# Sheaf associated to sheaf on basis

Statement: If we have a basis $B$ for a topological space $X$, then a sheaf defined on $B$ defines uniquely a sheaf on $X$.

I was wondering if the following proof is correct:

Let $\mathcal{F}$ be a sheaf on $B$, then this gives a presheaf $\mathcal{G}$ on $X$ if we give some abelian groups for open sets $U\subseteq X$ not in $B$ and the needed restriction maps, such that they satisfy the presheaf axioms. Then sheafification of $\mathcal{G}$ gives an unique sheaf $\mathcal{F}^+$ on $X$. It follows from the construction of the sheaf $\mathcal{F}^+$ that the abelian groups we chose for $\mathcal{G}$ have no influence on the construction because stalks on $B$ and $X$ are the same.

• You can't define $\mathscr{G}$ arbitrarily – after all, it has to be a presheaf, so the restriction maps have to compose correctly etc. – so it's not obvious that you can construct such a $\mathscr{G}$ in the first place. – Zhen Lin Aug 30 '14 at 11:42
• I agree that 'arbitrarily' is too strong. The abelian groups and restriction maps have to satisfy the presheaf axioms. – Ryan Aug 30 '14 at 12:03

Assume you want to have a sheaf of rings. Let $X=\{u,v\}$ be a set consisting of two points with the discrete topology. A base is given by $U:=\{u\}$ and $V:=\{v\}$. Declare $\mathscr F$ on this base by $\mathscr F(U)=\mathscr F(V)=\mathbb Z$, then you will not be able to decree $\mathscr G(X)=\mathbb Q$, because there is no ring homomorphism $\mathbb Q\to\mathbb Z$. If you want a sheaf of abelian groups, you will similarly not be able to decree $\mathscr G(X)$ as a cyclic group. This is just a very simple example, much more problems can arise as hinted to in the comments above. The point here is that the restriction maps are a very important part of the data that constitutes a presheaf.