\rho_{U,U_i}(s) = 0$ for all $i$ then $s = 0$.
4. (Local data can be glued.) Suppose $U = \cup_i U_i$. If there are $s_i \in \mathscr{F}(U_i)$ such that for any pair $i,j$ one has $\rho_{U,U_i\cap U_j}(s_i) = \rho_{U,U_i \cap U_j}(s_j)$ then there is a unique $s \in \mathscr{F}(U)$ for which $\rho_{U,U_i}(s) = s_i$ for all $i$.

Given a ringed space, to any $x \in X$ we define the stalk of the sheaf at $x$ to be $$ \mathscr{F}_x = \varinjlim_{x \in U} \mathscr{F}(U). $$ I.e we take the direct limit over all of the open sets of $U$ which contain the point $x$, which is a directed set.

To first define a presheaf on $X$, I need to associate to its collection of open sets $\tau(X)$, a preorder, and make it into a preorder category. I have two ways to do this, but neither of them seem to give me exactly what I want.

1. If I define $U \leq V \iff U \subseteq V$, then $\tau(X)$ is a preorder, and so corresponds to a category where there is an arrow $U \rightarrow V$ iff $U \subseteq V$. A contravariant functor out of this category would associate to $U \rightarrow V$, i.e $U \subseteq V$ a morphism $FV \rightarrow FU$, where because $V$ contains $U$ the intuition of restriction map makes sense. However, when we want to take the stalk of the sheaf at a point $x$, we use that it is a directed set, however with the order on $\tau(X)$, it is in some sense directed in the wrong way, because if $A,B$ are open sets then $A \cup B \geq A,B$ but we aren't ``zooming in'' to the point $x$ with this direction.
2. On the other hand, I could have said $U \leq V \iff U \supseteq V$. Now we are directed ``downwards to x'' as we would like, since for $A$ and $B$ we have $A \cap B \geq A,B$. But now the contravariance out of the preorder category associated to $\tau(X)$ looses the intuition of being restriction, because to $U \rightarrow V$ we associate a map $F(V) \rightarrow F(U)$ while $V \subseteq U$.

Is the data of a presheaf on this category rather a covariant functor? Is this all really not something worth worrying about and just remember which way ``makes sense''? Thanks in advance for the clarification, let me know if I can be more precise.

Answer 1

Your first approach is the correct one. The category is $O(X)$, which has objects the open sets of $X$ and a single arrow $U\to V$ when $U\subseteq V$.

Now a sheaf $F$ of rings on $X$ does indeed determine a contravariant functor on $X$, with a restriction map $F(V)\to F(U)$ associated to every arrow $U\to V$ in $O(X)$.

For stalks, note that for $x\in X$, the poset of open sets containing $x$ is co-directed: given $U$ and $V$ containing $x$, $U\cap V$ is an open set containing $x$ with arrows to both $U$ and $V$. But since functor $F$ is contravariant, applying $F$ to this codirected diagram in $O(X)$ gives a directed diagram in the category of rings. The stalk is the directed colimit of this diagram.