# Extending a $\mathcal{B}$-sheaf via inverse limits vs stalks.

Let $X$ be topological space and $\mathcal{B}$ a basis. A $\mathcal{B}$-sheaf is a contravariant functor $\mathcal{O}_{\mathcal{B}}$ from $\mathcal{B}$ to the category of abelian groups that satisfies the obvious notions of separation (locally zero implies zero) and sheaf-ness (compatible sections on cover of basic open sets can be glued together...). This $\mathcal{B}$-sheaf can be extended to a sheaf $\mathcal{O}_{X}$ on all open sets of $X$ by taking the inverse limit $\mathcal{O}_X (U) = \varprojlim \mathcal{O}_{\mathcal{B}}(V_i)$ over all basic $V_i$ contained in $U$. Here the restriction maps are defined naturally via the universal property and it is routinely verified that this is a sheaf.

However the 'stalks' $\mathcal{O}_{\mathcal{B},P}$ of the $\mathcal{B}$-sheaf (the direct limit of basic open set containing $P \in X$) are the same as the stalks of the sheaf we constructed above. Therefore couldn't we extend to the same sheaf of inverse limits by instead letting the sections $\mathcal{O} (U)$ be the $s: U \to \coprod \mathcal{O}_{\mathcal{B},P}$, maps from points into their 'stalks' that are locally flat, meaning locally induced by a given section on a given basic $V \subseteq U$? In short, is construction by inverse limits isomorphic to the standard construction of the associated sheaf for a presheaf, where instead we have a $\mathcal{B}$-sheaf?

• I think every sheaf is naturally isomophic with the sheaf of functions from points to stalks which are locally induced by a section, and this question is just a particular case of that. – basket Jan 17 '16 at 17:06
• I had answer in general here math.stackexchange.com/questions/1588584/…; if you would like, I can explain the proof in the case of the sheaves of Abelian groups. – Armando j18eos Feb 27 '16 at 10:06