Let $X$ be an algebraic variety, say over algebraically closed field $k$, $\mathcal{O}_{X}$ its's structure sheaf. Denote also $X_{1},...,X_{n}$ the irreducible components of $X$. Let $\mathcal{U}$ be the category of all Zariski dense open subsets of $X$ with inclusions as morphisms. Is it true that this category is cofiltered ? I define cofiltered category $C$ as a one satisfying the following two axioms: 1) for any objects $a,b$ there exist an object $c$ and a couple of morphisms $c\rightarrow a$ $c\rightarrow b$ 2) for two parallel morphisms $f: a\rightarrow b$ , $g:a\rightarrow b$ there exists a morphism $h:c\rightarrow a$ such that $fh=gh$.
Note that if the category of Zariski open subsets is cofiltered, than the set of vector spaces $\mathcal{O}_{X}(U)$ has a structure of filtered diagram in the category of vector spaces. What would be the $colim _{U} \mathcal{O}_{X}(U\in \mathcal{U})$?
Yes. The first axiom follows from the fact that an open subset is dense iff it is dense in each $X_n$, so an intersection of two dense open subsets is dense and you can take $c=a\cap b$. The second axiom is automatic since there is only one morphism between any two objects in this category.
To compute your colimit, note that the set $\bigcup_i \left(X_i\setminus \bigcup_{j\neq i} X_j\right)$ of all points that are only contained in one component is an open dense subset, so we may assume the $X_n$ are disjoint. We then have $\mathcal{O}_X(U)=\prod_i\mathcal{O}_X(U\cap X_i)$ for each $i$, and we can take the colimit on each coordinate separately. But the colimit on the $i$th coordinate is just the the stalk of $\mathcal{O}_{X_i}$ at the generic point of $X_i$, also known as the function field $K(X_i)$, so your colimit is $\prod_i K(X_i)$.