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.

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    $\begingroup$ 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. $\endgroup$ – Zhen Lin Aug 30 '14 at 11:42
  • $\begingroup$ I agree that 'arbitrarily' is too strong. The abelian groups and restriction maps have to satisfy the presheaf axioms. $\endgroup$ – 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.

  • $\begingroup$ Yes, the 'arbitrarily' part was indeed wrong. But if we choose abelian groups and restriction maps that satisfy the presheaf axioms, is my statement then true? Sorry for the confusion, I will edit my question. $\endgroup$ – Ryan Aug 31 '14 at 14:38
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    $\begingroup$ @Ryan, the entire meat of a such a proof is in showing that we CAN "choose" such a collection of abelian groups and maps. $\endgroup$ – JHance Aug 31 '14 at 14:43
  • $\begingroup$ Yes, but IF we can do this, then the statement is correct? $\endgroup$ – Ryan Aug 31 '14 at 15:00
  • $\begingroup$ Verifying the stalks are the same is somewhat non-trivial, but the bigger problem is that a sheaf is not determined by its stalks alone. See here. $\endgroup$ – Zhen Lin Aug 31 '14 at 15:11

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